# Characterizing total negativity and testing their interval hulls

Projesh Nath Choudhury  Department of Mathematics, Indian Institute of
Science, Bangalore 560012, India<EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

A matrix is called totally negative (totally non-positive) of order $k$, if
all its minors of size at most $k$ are negative (non-positive). The objective
of this article is to provide several novel characterizations of total
negativity via the (a) sign non-reversal property, (b) variation diminishing
property, and (c) Linear Complementarity Problem. More strongly, each of these
three characterizations uses a single test vector. As an application of the
sign non-reversal property, we study the interval hull of two rectangular
matrices. In particular, we identify two matrices $C^{\pm}(A,B)$ in the
interval hull of matrices $A$ and $B$ that test total negativity of order $k$,
simultaneously for the entire interval hull. We also show analogous
characterizations for totally non-positive matrices. These novel
characterizations may be considered similar in spirit to fundamental results
characterizing totally positive matrices by Brown–Johnstone–MacGibbon [J.
Amer. Statist. Assoc. 1981] (see also Gantmacher–Krein, 1950),
Choudhury–Kannan–Khare [Bull. London Math. Soc., in press] and Choudhury [2021
preprint]. Finally using a 1950 result of Gantmacher–Krein, we show that
totally negative/non-positive matrices can not be detected by (single) test
vectors from orthants other than the open bi-orthant that have coordinates
with alternating signs, via the sign non-reversal property or the variation
diminishing property.

###### Key words and phrases:

Totally negative matrix, totally non-positive matrix, sign non-reversal
property, variation diminishing property, Linear Complementarity Problem,
interval hull of matrices

###### 2020 Mathematics Subject Classification:

15B48,(primary), 15A24, 65G30, 90C33 (secondary)

section

###### Contents

  1. 1 Introduction and main results
  2. 2 Theorems A and B: Sign non-reversal property and interval hulls for totally negative matrices
  3. 3 Theorem C: Characterization of total negativity using variation diminution
  4. 4 Theorems D and E: Characterization of total negativity via the LCP
  5. 5 Characterizations for totally non-positive matrices
  6. 6 Other orthants cannot yield test vectors

##  1\. Introduction and main results

Given integers $m,n\geq k\geq 1$, a real matrix $A\in\mathbb{R}^{m\times n}$
is called totally negative of order $k$ if all minors of $A$ of order at most
$k$ are negative, and $A$ is called totally negative if $k=\min\\{m,n\\}$.
Similarly, one defines totally non-positive matrices (including order $k$).

In prior discussions of totally negative (or totally non-positive) matrices,
their analogy to totally positive/TP (or totally non-negative/TN) matrices,
which are matrices with all positive (or non-negative) minors, perpetually
comes up first (and which we refer to henceforth by the standard abbreviations
TP and TN, or $TP_{k}$ and $TN_{k}$ to distinguish them from the matrices
studied in the present paper.) Indeed, TP and TN are widely studied because of
their numerous applications in a variety of topics in mathematics, including,
analysis, combinatorics, cluster algebras, differential equations, Gabor
analysis, matrix theory, probability, and representation theory [2, 3, 4, 15,
16, 17, 19, 22, 26, 30, 31, 34, 35, 36]. In 1937, Gantmacher–Krein [18]
characterized TP matrices using positivity of the spectra of all square
submatrices. A similar characterization for totally negative matrices was
given by Fallat–van den Driessche in 2000 [14]. They proved that if $A$ is a
totally negative matrix then all the eigenvalues of each square submatrix of
$A$ are real, distinct, nonzero and exactly one eigenvalue is negative. (The
converse is immediate.) Similar spectral characterizations hold for TN and
totally non-positive matrices, using the density of TP and totally negative
matrices in these classes and the continuity of roots. Totally negative/non-
positive matrices are also studied in the context of matrix factorization and
spectral perturbation, see [6, 7, 23].

Another interesting property exhibited by TP and TN matrices is the variation
diminishing property. The variation diminishing property is also satisfied by
other matrices, e.g. $-A$ for $A$ TP/TN. This was explored by Motzkin [28] in
his PhD thesis in 1936, and he showed that the class of matrices satisfying
the variation diminishing property are precisely the sign-regular matrices –
in other words, there exists a sequence of signs $\varepsilon_{k}\in\\{\pm
1\\}$ such that every $k\times k$ minor of $A$ has sign $\varepsilon_{k}$, for
all $k\geq 1$. A systematic treatment of sign-regular matrices and kernels can
be found in Karlin’s comprehensive monograph [26].

The goal of this note is to provide similar characterization results for the
class of totally negative/totally non-positive matrices, i.e. ones for which
all minors are negative/non-positive. More precisely, we provide novel
characterizations of such matrices using (a) sign non-reversal and (b) Linear
Complementarity, to go along with Motzkin’s classical characterization using
(c) variation diminution. In addition, we strengthen all three tests (a), (b),
and (c) to work with only a single test vector, in parallel to our recent
works [8, 9] for TP matrices.

To state these results, we need some preliminary definitions, which are used
below without further reference.

######  Definition 1.1.

Let $m,n\geq 1$ be integers.

  1. (1)

A square matrix $A\in\mathbb{R}^{n\times n}$ has the sign non-reversal
property on a set of vectors $T\subseteq\mathbb{R}^{n}$, if for all vectors
$0\neq x\in T$, there exists a coordinate $i\in[1,n]$ such that
$x_{i}(Ax)_{i}>0$.

  2. (2)

A matrix $A\in\mathbb{R}^{n\times n}$ has the non-strict sign non-reversal
property on a set of vectors $T\subseteq\mathbb{R}^{n}$ if for all nonzero
vectors $x\in T$, there exists a coordinate $i\in[1,n]$ such that $x_{i}\neq
0$ and $x_{i}(Ax)_{i}\geq 0$.

  3. (3)

A contiguous submatrix is a submatrix whose rows and columns are indexed by
sets of consecutive integers.

  4. (4)

Let $\mathbb{R}^{n}_{\rm alt}\subset\mathbb{R}^{n}$ denote the set of real
vectors whose coordinates are nonzero and have alternating signs. Also define
$\mathbb{R}^{n}_{\rm+,-}$ to comprise the vectors in $\mathbb{R}^{n}$ with at
least one positive and one negative coordinate.

  5. (5)

Given a matrix $A\in\mathbb{R}^{n\times n}$ and $i,j\in[1,n]$, let $A^{ij}$
denote the determinant of the submatrix of $A$ of size $n-1$ formed by
deleting the $ith$ row and $jth$ column of $A$.

  6. (6)

If $n=1$, then define $A^{11}:=-1$ to be the determinant of the empty matrix.
(This, convention will only be used in Theorem E, Proposition 5.8 and their
proofs.)

  7. (7)

Let $\operatorname{adj}(A)$ denote the adjugate matrix of $A$ and let
$e^{i}\in\mathbb{R}^{n}$ denote the vector whose $i$-th component is $1$, and
other components are zero.

  8. (8)

We say that an array $X$ is $\geq 0$ (respectively $X>0,~{}X\leq 0,~{}X<0$) if
all coordinates of $X$ are $\geq 0$ (respectively $>0,~{}\leq 0,~{}<0$).

  9. (9)

Define the vector $d^{[n]}:=(1,-1,\dots,(-1)^{n-1})^{T}\in\mathbb{R}^{n}$.

  10. (10)

Given $z=(z_{1},\dots,z_{n})^{T}\in\\{\pm 1\\}^{n}$, define the diagonal
matrix $D_{z}$ whose $i$th diagonal entry is $z_{i}$.

  11. (11)

Given two matrices $A,B\in\mathbb{R}^{m\times n}$, and tuples of signs
$z\in\\{\pm 1\\}^{m},\tilde{z}\in\\{\pm 1\\}^{n}$, define the $m\times n$
matrices $|A|$, $I_{z,\tilde{z}}(A,B)$, and $C^{\pm}(A,B)$ via:

$|A|_{ij}:=|a_{ij}|,\qquad
I_{z,\tilde{z}}(A,B):=\frac{A+B}{2}-D_{z}\frac{|A-B|}{2}D_{\tilde{z}},\qquad
C^{\pm}(A,B):=I_{d^{[m]},\;\pm d^{[n]}}(A,B).$

  12. (12)

Given a vector $x\in\mathbb{R}^{n}$, let $S^{-}(x)$ denote the number of sign
changes in $x$ after removing all zero entries. Next, assign a value of $\pm
1$ to each zero entry of $x$, and let $S^{+}(x)$ denote the maximum possible
number of sign changes in the resulting sequence. For $0\in\mathbb{R}^{n}$, we
define $S^{+}(0):=n$ and $S^{-}(0):=0$.

Now we state our main results only for totally negative matrices. The
penultimate section contains the analogous theorems for totally non-positive
matrices. Our first result provides a characterization of total negativity in
terms of sign non-reversal phenomena.

######  Theorem A.

Let $m,n\geq k\geq 2$ be integers, and $A\in\mathbb{R}^{m\times n}$ such that
$A<0$. The following statements are equivalent:

  1. (1)

The matrix $A$ is totally negative of order $k$.

  2. (2)

Every square submatrix of $A$ of size $r\in[2,k]$ has the sign non-reversal
property on $\mathbb{R}^{r}_{\rm+,-}$.

  3. (3)

Every contiguous square submatrix of $A$ of size $r\in[2,k]$ has the sign non-
reversal property on $\mathbb{R}^{r}_{\rm alt}$.

In fact, these conditions are further equivalent to sign non-reversal at a
single vector:

  4. (4)

For every contiguous submatrix $A_{r}$ of $A$ of size $r\in[2,k]$ and for any
fixed nonzero vector
$\alpha:=(\alpha_{1},-\alpha_{2},\ldots,(-1)^{r-1}\alpha_{r})^{T}\in\mathbb{R}^{r}$
with all $\alpha_{i}\geq 0$ (or $\leq 0$), define the vector

$x^{A_{r}}:=\operatorname{adj}(A_{r})\alpha.$ (1.1)

Then $A_{r}$ has the sign non-reversal property for $x^{A_{r}}$.

######  Remark 1.2.

A careful look at the characterization of TP in [8, 9] reveals the importance
of the signs of the entries of $A$. Indeed, if $A>0$ in Theorem A, the
assertions $(3),(4)$ are instead equivalent to $TP_{k}$.

As an application of Theorem A, we simultaneously detect the total negativity
of an entire interval hull of matrices by reducing it to a set of two test
matrices. Recall that the interval hull of two matrices
$A,B\in\mathbb{R}^{m\times n}$, denoted by $\mathbb{I}(A,B)$, is defined as
follows:

$\mathbb{I}(A,B)=\\{C\in\mathbb{R}^{m\times
n}:c_{ij}=t_{ij}a_{ij}+(1-t_{ij})b_{ij},t_{ij}\in[0,1]\\}.$ (1.2)

If $A\neq B$, then $\mathbb{I}(A,B)$ is an uncountable set. We say that an
interval hull is totally negative of order $k$ if all the matrices in it are
totally negative of order $k$. For more details about interval hulls of
matrices, we refer to [21]. One of the interesting questions, related to
interval hulls of matrices, is to find a minimal test set which would
determine if an entire interval hull $\mathbb{I}(A,B)$ is totally negative. In
[20], Garloff answered this question when (i) the interval hull consists of
square matrices ($m=n$), and (ii) the order of total negativity equals the
size of the matrices ($k=n$). In the next result, we drop these two
constraints and answer this question completely.

######  Theorem B.

Let $m,n\geq k\geq 1$ be integers, and $A,B\in\mathbb{R}^{m\times n}$. Then
$\mathbb{I}(A,B)$ is totally negative of order $k$ if and only if the matrices
$C^{\pm}(A,B)$ are totally negative of order $k$.

######  Remark 1.3.

Note that $C^{+}(A,B)$ and $C^{-}(A,B)$ are independent of the choice of $k$.

We now turn our attention to one of the most well known and widely used
properties of TP matrices: their variation diminution on the test set of all
nonzero real vectors [5]. Recently in [8], we characterized total positivity
via the variation diminishing property by providing a finite set of test
vectors – in fact a single vector for each contiguous submatrix. Our next main
result similarly characterizes total negativity using the variation
diminishing property.

######  Theorem C.

Let $m,n\geq 2$ be integers. Given a real $m\times n$ matrix $A$ with $A<0$,
the following statements are equivalent:

  1. (1)

$A$ is totally negative.

  2. (2)

For all $x\in\mathbb{R}^{n}_{\rm+,-}$, $S^{+}(Ax)\leq S^{-}(x)$. Moreover, if
equality holds and $Ax\neq 0$, then the first (last) component of $Ax$ (if
zero, the unique sign given in computing $S^{+}(Ax)$) has the same sign as the
first (last) component of $x$.

  3. (3)

Let $k=\min\\{m,n\\}$ and $r\in[2,k]$. For every $r\times r$ contiguous
submatrix $A_{r}$ of $A$ and for any choice of nonzero vector
$\alpha:=(\alpha_{1},-\alpha_{2},\ldots,(-1)^{r-1}\alpha_{r})^{T}\in\mathbb{R}^{r}$
with all $\alpha_{i}\geq 0$ (or $\leq 0$), define the vector

$x^{A_{r}}:=\operatorname{adj}(A_{r})\alpha.$ (1.3)

Then $S^{+}(A_{r}x^{A_{r}})\leq S^{-}(x^{A_{r}})$. If equality occurs here,
then the first (last) component of $A_{r}x^{A_{r}}$ (if zero, the unique sign
given in computing $S^{+}(A_{r}x^{A_{r}})$) has the same sign as the first
(last) nonzero component of $x^{A_{r}}$.

In the last part of our discussion of total negativity, we characterize
totally negative matrices using the Linear Complementarity Problem (LCP).
Given a real square matrix $A\in\mathbb{R}^{n\times n}$ and a vector
$q\in\mathbb{R}^{n}$, the Linear Complementarity Problem, denoted by
LCP$(A,q)$ asks to determine, if possible, a vector $x\in\mathbb{R}^{n}$ such
that

$x\geq 0,\quad y=Ax+q\geq 0,\quad and\quad x^{T}y=0.$ (1.4)

Any $x$ that satisfies the above three conditions is called a
(complementarity) solution of LCP$(A,q)$. Let SOL$(A,q)$ denote the set of all
solutions of LCP$(A,q)$. The Linear Complementarity Problem has important
applications in diverse areas in mathematics, including bimatrix games, convex
quadratic programming, economics, fluid mechanics, and variational inequality
[11, 12, 13, 27]. A plethora of matrix classes have been studied in connection
with the Linear Complementarity Problem, for example, $P$-matrices [24, 33],
$N$-matrices [29], and recently, TP matrices [8]. For more details about
Linear Complementarity Problems and their applications, we refer to [10, 12,
25]. Our final main results characterizes total negativity via the solution
set of the LCP.

######  Theorem D.

Let $m,n\geq k\geq 1$ be integers. Given $A\in\mathbb{R}^{n\times n}$, the
following statements are equivalent:

  1. (1)

The matrix $A$ is totally negative of order $k$.

  2. (2)

For every square submatrix $A_{r}$ of $A$ of size $r\in[1,k]$, LCP$(A_{r},q)$
has exactly two solutions for all $q\in\mathbb{R}^{r}$ with $q>0$.

  3. (3)

For every contiguous square submatrix $A_{r}$ of $A$ of size $r\in[1,k]$,
LCP$(A_{r},q)$ has exactly two solutions for all $q\in\mathbb{R}^{r}$ with
$q>0$.

  4. (4)

For every contiguous square submatrix $A_{r}$ of $A$ of size $r\in[2,k]$ and
for all $q\in\mathbb{R}^{r}$ with $q>0$, SOL$(A_{r},q)$ does not contain two
vectors with sign pattern $\begin{pmatrix}+\\\ 0\\\ +\\\ 0\\\
\vdots\end{pmatrix}$ and $\begin{pmatrix}0\\\ +\\\ 0\\\ +\\\
\vdots\end{pmatrix}$ simultaneously. Moreover for $r=1$ and all
$i\in[1,m],~{}j\in[1,n]$, LCP$((a_{ij})_{1\times 1},q)$ has exactly two
solutions for some scalar $q>0$.

We improve this result by providing a characterization of totally negative
matrices in terms of the LCP at certain single vector, for each contiguous
submatrix.

######  Theorem E.

Let $m,n\geq k\geq 1$ be integers. Given $A\in\mathbb{R}^{m\times n}$, the
following statements are equivalent.

  1. (1)

The matrix $A$ is totally negative of order $k$.

  2. (2)

For every $r\in[1,k]$ and contiguous $r\times r$ submatrix $A_{r}$ of $A$,
define the vectors

$x^{A_{r}}:=(A^{11}_{r},0,A^{13}_{r},0,\ldots)^{T},\qquad
q^{A_{r}}:=A_{r}x^{A_{r}}.$ (1.5)

Then $0$ and $-x^{A_{r}}$ are the only solutions of LCP$(A_{r},q^{A_{r}})$.

We end by explaining the organization of the paper. The next three sections
contain the proofs of our main results above: the sign non-reversal and
interval hull result; the variation diminishing property; and the results
involving the LCP. Also note that it is natural to ask for ‘totally non-
positive’ analogues of the above main results, in the spirit of TN analogues
of results for TP matrices. In the penultimate section of the paper, we indeed
provide these – see Theorems 5.2, 5.3, 5.5, and Propositions 5.7 and 5.8,
respectively. In the final section, we show that test vectors from open
orthants other than the alternating bi-orthant cannot be used in the above
characterizations of total negativity/non-positivity.

##  2\. Theorems A and B: Sign non-reversal property and interval hulls for
totally negative matrices

In this section we prove Theorems A and B. To prove Theorem A, we begin with
two basic results. The first is a 1968 result of Karlin for totally negative
matrices (in fact strictly sign-regular matrices) which was first analogously
proved by Fekete in 1912 for TP matrices, and subsequently extended by
Schoenberg to $TP_{k}$ in 1955.

######  Theorem 2.1 (Karlin [26]).

Let $m,n\geq k\geq 1$ be integers. Then $A\in\mathbb{R}^{m\times n}$ is
totally negative of order $k$ if and only if every $r\times r$ contiguous
submatrix of $A$ has negative determinant, for $r\in[1,k]$.

The next preliminary result establishes the sign non-reversal property for
matrices with negative principal minors (these are known as $N$-matrices):

######  Theorem 2.2.

[29] Let $n\geq 2$ be an integer and $A\in\mathbb{R}^{n\times n}$ such that
$A<0$. Then $A$ has all principal minors negative if and only if for all
$x\in\mathbb{R}^{n}_{\rm+,-}$, there exists $i\in[1,n]$ such that
$x_{i}(Ax)_{i}>0$.

###### Proof of Theorem A.

That $(1)\implies(2)$ follows from Theorem 2.2; and that
$(2)\implies(3)\implies(4)$ is immediate. Finally, suppose $(4)$ holds. By
Theorem 2.1, it suffices to show the determinants of all $r\times r$
contiguous submatrices of $A$ are negative, for $r\in[1,k]$. We prove this by
induction on $r$, where the base case holds by the hypothesis. Fix $r\in[2,k]$
and assume that all $(r-1)\times(r-1)$ and smaller contiguous minors of $A$
are negative. Let $A_{r}$ be an $r\times r$ contiguous submatrix of $A$. By
the induction hypothesis, all the proper contiguous minors of $A_{r}$ are
negative. In fact, all the proper minors of $A_{r}$ are negative by Theorem
2.1.

For any choice of nonzero vector
$\alpha:=(\alpha_{1},-\alpha_{2},\ldots,(-1)^{r-1}\alpha_{r})^{T}$ with all
$\alpha_{i}\geq 0$ (or $\leq 0$), define the vector $x^{A_{r}}$ as in (1.1).
Then

$x^{A_{r}}_{i}=\sum_{j=1}^{r}(-1)^{j-1}\alpha_{j}A^{ji}_{r}$ (2.1)

and if the $\alpha_{i}$s are non-negative (or non-positive), then the summand
on the right is negative (or positive) for odd $i$ and positive (or negative)
for even $i$. Thus $x^{A_{r}}\in\mathbb{R}^{r}_{\rm alt}$, and

$A_{r}x^{A_{r}}=(\det A_{r})\alpha.$ (2.2)

By the hypothesis, (2.1), and (2.2), we have $i_{0}\in[1,r]$ such that

$0<x^{A_{r}}_{i_{0}}(A_{r}x^{A_{r}})_{i_{0}}=(-1)^{i_{0}-1}(\det
A_{r})\alpha_{i_{0}}x^{A_{r}}_{i_{0}}.$ (2.3)

From this, it follows that $\det A_{r}<0$, which completes the induction step.
∎

To prove Theorem B, we need two preliminary lemmas; the first is
straightforward.

######  Lemma 2.3.

Let $m,n\geq 1$ and $A,B\in\mathbb{R}^{m\times n}$. Then
$I_{z,\tilde{z}}(A,B)\in\mathbb{I}(A,B)$ for all $z\in\\{\pm
1\\}^{m},\tilde{z}\in\\{\pm 1\\}^{n}$.

######  Lemma 2.4.

[32] Let $A,B\in\mathbb{R}^{n\times n}$ and $x\in\mathbb{R}^{n}$. Let the
vector $z\in\\{\pm 1\\}^{n}$ be such that $z_{i}=1$ if $x_{i}\geq 0$ and
$z_{i}=-1$ if $x_{i}<0$. If $C\in\mathbb{I}(A,B)$, then

$x_{i}(Cx)_{i}\geq x_{i}(I_{z,z}(A,B)x)_{i},\quad\forall i\in[1,n].$

With these preliminaries at hand, we now prove Theorem B.

###### Proof of Theorem B.

The result is immediate for $k=1$, so we henceforth suppose $m,n\geq k\geq 2$.
Let $\mathbb{I}(A,B)$ be totally negative of order $k$. By Lemma 2.3,
$C^{\pm}(A,B)$ are totally negative of order $k$.

Conversely, suppose $C^{\pm}(A,B)$ are totally negative of order $k$, and let
$M\in\mathbb{I}(A,B)$. Then $M<0$, since $C^{+}(A,B),C^{-}(A,B)<0$. Fix
$r\in[2,k]$ and a vector $x\in\mathbb{R}^{r}_{\rm alt}$. Let
$M^{\prime}=M_{J\times K}$ be an $r\times r$ contiguous submatrix of $M$,
where $J\subseteq[1,m]$ and $K\subset[1,n]$ are contiguous sets of indices
with $|J|=|K|=r$. By Theorem A(3), it suffices to show that there exists a
coordinate $i\in[1,r]$ such that $x_{i}(M^{\prime}x)_{i}>0$.

To proceed further, we need some notation. Let $A^{\prime},B^{\prime}$ be
contiguous submatrices of $A,B$ respectively, with the rows and columns
indexed by $J$ and $K$ respectively. Since
$M^{\prime}\in\mathbb{I}(A^{\prime},B^{\prime})$, Lemma 2.4 implies there
exists $z^{\prime}\in\\{\pm 1\\}^{r}\cap\mathbb{R}^{r}_{\rm alt}$ such that

$x_{i}(M^{\prime}x)_{i}\geq
x_{i}(I_{z^{\prime},z^{\prime}}(A^{\prime},B^{\prime})x)_{i},\qquad\forall
i\in[1,r].$ (2.4)

Since $M^{\prime}=M_{J\times K}$, we extend $z^{\prime}$ to
$z\in\mathbb{R}^{m}_{\rm alt}\cap\\{\pm 1\\}^{m}$
($\tilde{z}\in\mathbb{R}^{n}_{\rm alt}\cap\\{\pm 1\\}^{n}$) by embedding in
contiguous positions $J\subseteq[1,m]$ ($K\subseteq[1,n]$) and uniquely
padding by $\pm 1$ elsewhere. Then
$I_{z^{\prime},z^{\prime}}(A^{\prime},B^{\prime})$ is a $r\times r$ contiguous
submatrix of

$I_{z,\;\tilde{z}}(A,B):=\frac{A+B}{2}-D_{z}\frac{|A-B|}{2}D_{\tilde{z}}\in\\{C^{+}(A,B),C^{-}(A,B)\\}.$

By hypothesis, this matrix is totally negative of order $k$. Thus
$I_{z^{\prime},z^{\prime}}(A^{\prime},B^{\prime})$ is totally negative. Using
Theorem A(3) and (2.4), for some $i\in[1,r]$, we have

$x_{i}(M^{\prime}x)_{i}\geq
x_{i}(I_{z^{\prime},z^{\prime}}(A^{\prime},B^{\prime})x)_{i}>0.$

Thus $M^{\prime}$ has the sign non-reversal property on $\mathbb{R}^{r}_{\rm
alt}$. Since $M<0$, by Theorem A, $M$ is totally negative of order $k$. ∎

##  3\. Theorem C: Characterization of total negativity using variation
diminution

We next prove Theorem C – i.e., to provide a characterization of totally
negative matrices using variation diminution.

###### Proof of Theorem C.

We begin by showing $(1)\implies(2)$. Let $A\in\mathbb{R}^{m\times n}$ be a
totally negative matrix and $x\in\mathbb{R}^{n}_{\rm+,-}$ with
$1<S^{-}(x)=k\leq n-1$. Then $x$ can be partitioned into $k+1$ components of
contiguous coordinates with like signs:

$(x_{1},\ldots,x_{s_{1}}),~{}~{}(x_{s_{1}+1},\ldots,x_{s_{2}}),~{}~{}\ldots(x_{s_{k}+1},\ldots,x_{n}),$
(3.1)

with at least one coordinate in each component nonzero and all nonzero
coordinates in the $i$th component having the same sign $(-1)^{i-1}$, without
loss of generality. Moreover, we set $s_{0}=0$ and $s_{k+1}=n$. Denote the
columns of $A$ by $a^{1},\ldots,a^{n}\in\mathbb{R}^{n}$, and define

$b^{i}:=\sum\limits_{j=s_{i-1}+1}^{s_{i}}|x_{j}|a^{j},\hbox{~{}for~{}}i\in[1,{k+1}].$

One can verify that the matrix
$B:=[b^{1},\ldots,b^{k+1}]\in\mathbb{R}^{m\times(k+1)}$ is totally negative
and $Bd^{[k+1]}=Ax$.

With this information in hand, we now prove $S^{+}(Ax)\leq S^{-}(x)=k$. For
ease of exposition, we split the remainder of the proof into two cases.

Case 1. $m\leq{k+1}.$

If $Ax\neq 0$, then $S^{+}(Ax)\leq{m-1}\leq S^{-}(x)$. If $Ax=0$, then $m\leq
k$, since $Ax=Bd^{[k+1]}$ and $B$ is invertible if $m=k+1$. Thus
$S^{+}(Ax)=m\leq S^{-}(x)$.

Case 2. $m>{k+1}.$

Set $y:=Ax=Bd^{[k+1]}$. If $S^{+}(Ax)>k$, then there exist indices
$i_{1}<i_{2}<\cdots<i_{k+2}\in[1,m]$ and a sign $\epsilon\in\\{+,-\\}$ such
that $(-1)^{r-1}\epsilon y_{i_{r}}\geq 0$ for $r\in[1,~{}{k+2}]$. Since $B$ is
totally negative, at least two of the $y_{i_{r}}$ are nonzero. Let
$I=\\{i_{1},\ldots,i_{k+2}\\}$ and define the $(k+2)\times(k+2)$ matrix

$M:=[y_{I}|B_{I\times[1,~{}k+1]}].$

Then $\det M=0$, since the first column of $M$ is a linear combination of the
rest. Moreover, expanding along the first column gives

$0=\sum_{r=1}^{k+2}(-1)^{r-1}y_{i_{r}}\det
B_{{I\setminus\\{i_{r}\\}}\times[1,~{}k+1]},$ (3.2)

a contradiction, since $B$ is totally negative, all $(-1)^{r-1}y_{i_{r}}$ have
the same sign, and at least two $y_{i_{r}}$ are nonzero. Thus $S^{+}(Ax)\leq
k=S^{-}(x)$.

It remains to prove the second part of the assertion $(2)$. We continue our
discussion using the notation in the preceding analysis. We claim that, if
$S^{+}(Ax)=S^{-}(x)=k$ with $Ax\neq 0$, and $(-1)^{r-1}\epsilon y_{i_{r}}\geq
0$ for $r\in[1,~{}{k+1}]$ – as opposed to $[1,~{}k+2]$, then $\epsilon=1$.
Since $B$ is totally negative and $Bd^{[k+1]}=Ax$, so the submatrix
$B_{I\times[1,k+1]}$ is invertible and $B_{I\times[1,~{}k+1]}d^{[k+1]}=y_{I}$.
By Cramer’s rule, the first coordinate of $d^{[k+1]}$ is

$1=\frac{\det[y_{I}|B_{I\times[2,~{}k+1]}]}{\det B_{I\times[1,~{}k+1]}}.$

Multiplying both sides by $\epsilon\det B_{I\times[1,~{}k+1]}$ and expanding
the numerator along the first column, we have

$\epsilon\det B_{I\times[1,~{}k+1]}=\sum_{r=1}^{k+1}(-1)^{r-1}\epsilon
y_{i_{r}}\det B_{I\setminus\\{i_{r}\\}\times[2,~{}{k+1}]}.$ (3.3)

Since the summation on the right side is negative and $B$ is totally negative,
we obtain $\epsilon=1$.

We next show that $(2)\implies(3)$. Fix $r\in[2,k]$ and suppose
$A_{r}=A_{I\times J}$ is an $r\times r$ contiguous submatrix of $A$, for
contiguous sets of indices $I\subseteq[1,m]$ and $J\subseteq[1,n]$ with
$\lvert I\rvert=\lvert J\rvert=r$. Define $x^{A_{r}}$ as in (1.3); note this
lies in $\mathbb{R}^{r}_{\rm+,-}$ by (2). We define
$x\in\mathbb{R}^{n}_{\rm+,-}$ to have $x^{A_{r}}$ in contiguous positions
$J\subseteq[1,m]$ and $0$ elsewhere. Then

$S^{+}(A_{r}x^{A_{r}})\leq S^{+}(Ax)\quad\hbox{and}\quad
S^{-}(x)=S^{-}(x^{A_{r}}).$ (3.4)

By $(2)$, we have $S^{+}(A_{r}x^{A_{r}})\leq S^{-}(x^{A_{r}})$.

Next, suppose that $S^{+}(A_{r}x^{A_{r}})=S^{-}(x^{A_{r}})$. Then
$S^{+}(Ax)=S^{-}(x)$. Let the first and last nonzero component of $x^{A_{r}}$
occur in positions $s,t\in[1,r]$, respectively. List the indices in $I$ by
$i_{1}<i_{2}<\ldots<i_{r}$. By (2), it follows that all coordinates of $Ax$ in
positions $1,2,\ldots,i_{s}$ (respectively, in positions $i_{t},\cdots,m$) are
all nonzero, and with the same sign as that of
$x^{A_{r}}_{s}~{}(\hbox{respectively}~{}x^{A_{r}}_{t})$. This shows
$(2)\implies(3)$.

Finally, we show that $(3)\implies(1)$. By Theorem 2.1, it suffices to show
for all $r\times r$ ($r\in[2,k]$) contiguous submatrices $A_{r}$ of $A$ that
$\det A_{r}<0$. The proof is by induction on $r\in[1,k]$. The base case holds
by the hypothesis.

Let $A_{r}$ be an $r\times r$ contiguous submatrix of $A$ with $r\in[2,k]$ and
suppose that all contiguous minors of $A$ of size at most $(r-1)$ are
negative. The same holds for all proper minors of $A$ by Theorem 2.1. Define
the vector $x^{A_{r}}$ as in (1.3). Then

$x^{A_{r}}_{i}=\sum_{j=1}^{r}(-1)^{j-1}\alpha_{j}A^{ji}_{r}$ (3.5)

and if the $\alpha_{i}$s are non-negative (or non-positive), then the
summation on the right is negative (or positive) for odd $i$ and positive (or
negative) for even $i$. It follows that

$x^{A_{r}}\in\mathbb{R}^{r}_{\rm alt},\qquad
A_{r}x^{A_{r}}=(\det{A_{r}})\alpha.$ (3.6)

We first claim that $A_{r}$ is invertible. Suppose instead that $A_{r}$ is
singular. Then $r=S^{+}(A_{r}x^{A_{r}})>S^{-}(x^{A_{r}})=r-1$, a contradiction
by $(3)$, and hence $A_{r}$ is invertible as claimed.

Next we show that $\det{A_{r}}<0$. By (3.6), we have

$r-1=S^{+}(A_{r}x^{A_{r}})=S^{-}(x^{A_{r}}),$

since the condition on the $\alpha_{i}$s implies that $S^{+}(\alpha)=r-1$.
Also, the sign of the first (last) component of $A_{r}x^{A_{r}}$ (if zero, the
sign that attains $S^{+}(A_{r}x^{A_{r}})$) is positive or negative if the
$\alpha_{i}$s are non-negative or non-positive respectively. The hypothesis in
$(3)$ and (3.5) imply that $\det{A_{r}}<0$. This completes the induction step
and also the proof. ∎

##  4\. Theorems D and E: Characterization of total negativity via the LCP

In this section, we show Theorems D and E. To proceed, we recall a 1990 result
which establishes a connection between the LCP and matrices with negative
principal minors.

######  Theorem 4.1.

[29] A matrix $A\in\mathbb{R}^{n\times n}$ with $A<0$ has all principal minors
negative if and only if LCP$(A,q)$ has exactly two solutions for all
$q\in\mathbb{R}^{n}$ with $q>0$.

###### Proof of Theorem D.

That $(1)\implies(2)$ follows from Theorem 4.1, while
$(2)\implies(3)\implies(4)$ is immediate. We now show $(4)\implies(1)$. We
first claim that $A<0$. Indeed, if $a_{ij}\geq 0$ for some $i\in[1,m]$ and
$j\in[1,n]$, then $(0)$ is the only solution of LCP$((a_{ij})_{1\times 1},q)$
for any scalar $q>0$.

Next, we show that the determinants of all $r\times r$ submatrices of $A$ are
negative, for $r\in[2,k]$. By Theorem A, it suffices to show for that every
contiguous square submatrix of $A$ of size $r\in[2,k]$ has the sign non-
reversal property on $\mathbb{R}^{r}_{\rm alt}$. Let $r\in[2,k]$ and $A_{r}$
be an $r\times r$ contiguous submatrix of $A$ which does not satisfy this
property. Then there exists $x\in\mathbb{R}^{r}_{\rm alt}$ such that
$x_{i}(A_{r}x)_{i}\leq 0$ for all $i\in[1,r]$. Let $A_{r}x=v$. Define
$x^{\pm}:=\frac{1}{2}(|x|\pm x)$ and $v^{\pm}:=\frac{1}{2}(|v|\pm v)$. Then
$x^{+}$, $x^{-}$ have sign patterns $\begin{pmatrix}+\\\ 0\\\ +\\\ 0\\\
\vdots\end{pmatrix}$, $\begin{pmatrix}0\\\ +\\\ 0\\\ +\\\ \vdots\end{pmatrix}$
respectively and ${(x^{+})}^{T}v^{+}={(x^{-})}^{T}v^{-}=0$, since
$x_{i}v_{i}\leq 0$ for all $i\in[1,r]$. Define

$q:=v^{+}-A_{r}x^{+}=v^{-}-A_{r}x^{-}.$ (4.1)

Then $q>0$, since $A<0$. Thus LCP$(A_{r},q)$ has solutions with sign patterns
$\begin{pmatrix}+\\\ 0\\\ +\\\ 0\\\ \vdots\end{pmatrix}$ and
$\begin{pmatrix}0\\\ +\\\ 0\\\ +\\\ \vdots\end{pmatrix}$, a contradiction.
Hence every contiguous square submatrix of $A$ of size $r\in[2,k]$ has the
sign non-reversal property on $\mathbb{R}^{r}_{\rm alt}$ and the result
follows. ∎

###### Proof of Theorem E.

That $(1)\implies(2)$ follows from Theorem D. To show $(2)\implies(1)$, by
Theorem 2.1, it suffices to show, for $r\in[1,k]$, that all contiguous minors
of $A$ of size $r$ are negative. The proof is by induction on $r$. Let $r=1$
and $A_{1}=(a_{ij})$ for some $i\in[1,m]$ and $j\in[1,n]$. Then
$x^{A_{1}}=(-1)$ by convention, and $q^{A_{1}}=-(a_{ij})$. If $a_{ij}=0$, then
LCP$(A_{1},q^{A_{1}})$ has infinitely many solutions, while if $a_{ij}>0$,
then LCP$(A_{1},q^{A_{1}})$ has only one solution, a contradiction. Thus all
entries of $A$ are negative.

For the induction step, suppose $A_{r}$ is an $r\times r$ contiguous submatrix
of $A$, with $r\in[2,k]$ and all contiguous minors of $A$ of size at most
$r-1$ are negative. By Theorem 2.1, all proper minors of $A$ are negative.
Define

$x^{A_{r}}:=(A^{11}_{r},0,A^{13}_{r},0,\ldots)^{T},\qquad
z^{A_{r}}:=(0,A^{12}_{r},0,A^{14}_{r},\ldots)^{T},\qquad
q^{A_{r}}:=A_{r}x^{A_{r}}.$ (4.2)

Then $x^{A_{r}},~{}z^{A_{r}}\leq 0$ and $q^{A_{r}}>0$. Thus $\bf{0}$ and
$-x^{A_{r}}$ are two solutions of LCP$(A_{r},q^{A_{r}})$.

We next claim that $A_{r}$ is invertible. If not, then
$Ax^{A_{r}}=Az^{A_{r}}$. So $-z^{A_{r}}$ is also a solution of
LCP$(A_{r},q^{A_{r}})$, a contradiction. This shows the claim.

Finally we show that $\det{A_{r}}<0$. Indeed, suppose $\det{A_{r}}>0$. Then

$y=-A_{r}z^{A_{r}}+q^{A_{r}}=(\det{A_{r}})e^{1}\geq 0\hbox{~{}~{}~{}
and~{}~{}~{}}y^{T}z^{A_{r}}=0$.

Thus $-z^{A_{r}}$ is also a solution of LCP$(A_{r},q^{A_{r}})$, a
contradiction by the hypothesis (2). Thus $\det{A_{r}}<0$. This completes the
induction step and hence the proof. ∎

######  Remark 4.2.

In Theorem E, in place of the vector
$x^{A_{r}}=(A^{11}_{r},0,A^{13}_{r},0,\ldots)^{T}$, one can take
${x^{i}}^{A_{r}}:=(A^{i1}_{r},0,A^{i3}_{r},0,\ldots)^{T}$ for odd $i\in[1,r]$,
or a positive linear combination of these vectors. The result still holds with
a similar proof, where we define $q^{A_{r}}$ similarly as in (1.5).

##  5\. Characterizations for totally non-positive matrices

Our final section presents the counterparts of the above main results for
totally non-positive matrices as promised in the introduction. We first derive
the sign non-reversal property and interval hull tests for totally non-
positive matrices. For this, we require a 1950 density result of
Gantmacher–Krein for totally negative matrices (in fact strictly sign regular
matrices).

######  Theorem 5.1.

[19] Given integers $m,n\geq k\geq 1$, the set of $m\times n$ totally negative
matrices of order $k$ is dense in the set of $m\times n$ totally non-positive
matrices of order $k$.

######  Theorem 5.2.

Let $m,n\geq k\geq 2$ be integers. Given $A\in\mathbb{R}^{m\times n}$ with
$A\leq 0$, the following statements are equivalent:

  1. (1)

The matrix $A$ is totally non-positive of order $k$.

  2. (2)

Every square submatrix of $A$ of size $r\in[2,k]$ has the non-strict sign non-
reversal property on $\mathbb{R}^{r}_{\rm+,-}$.

  3. (3)

Every square submatrix of $A$ of size $r\in[2,k]$ has the non-strict sign non-
reversal property on $\mathbb{R}^{r}_{\rm alt}$.

  4. (4)

For every $r\in[2,k]$ and $r\times r$ submatrix $A_{r}$ of $A$, the matrix
$A_{r}$ has the non-strict sign non-reversal property for the vector
$x^{A_{r}}=\operatorname{adj}(A_{r})\alpha$, for any choice of
$\alpha\in\mathbb{R}^{r}_{\rm alt}$.

###### Proof.

First suppose $A\in\mathbb{R}^{m\times n}$ is totally non-positive of order
$k$. By Theorem 5.1, there exists a sequence $A^{(l)}$ of totally negative
matrices of order $k$ such that $A^{(l)}\to A$ entrywise. Fix an $r\times r$
submatrix $A_{r}$ of $A$ for $r\in[2,k]$, and let $A^{(l)}_{r}$ be the
submatrix of $A^{(l)}$ indexed by the same rows and columns of $A_{r}$. Now
fix a vector $x\in\mathbb{R}^{r}_{\rm+,-}$, and let $I\subseteq[1,r]$ index
the nonzero coordinates of $x$. Since $A^{(l)}_{r}$ is totally negative, so by
Theorem A(2) there exists $i_{l}\in I$ such that
$x_{i_{l}}(A^{(l)}_{r}x)_{i_{l}}>0$. Since $I$ is finite, there exists an
increasing subsequence of positive integers $l_{s}$ such that
$i_{l_{s}}=i_{0}$ for some $i_{0}\in I$ and for all $s\geq 1$. Hence

$x_{i_{0}}(A_{r}x)_{i_{0}}=\lim_{s\to\infty}x_{i_{l_{s}}}(A^{(l_{s})}_{r}x)_{i_{l_{s}}}\geq
0,\qquad x_{i_{0}}\neq 0.$

This completes the proof of $(1)\implies(2)$.

Next, that $(2)\implies(3)\implies(4)$ is immediate. Now assume (4) holds. We
show by induction on $r\in[1,k]$ that the determinant of every $r\times r$
submatrix of $A$ is non-positive. The base case $r=1$ follows from the
hypothesis. For the induction step, suppose $A_{r}$ is a square submatrix of
$A$ of size $r\in[2,k]$ and all the proper minors of $A$ are non-positive. If
$\det A_{r}=0$ then we are done; else $\det(A_{r})\neq 0$. Let
$\alpha=(\alpha_{1},\ldots,\alpha_{r})^{T}\in\mathbb{R}^{r}_{\rm alt}$ and
define the vector $x^{A_{r}}:=\operatorname{adj}(A_{r})\alpha$. Since no row
of $\operatorname{adj}(A_{r})$ is zero, by (2.1),
$x^{A_{r}}\in\mathbb{R}^{r}_{\rm alt}$ and $x^{A_{r}}_{i}<0~{}(>0)$ if
$\alpha_{i}>0~{}(<0)$ for all $i\in[1,r]$. Now a similar calculation as in
(2.3), implies that for some $i_{0}\in[1,r]$, we have

$0\leq x^{A_{r}}_{i_{0}}(A_{r}x^{A_{r}})_{i_{0}}=(\det
A_{r})\alpha_{i_{0}}x^{A_{r}}_{i_{0}}.$

Since $x^{A_{r}}_{i_{0}}$ and $\alpha_{i_{0}}$ have from opposite signs, it
follows that $\det A_{r}<0$, which completes the proof by induction. ∎

Given Theorem 5.2, which is a totally non-positive of order $k$ analogue of
Theorem A, a natural question is to seek a similar totally non-positive (of
order $k$) analogue of Theorem B. In [1], the authors show that under certain
technical constraints, a minimal test set of two matrices suffices to check
the total non-negativity of the entire interval hull. Our next result removes
these technical constraints from [1], and holds for totally non-positive (of
order $k$) interval hulls for arbitrary $k\geq 1$ – at the cost of working
with a larger (but nevertheless finite) test set:

######  Theorem 5.3.

Let $m,n\geq k\geq 1$ be integers, and $A,B\in\mathbb{R}^{m\times n}$. Then
$\mathbb{I}(A,B)$ is totally non-positive of order $k$ if and only if the
matrices $\\{I_{z,\tilde{z}}(A,B):z\in\\{\pm 1\\}^{m},\tilde{z}\in\\{\pm
1\\}^{n}\\}$ are all totally non-positive of order $k$.

######  Remark 5.4.

As in Remark 1.3, note that this set of test vectors is independent of the
choice $k$.

###### Proof.

We assume $m,n\geq k\geq 2$, since the result is immediate for $k=1$. If
$\mathbb{I}(A,B)$ is totally non-positive of order $k$, then so are
$I_{z,\tilde{z}}(A,B)$ by Lemma 2.3.

To show the converse, we repeat the proof of Theorem B, but we work with
arbitrary submatrices $M^{\prime}$ of $M\in\mathbb{I}(A,B)$ of size
$r\in[2,k]$ instead of contiguous submatrices. We once again have (2.4), but
the matrix $I_{z^{\prime},z^{\prime}}(A^{\prime},B^{\prime})$ is a submatrix
of $I_{z,\tilde{z}}(A,B)$ for some $z\in\\{\pm 1\\}^{m},~{}\tilde{z}\in\\{\pm
1\\}^{n}$, which is totally non-positive of order $k$ by the hypothesis. The
remainder of the proof is unchanged, except for the fact that we use Theorem
5.2 in place of Theorem A and the matrix $M$ is non-positive (since
$I_{z,z^{\prime}}(A,B)$ are non-positive). ∎

Analogous to Theorem C, we next provide a characterization in terms of
variation diminution for totally non-positive matrices.

######  Theorem 5.5.

Given a real $m\times n$ matrix $A\leq 0$ and $m,n\geq 2$, the following
statements are equivalent:

  1. (1)

$A$ is totally non-positive.

  2. (2)

For all $x\in\mathbb{R}^{n}_{\rm+,-}$, $S^{-}(Ax)\leq S^{-}(x)$. Moreover, if
equality occurs and $Ax\neq 0$, then the first (last) nonzero component of
$Ax$ has the same sign as the first (last) nonzero component of $x$.

  3. (3)

Let $k=\min\\{m,n\\}$. For every square submatrix $A_{r}$ of $A$ of size
$r\in[2,k]$ define the vector

$y^{A_{r}}:=\operatorname{adj}(A_{r})\alpha,\hbox{for
some}~{}\alpha\in\mathbb{R}^{r}_{\rm alt}.$ (5.1)

Then $S^{-}(A_{r}y^{A_{r}})\leq S^{-}(y^{A_{r}})$. If equality occurs here and
$A_{r}y^{A_{r}}\neq 0$, then the first (last) nonzero component of
$A_{r}y^{A_{r}}$ has the same sign as the first (last) nonzero component of
$y^{A_{r}}$.

Apart from Gantmacher–Krein’s density Theorem 5.1, the proofs require another
preliminary lemma on sign changes of limits of vectors; see e.g. [30] for the
proof.

######  Lemma 5.6.

Given $x=(x_{1},\ldots,x_{n})^{T}\in\mathbb{R}^{n}\setminus\\{0\\}$, define
the vector
$\overline{x}:=(x_{1},-x_{2},x_{3},\ldots,(-1)^{n-1}x_{n})\in\mathbb{R}^{n}$.
Then

$S^{+}(x)+S^{-}(\overline{x})=n-1.$

Moreover, if $\lim\limits_{p\to\infty}x_{p}=x$, then

$\liminf\limits_{p\to\infty}S^{-}(x_{p})\geq
S^{-}(x),\qquad\limsup\limits_{p\to\infty}S^{+}(x_{p})\leq S^{+}(x).$

We now prove the above characterization of totally non-positive matrices.

###### Proof.

We first show that $(1)\implies(2)$. Since $A$ is totally non-positive, by
Gantmacher–Krein’s density Theorem 5.1, there exists a sequence $(A_{p})$ of
totally negative matrices such that

$\lim\limits_{p\to\infty}A_{p}=A.$

Let $x\in\mathbb{R}^{n}_{\rm+,-}$. By Theorem C and Lemma 5.6, we have

$S^{-}(Ax)\leq\liminf\limits_{p\to\infty}S^{-}(A_{p}x)\leq\liminf\limits_{p\to\infty}S^{+}(A_{p}x)\leq\liminf\limits_{p\to\infty}S^{-}(x)=S^{-}(x).$

Next, assume that $S^{-}(Ax)=S^{-}(x)=k$ and $Ax\neq 0$. Then for all
sufficiently large $p$, by Theorem C and Lemma 5.6,

$S^{-}(Ax)\leq S^{-}(A_{p}x)\leq S^{+}(A_{p}x)\leq S^{-}(x)=k.$

Thus $S^{-}(A_{p}x)=S^{+}(A_{p}x)=S^{-}(x)=k$ for $p$ large enough. This
implies the sign patterns in $A_{p}x$ (for large $p$) does not depend on the
zero entries. Thus the nonzero sign patterns of $A_{p}x$ agree with those of
$Ax$. Also, by Theorem C, the sign patterns of $x$ agree with the sign
patterns of $A_{p}x$. Hence the same holds for the sign patterns of $Ax$ and
$x$.

To show $(2)\implies(3)$, repeat the proof of Theorem C, but work with
arbitrary submatrices $A_{r}$ of $A$ of size $r\in[2,k]$ and the vector
$y^{A_{r}}=\operatorname{adj}(A_{r})\alpha$, where
$\alpha\in\mathbb{R}^{r}_{\rm alt}$ is arbitrarily chosen.

Finally, we show that $(3)\implies(1)$. The claim that $\det{A_{r}}\leq 0$ for
all $r\times r$ submatrices $A_{r}$ of $A$, is shown by induction on
$r\in[1,k]$. The base case $r=1$ holds by the hypothesis.

For the induction step, $A_{r}$ is an $r\times r$ submatrix of $A$ and all
proper minors of $A_{r}$ are non-positive. If $\det{A_{r}}=0$ then there is
nothing to prove, so suppose $\det{A_{r}}\neq 0$. Let
$\alpha=(\alpha_{1},\alpha_{2}\ldots,\alpha_{r})^{T}\in\mathbb{R}^{r}_{\rm
alt}$ and define $y^{A_{r}}$ as in (5.1). Then no row of
$\operatorname{adj}(A_{r})$ is zero and $y^{A_{r}}\in\mathbb{R}^{r}_{\rm
alt}$. Also, $y^{A_{r}}_{i}$ is negative (or positive) if $\alpha_{i}$ is
positive (or negative) for all $i\in[1,r]$ (by (3.5)). Since
$A_{r}y^{A_{r}}=(\det{A_{r}})\alpha$, we have

$S^{-}(A_{r}y^{A_{r}})=S^{-}(y^{A_{r}})=r-1.$

By the last part of the hypothesis (3), $\det{A_{r}}<0$ and so we are done by
induction. ∎

For the sake of completeness, we conclude by showing sufficient condition for
totally non-positive matrices using the Linear Complementarity Problem.

######  Proposition 5.7.

Let $m,n\geq k\geq 2$ be integers and let $A\in\mathbb{R}^{m\times n}$ with
$A\leq 0$. Then $A$ is totally non-positive of order $k$ if for every $r\times
r$ submatrix $A_{r}$ of $A$ and for all $q\in\mathbb{R}^{n}$ with $q\geq 0$,
if $z^{1}=(x_{1},0,x_{3},\ldots)^{T}$ and $z^{2}=(0,x_{2},0,x_{4},\ldots)^{T}$
(where all $x_{i}>0$) are two solutions of LCP$(A_{r},q)$, then
$A_{r}z^{1}=A_{r}z^{2}$.

###### Proof.

By Theorem 5.2, it is enough to show that every $r\times r$ submatrix $A_{r}$
of $A$ has the non-strict sign non-reversal property with respect to
$\mathbb{R}^{r}_{\rm alt}$, for $r\in[2,k]$. Fix $r\in[2,k]$. Suppose that
there exist an $r\times r$ submatrix $A_{r}$ of $A$ and a vector
$x\in\mathbb{R}^{r}_{\rm alt}$ such that $x_{i}(A_{r}x)_{i}<0$ for all
$i\in[1,r]$. Defining $x^{+},x^{-},v^{+},v^{-}$ and $q$ as in the proof of
Theorem D, we conclude that $q\geq 0$ and $x^{+},x^{-}$ are two solutions of
LCP$(A_{r},q)$. Also, $x^{+},~{}x^{-}$ are with sign patterns
$\begin{pmatrix}+\\\ 0\\\ +\\\ 0\\\ \vdots\end{pmatrix}$ and
$\begin{pmatrix}0\\\ +\\\ 0\\\ +\\\ \vdots\end{pmatrix}$ respectively. By
(4.1), we have

$A_{r}x^{+}+q=v^{+}\neq v^{-}=A_{r}x^{-}+q,$

since $Ax\in\mathbb{R}^{r}_{\rm alt}$. Thus $A_{r}x^{+}\neq A_{r}x^{-}$, a
contradiction. Hence $A$ is totally non-positive of order $k$. ∎

The next result provides an improvement, via Linear Complementarity at a
single vector $q$ (for each square submatrix of $A$).

######  Proposition 5.8.

Let $m,n\geq k\geq 1$ be integers and $A\in\mathbb{R}^{m\times n}$. Then $A$
is totally non-positive of order $k$ if for every square submatrix $A_{r}$ of
$A$ of size $r\in[1,k]$ and $q^{A_{r}}$ as defined in (1.5), the following
holds:

  1. (i)

$\bf{0}$ is the solution of LCP$(A_{r},q^{A_{r}})$.

  2. (1)

If $z^{1}$ and $z^{2}$ are two nonzero solutions of LCP$(A_{r},q^{A_{r}})$
then $A_{r}z^{1}=A_{r}z^{2}$.

###### Proof.

We prove by induction on $r\in[1,k]$ that $\det{A_{r}}\leq 0$ for all $r\times
r$ submatrices $A_{r}$ of $A$. Let $r=1$ and $A_{1}=(a_{ij})$ for some
$i\in[1,m],~{}j\in[1,n]$. Then $x^{A_{1}}=(-1)$ by convention, and
$q^{A_{1}}=-(a_{ij})$. If $a_{ij}>0$ then $\bf{0}$ is not a solution of
LCP$(A_{1},q^{A_{1}})$. Thus $A\leq 0$.

Let $r\in[2,k]$ and suppose that all the $(r-1)\times(r-1)$ and smaller minors
of $A$ are non-positive, and let $A_{r}$ be a submatrix of $A$ of size $r$. If
$\det{A_{r}}=0$, then we are done, else assume $A_{r}$ is non-singular. If
$\det{A_{r}}>0$, repeating the proof of Theorem E, once again we have
$-x^{A_{r}}$ and $-z^{A_{r}}$ (as defined in (4.2)) are two distinct nonzero
solutions of LCP$(A_{r},q^{A_{r}})$, but $Ax^{A_{r}}\neq Az^{A_{r}}$. Thus
$\det{A_{r}}<0$ and hence $A$ is totally non-positive of order $k$. ∎

######  Example 5.9.

The converse of the previous theorem does not hold. We explain this with an
example. Consider the totally non-positive matrix $A=\begin{pmatrix}0&0&0\\\
-1&-3&-3\\\ -1&-1&-1\end{pmatrix}$. Then $q=\begin{pmatrix}0\\\ 1\\\
1\end{pmatrix}$ and $z^{1}=\begin{pmatrix}1\\\ 0\\\ 0\end{pmatrix}$,
$z^{2}=\begin{pmatrix}t\\\ 0\\\ 0\end{pmatrix}$, where $t>1$, are two
solutions of LCP$(A,q)$, but $Az^{1}\neq Az^{2}$.

##  6\. Other orthants cannot yield test vectors

In the preceding sections, we have characterized a negative/non-positive
matrix to be totally negative/totally non-positive of order $k$, by its square
submatrices $A_{r}$ with $r\in[2,k]$ satisfying the sign non-reversal or
variation diminishing property on all of the open bi-orthant
$\mathbb{R}^{r}_{\rm alt}$ – or equivalently, on single test vectors which are
drawn from this bi-orthant. Our next result explains that these results are
‘best possible’ in the following sense: if
$x\in\mathbb{R}^{r}_{\rm+,-}\setminus\mathbb{R}^{r}_{\rm alt}$ with all
$x_{i}\neq 0$ (i.e., there are two successive $x_{i}$ of the same sign), then
every $r\times r$ totally negative matrix of order $r-1$ satisfies the sign
non-reversal property and the variation diminishing property for $x$. Thus the
above characterizations cannot hold with test vectors in any open bi-orthant
other than in $\mathbb{R}^{r}_{\rm alt}$, since one shows that there exist
totally negative matrices with positive determinants (see e.g. [19]).

######  Theorem 6.1.

Let $A_{r}\in\mathbb{R}^{r\times r}$ be totally negative matrix of order $r-1$
and let $x\in\mathbb{R}^{r}_{\rm+,-}\setminus\mathbb{R}^{r}_{\rm alt}$ with
all $x_{i}$ are nonzero. Then:

  1. (1)

There exists a coordinate $i\in[1,r]$ such that $x_{i}(A_{r}x)_{i}>0$. In
other words, $A$ satisfies the sign non-reversal property.

  2. (2)

$S^{+}(A_{r}x)\leq S^{-}(x)$. Moreover, if equality occurs, then the first
(last) nonzero component of $Ax$ (if zero, the unique sign required to
determine $S^{+}(Ax)$) has the same sign as the first (last) component of $x$.
In other words, $A$ satisfies the variation diminishing property.

###### Proof.

Suppose $x\in\mathbb{R}^{r}_{\rm+,-}\setminus\mathbb{R}^{r}_{\rm alt}$ with
all $x_{i}\neq 0$. One can partition $x$ as in (3.1) and the subsequent lines,
with $k$ replaced by some $p\leq r-2$. By the argument following (3.1), the
matrix $B\in\mathbb{R}^{r\times{(p+1)}}$ is totally negative and
$Ax=Bd^{[p+1]}$. We define $y:=Ax=Bd^{[p+1]}$.

To show $(1)$, suppose for contradiction that $x_{i}y_{i}\leq 0$ for all
$i\in[1,r]$. Now consider the index set

$I=\\{i_{1},\ldots,i_{k+1}\\},\hbox{ where }i_{t}\in[s_{t-1}+1,s_{t}]$.

Then the square submatrix $B_{I\times[p+1]}$ is totally negative and it
reverses the signs of $d^{[p+1]}$, since $B_{I\times[k+1]}d^{[k+1]}=y_{I}$.
This gives the desired contradiction by Theorem A.

The proof of $(2)$ is similar to that of case $(2)$ of Theorem C. ∎

An analogue of Theorem 6.1 also holds for the class of totally non-positive
matrices:

######  Proposition 6.2.

If $A\in\mathbb{R}^{r\times r}$ is a totally non-positive matrix of order
$r-1$ and the vector $x$ is defined as in Theorem 6.1, then:

  1. (1)

There exists a coordinate $i\in[1,r]$ such that $x_{i}\neq 0$ and
$x_{i}(A_{r}x)_{i}\geq 0$.

  2. (2)

$S^{-}(A_{r}x)\leq S^{-}(x)$. Moreover, if equality occurs and $Ax\neq 0$,
then the first (last) nonzero component of $Ax$ has the same sign as the first
(last) nonzero component of $x$.

###### Proof.

Repeat the proofs of Theorems 5.2 and 5.5, except for the use of Theorem 6.1
in place of Theorems A and C respectively. ∎

We conclude with a similar observation about the LCP: Theorem D shows that for
a totally negative matrix $A$ of order $k$, the solution sets SOL$(A_{r},q)$
with $r\in[2,k]$ do not contain two vectors with alternately positive and zero
coordinates (and disjoint supports) simultaneously, for all
$q\in\mathbb{R}^{r}$ with $q>0$. Our final result shows that if
$A\in\mathbb{R}^{r\times r}$ is totally negative of order $k$, the same holds
when ‘alternating’ is replaced by ‘not always alternating’. Hence by the
existence result alluded to in the first paragraph of this section, totally
negative matrices can not be detected by the solution sets SOL$(A,q)$ of LCP,
which do not contain two vectors with disjoint supports, at least one of which
has two consecutive positive coordinates. Thus, the LCP-characterization of
total negativity also distinguishes the above ‘alternation’.

######  Proposition 6.3.

If $A\in\mathbb{R}^{r\times r}$ with $r\geq 2$ is a totally negative matrix of
order $r-1$, then SOL$(A_{r},q)$ with $q>0$ does not simultaneously contain
two vectors which have disjoint supports, and at least one of which has two
consecutive positive coordinates.

The proof is analogous to Theorem D using Theorem 6.1 $(1)$.

## Acknowledgments

I thank Jürgen Garloff for pointing me to the work of Gantmacher–Krein for
existence and density results involving totally negative matrices. I also
thank Apoorva Khare for a detailed reading of an earlier draft and for
providing valuable feedback. This work is supported by National Post-Doctoral
Fellowship (PDF/2019/000275) from SERB (Govt. of India).

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11institutetext:  IKERBASQUE, Basque Foundation for Science, E-48009 Bilbao,
Bizkaia, Spain
Departmento de Química-Física, Universidad del País Vasco, UPV/EHU, Leioa,
Bizkaia, Spain
Foundations of quantum mechanics Basque Center for Applied Mathematics (BCAM),
Alameda de Mazarredo, 14, 48009 Bilbao, Bizkaia, Spain

Entanglement and quantum nonlocality

#  Wigner’s friends, tunnelling times and Feynman’s only mystery of quantum
mechanics

D. Sokolovski 1122 and E. Akhmatskaya 2233112233

###### Abstract

Recent developments in elementary quantum mechanics have seen a number of
extraordinary claims regarding quantum behaviour, and even questioning
internal consistency of the theory. These are, we argue, different disguises
of what Feynman described as quantum theory’s only mystery

###### pacs:

03.65.Ta

###### pacs:

03.65.UD

> Real mystics don’t hide mysteries, they reveal them. They set a thing in
> broad daylight, and when you’ve seen it it’s still a mystery. But the
> mystagogues hide a thing in darkness and secrecy, and when you find it, it’s
> a platitude.
>
> G. K. Chesterton in The arrow of heaven

> I will take just one this experiment, which has been designed to contain all
> of the mysteries of quantum mechanics…. Any other situation in quantum
> mechanics, it turns out, can always be explained by saying “You remember the
> case of the experiment with the two holes?”
>
> R.P. Feynman in Character of Physical Law

##  1 Introduction

We would like to begin by assuring the reader that the first item in the
epigraph was not chosen in order to be offensive. However, the two quotes suit
the purpose of this article so well that it would be a pity not to include
them. Since Wigner’s surprising suggestion [1] that the laws of quantum
mechanics may need to be modified to accommodate human consciousness, there
have been many other eye-catching claims, associated with quantum behaviour.
The list includes, to name a few, particles being in several places at the
same time [2], electrons “disembodied of their charge” [3], quantum “Cheshire
cats” [4], photons with “discontinuous trajectories” [5], observer-dependent
facts [6],[7], doubts about the internal consistency of elementary quantum
theory [8] and the conflict of “faster-than-light tunnelling” with special
relativity [9]. A recent claim that it takes a finite amount of time to tunnel
across a potential barrier can be found in [10].
Feynman, for his part, maintained that all rarities of quantum mechanics can
be traced back to the double slit experiment. So are these quantum “paradoxes”
just different form of the familiar double-slit conundrum? We show that most,
if not all of them, are.

##  2 The double-slit experiment. Feynman’s rules

It is worth recalling the description, given by Feynman in his undergraduate
text [11]. An electron, emitted by a source (s), can reach a point on a
screen, $x$, by passing via two slits, labelled $1$ and $2$. If it is
impossible in principle to establish which of the two paths was taken, the
probability of arriving at $x$ is given by the absolute square of the sum of
the path amplitudes, $P(x)=|A(x\leftarrow 1\leftarrow s)+A(x\leftarrow
2\leftarrow s)|^{2}$. If it is possible, even in principle, the path
probabilities are added, $P(x)=|A(x\leftarrow 1\leftarrow
s)|^{2}+|A(x\leftarrow 2\leftarrow s)|^{2}$. The fact that these two
situations are mutually exclusive constitutes the Uncertainty Principle (UP).

A useful illustration can be provided by considering a two-level system (S) (a
spin-$1/2$) and two probes $\overline{D}$ and $D$, used to measure the spin’s
condition at $t_{1}$ and $t_{2}>t_{1}$, respectively. The joint initial state
at some $t_{0}<t_{1}$ is
$|\Psi(0){\rangle}=|D(0){\rangle}\otimes|\overline{D}(0){\rangle}\otimes|s_{0}{\rangle}$,
and just after $t_{1}$ we have

$\displaystyle|\Psi(t_{1}){\rangle}=|D(0){\rangle}\otimes[{\langle}up|s_{0}{\rangle}|\overline{D}(up){\rangle}\otimes|up{\rangle}+$
(1)
$\displaystyle{\langle}down|s_{0}{\rangle}|\overline{D}(down){\rangle}\otimes|down{\rangle}],$

where $|up{\rangle}$ and $|down{\rangle}$ form a measurement basis in the
spin’s Hilbert space, and we assumed, for simplicity, only the spin has its
own dynamics.

[angle=0,width=6.0 cm, height= 3cm]FIG1.pdf

Figure 1: A primitive double slit experiment. A spin can reach its final
states via four virtual paths.

If the second measurement engages only the spin, by using a measurement basis,

$\displaystyle|ok{\rangle}=\alpha|up{\rangle}+\beta|down{\rangle},\quad|fail{\rangle}=\gamma|up{\rangle}+\delta|down{\rangle}$
(2)

the probability for each of the four possible outcomes is easily found to be

$\displaystyle P(i,j)=|A^{S}(j\leftarrow i\leftarrow s_{0})|^{2},$ (3)
$\displaystyle i=up,down,\quad j=ok,fail,$

where

$\displaystyle A^{S}(j\leftarrow i\leftarrow
s_{0})\equiv{\langle}j|\hat{U}^{S}(t_{2},t_{1})|i{\rangle}{\langle}i|\hat{U}^{S}(t_{1},t_{0})|s_{0}{\rangle},$
(4)

and $\hat{U}^{S}(t^{\prime},t)$ is the spin’s evolution operator. If the
second measurement engages both the spin and the probe $\overline{D}$, by
using a basis containing the states

$\displaystyle|ok{\rangle}=\alpha|\overline{D}(up){\rangle}\otimes|up{\rangle}+\beta|\overline{D}(down){\rangle}\otimes|down{\rangle},$
(5)
$\displaystyle|fail{\rangle}=\gamma|\overline{D}(up){\rangle}\otimes|up{\rangle}+\delta|\overline{D}(down){\rangle}\otimes|down{\rangle},$

the probabilities for the two corresponding outcomes are given by

$\displaystyle P(j)=|A^{S}(j\leftarrow up\leftarrow s_{0})+A^{S}(j\leftarrow
down\leftarrow s_{0})|^{2},$ (6) $\displaystyle j=ok,fail.$

In the first case, system’s past can be determined by inspecting the state of
the probe $\overline{D}$ after the experiment is finished. Accordingly, the
four scenarios in Fig. 1 can be distinguished, and endowed with the
probabilities in Eq.(3). In the second case, the record carried by
$\overline{D}$ has been erased and, according to the UP, it should be
impossible to say whether the spin’s condition at $t_{1}$ was $up$ or $down$.
A reader, worried about the collapse of the wave function, may find some
comfort in noting that the calculation of probabilities is reduced to
evaluation of matrix elements of unitary operators in the system’s Hilbert
space, and the collapse problem is not mentioned at all.

The UP rule does not rely on human experience. Mere existence of a photon,
scattered near one of the slits, and carrying a record of the electron’s past,
is enough to destroy the interference pattern, even if the photon is never
observed.
Feynman’s advice was to accept the above rules as the definitive content of
the quantum theory, and avoid in this way the “blind alley”, reserved for
those who ask “how can it be like that?”[13]. We follow the advice, and ask
instead a simpler question, are the recently discovered quantum “paradoxes”
mere illustrations of the Feynman’s rules? Some of them, such as those in
[2-5], certainly appear to be just that. With one of the two holes closed, an
electron is able to arrive where it could not arrive with both of them open.
For this reason, the amplitudes should not be used to prove that the electron
was at a particular place in the past [12]. This explains the “quantum
paradoxes” constructed in [2]-[5] (for more details see [14]). Would it also
be true for the cases discussed in [6]-[10]?

##  3 The question of consciousness

Recent work on various versions of the Wigner’s friend scenario [6]-[8] rarely
mentions the consciousness of the participants explicitly. However, the
original Wigner’s analysis [1] was motivated by his concern about the role
played by consciousness in quantum measurements, and we must attend to the
question before we proceed. According to Wigner [1], Eq.(1) is valid when it
describes an inanimate object, but is unacceptable if
$|\overline{D}(up/down){\rangle}$ is understood as a state of conscious
Wigner’s friend, ($F$), who has just seen an outcome $up$ or $down$. Would the
Feynman’s rules of the previous section apply in the presence of intelligent
Observers?
There are at least three ways to look at an Observer’s (e.g. $F$’, $W$’),
consciousness, while maintaining that quantum mechanics should apply to all
physical objects, regardless of their size and complexity.
(a) Condition of a consciousness can be deemed to be fully determined by that
of the inanimate “physico-chemical substrate” [1] of an Observer’s organism.
Then Wigner’s objection can be dismissed by noting that in Eqs.(1) the states
$|\overline{D}(up){\rangle}$ and $|\overline{D}(down){\rangle}$ are used for
calculating the odds on $F$ seeing $up$ or $down$, but never both $up$ and
$down$ at the same time. However, the difficulty, resolved in this manner,
returns when one considers Eqs.(5)-(6). Then an outcome $ok$, seen by Wigner
($W$), who performs the second measurement, would leave the joint system,
which now includes the friend $F$, in an undesirable state of “suspended
animation” [1],
$\alpha|\overline{D}(up){\rangle}\otimes|up{\rangle}+\beta|\overline{D}(down){\rangle}\otimes|down{\rangle}$.
(b) A different problem arises if one were to try placing Observer’s “extra-
observational intellectual inner life” [27] outside the scope of quantum
theory. Now, after having seen an $up$, $F$ becomes permanently aware of
his/her outcome and this information resides in a place obscure to quantum
reasoning. Then if $W$ duly entangles all material records produced by $F$,
his probability, $P(ok)$ in Eq.(6) should contain an interference term
$2\text{Re}[A^{S^{*}}(ok\leftarrow up\leftarrow s_{0})A^{S}(ok\leftarrow
down\leftarrow s_{0})]$. But if $F$ were able to declare that his/her outcome
was $up$, the Uncertainty Principle would be violated, since $W$’s
interference picture would have to co-exist with $F$’s “which way?”
information.
(c) The third possibility [15], [16] is a compromise between the first two,
and is consistent with Feynman’s analysis of the double slit experiment. As in
(b), $F$’s consciousness should not be analysed by means of quantum theory.
However, unlike in (b), an Observer is not permanently aware of his/her
outcome but needs, when necessary, to consult a material record such as the
one kept by his/her memory or in a note, accessible to human senses. A mere
act of perception on the part of an Observer adds nothing to quantum analysis,
but a new record produced as a result (recall Feynman’s photon destroying the
double slit interference), needs to be taken into account. As in (a) only
material objects need to be subjects of a quantum analysis.
Accepting (c) has further implications. In particular, at any given time, the
facts about the outcomes of past experiments are contained in the set of
material records. The facts are objective in the sense of being equally valid
to all intelligent agents. We note that Feynman [11] goes even further in
suggesting that facts do not need to be verified, in order to make the
presence of records felt. For example, the mere existence of an additional
spin, prepared by $F$ in a state $|up{\rangle}$ after seeing an $up$, would
remove the interference term from $W$’s probabilities in Eq.(6) even if this
second spin were to be sent to the Alfa Centauri, and never seen by anyone
again. A record can be created, as well as destroyed, in which case the
information it contains may be irretrievably lost. Observers are free to
decide which experiments to make, and can sometimes use the arrangements
already provided by Nature, as would happen, for example, with Wheeler’s
“cosmic interferometer” [17]. Finally, as suggested by Wigner, Observer’s
consciousnesses “never seem to interact with each other directly, but only via
the physical world”[1], i.e. via the world of meaningful records.
Throughout the rest of the paper, we will continue to assume that quantum
theory is valid for all material objects, and leave the Observers all but
outside our discussion in accordance with (c). Next we turn to the extended
Wigner’s Friend scenario, recently proposed in [8] in order to expose alleged
inconsistencies of the elementary quantum mechanics.

##  4 Extended Wigner’s friend scenarios and quantum postmodernism

With what has just been said, the original Wigner friend scenario fits the
description of the double-slit experiment (1), where the states
$|\overline{D}(up/down){\rangle}$ refer to all material records produced by
$F$. One can, however, talk himself into a logical contradiction with the help
of the wave function in Eq.(1). Indeed, measuring the spin, $F$ obtains a
definite outcome $up$ with the probability (to shorten the notations, we put
$\hat{U}^{S}(t^{\prime},t)=1$)
${\langle}\Psi(t_{1})|up{\rangle}{\langle}up|\Psi(t_{1}){\rangle}=|{\langle}up|s_{0}{\rangle}|^{2}$.
On the other hand, the probability of $W$’s outcome $ok$ in Eq.(6),
${\langle}\Psi(t_{1})|ok{\rangle}{\langle}ok|\Psi(t_{1}){\rangle}$, contains
an interference term
$2\text{Re}[{\langle}s_{0}|down{\rangle}{\langle}down|ok{\rangle}{\langle}ok|up{\rangle}{\langle}up|s_{0}{\rangle}]$.
This should leave the question “$up$ or $down$?” without an answer, yet $F$
appears to know that the answer was $up$. The reader may be able to see
through this apparent contradiction. We will give a full explanation after
considering the case where a similar “contradiction” appears in a yet more
dramatic form.
In what has become known as the extended Wigner’s friend scenario (EWFS) [8],
two two-level systems, called “the coin” (C), and “the spin” (S),
respectively, are prepared in a product state at $t=t_{0}$. At $t_{1}$ an
Observer $\overline{F}$ measures the coin in a basis $|heads/tails{\rangle}$,
and at $\tau>t_{1}$ the coin is entangled with the spin. At a $t_{2}>\tau$,
$F$ measures the spin in a basis $|up/down{\rangle}$. The experiment is
finished when at $t_{3}>t_{2}$ external Observers, $\overline{W}$ and $W$,
measure the entire material content of $\overline{F}$’s and $F$’s labs using
bases

$\displaystyle|\overline{Fail}/\overline{Ok}{\rangle}=$ (7)
$\displaystyle[|\overline{D}(heads){\rangle}\otimes|heads{\rangle}\pm|\overline{D}(tails){\rangle}\otimes|tails{\rangle}]/\sqrt{2},$
$\displaystyle|{Fail/Ok}{\rangle}=$
$\displaystyle[|D(up){\rangle}\otimes|up{\rangle}\pm|D(down){\rangle}\otimes|down{\rangle}]/\sqrt{2}.$

The initial state and the interaction at $t=\tau$ are chosen so that the state
of $\overline{F}$’s and $F$’s labs just before $\overline{W}$ and $W$ make
their measurements is given by

$\displaystyle|\Phi{\rangle}=[|\overline{D}(tails){\rangle}\otimes|tails{\rangle}\otimes|D(up){{\rangle}}\otimes|up{\rangle}\quad\quad\quad$
(8)
$\displaystyle+|\overline{D}(heads){\rangle}\otimes|heads{\rangle}\otimes|D(down){{\rangle}}\otimes|down{\rangle}\quad$
$\displaystyle+|\overline{D}(tails){\rangle}\otimes|tails{\rangle}\otimes|D(down){{\rangle}}\otimes|down{\rangle}]/\sqrt{3},$

and can be used to evaluate four different probabilities. The likelihood of
$\overline{F}$ and $F$ seeng the outcomes $heads$ and $up$ is zero,

$\displaystyle
P(heads,up)={\langle}\Phi|heads{\rangle}{\langle}heads|\otimes|up{\rangle}{\langle}up|\Phi{\rangle}=0.$
(9)

A simple calculation shows that also vanish the probabilities of
$\overline{F}$ and $W$ seeing $tails$ and $Ok$, and of $F$ and $\overline{W}$
seeing $down$ and $\overline{Ok}$,

$\displaystyle
P(tails,Ok)={\langle}\Phi|tails{\rangle}{\langle}tails|\otimes|Ok{\rangle}{\langle}Ok|\Phi{\rangle}=0,$
(10) $\displaystyle
P(down,\overline{Ok})={\langle}\Phi|down{\rangle}{\langle}down|\otimes|\overline{Ok}{\rangle}{\langle}\overline{Ok}|\Phi{\rangle}=0.$
(11)

Equations (9)-(11) suggest that $\overline{W}$ and $W$ will never see
$\overline{Ok}$ and $Ok$, at the same time. Yet, a similar calculation shows
that the outcomes $\overline{Ok}$ and $Ok$ will occur together in about $1/12$
of all trials,

$\displaystyle
P(\overline{Ok},Ok)={\langle}\Phi|{Ok}{\rangle}{\langle}{Ok}|\otimes|\overline{Ok}{\rangle}{\langle}\overline{Ok}|\Phi{\rangle}=1/12.$
(12)

All four probabilities in Eqs.(9)-(12) are legitimate results, and the
apparent contradiction needs to be resolved in one way or another.
Frauchiger and Renner [8] were quick to tell their readers that this is where
quantum mechanics looses the plot, by making too many conflicting predictions
where only one is required.
Alternatively, one may assume that all “facts”, as expressed by Eqs.(9)-(12)
are indeed valid at the same time, but not to everyone. In other words, they
refer to private perceptions of the participants, held in “sealed”
laboratories. With the labs isolated from each other, there is no danger of
comparing the conflicting outcomes, and all perceived results are equally
valid or, if one prefers, equally invalid. The view that quantum theory may
only describe such “observer-dependent” facts was proposed by Brukner [6] and
found further support, e.g., in [7].
There is, however, no need for a radical departure from the standard textbook
rules [11]. The “contradiction”, discussed in the first paragraph of this
Section, is a spurious one. The probabilities in Eqs.(3) and (6) refer to two
mutually exclusive scenarios, in which $W$ either erases all records produced
by $F$, or preserves them. Like the proverbial cake, a record cannot be both
present and destroyed, and the results (3) and (6) should never be played
against each other (we would like to avoid using an over-used term “contextual
paradox”). The wave function (1) just before $W$’s measurement contains no
information about course of action $W$ is about to take, and contains the
answers for each of the $W$’s arrangements. It remains one’s own
responsibility to decide which one to use.

[angle=0,width=6.5 cm, height= 4cm]FIG2.pdf

Figure 2: Twelve virtual paths of a joint system $\\{coin+spin\\}$ in extended
Wigner’s friend scenario.

Precisely the same happens when the number of participants is increased to
four. With the choice of the initial state, and the coupling between the coin
and the spin, the system maps onto a three-slit setup, where each of the four
“points on the screen” can be reached via three virtual paths (see Fig.2 and
Ref. [18] for more details). Now $\overline{W}$’s and $W$’s may either erase
or preserve $\overline{F}$’s and $F$ records, respectively. There are four
exclusive scenarios, and one easily finds [18] that
i) Eq. (9) holds true if both records are preserved,
ii) Eq. (10) is valid, provided only $\overline{F}$’s record survives,
ii) Eq. (11) is valid, provided only $F$’s record survives.
Finally, Eq.(12) applies when both records are erased, and all information
about the past is irretrievably lost. Again, there is no contradiction if one
follows Feynman’s rules of Ref.[11]. Next we consider a different case where
Feynman’s analysis of the double-slit conundrum also plays a crucial role.

##  5 The search for the “tunnelling time”

The amount of time it takes a classical particle to cross a given region of
space is a useful quantity. In quantum tunnelling, a particle enters and
leaves the barrier region, so it is only natural to assume that it spends
there some duration $\tau$. In a recent Nature publication [10] the authors
put a number on the duration spent by a tunnelling atom in the barrier, and
considered the issue “resolved”.
Well, not quite, as we will show next. There are at least two approaches
measuring the duration spent by a quantum particle in a given region of space
$\Omega$. One, originally proposed by Baz’ [19], and recently used in [10],
consists of equipping the particle with a magnetic moment (spin) which rotates
in a small magnetic field, confined to the $\Omega$ . For a classical
particle, a ratio of the rotation angle $\varphi$ to the Larmor frequency
$\omega_{L}$ yields the desired result $\tau_{cl}=\varphi/\omega_{L}$.
For a quantum particle, making transition between initial and final states
$|\psi{\rangle}$ and $|\phi{\rangle}$ over a time $T$, one finds many possible
durations, and is able to define the corresponding probability amplitudes
$A(\phi\leftarrow\tau\leftarrow\psi)$ [22] by summing the amplitudes
$\exp\\{iS[x(t)]\\}$ [20] over the Feynman paths, spending in $\Omega$
precisely $\tau$ seconds. With this, a non-relativistic transition amplitude
can be seen to result from interference between all allowed durations

$\displaystyle{\langle}\phi|\hat{U}(T)|\psi{\rangle}=\int_{0}^{T}A(\phi\leftarrow\tau\leftarrow\psi)d\tau.$
(13)

With many virtual durations in place, the final state of the spin, travelling
with the particle, will be a superposition
$\int_{0}^{T}|\varphi=\omega_{L}\tau{\rangle}A(\tau\leftarrow\psi)d\tau$ where
$|\varphi{\rangle}$ stands for the initial spin’s state, rotated by an angle
$\varphi$ around the direction of the field. Notably, a sum of such rotations
is not a new rotation (there is no single angle to divide by $\omega_{L}$),
and the Larmor clock method meets with its main difficulty (see Fig.3b).

[angle=0,width=7.0 cm, height= 5.5cm]FIG3.pdf

Figure 3: a) A non-relativistic quantum particle can reach its final state
$|\phi{\rangle}$ by spending any virtual duration $0\leq\tau\leq T$ in a
specified region of space $\Omega$. b) In each virtual scenario, a spin-$1/2$
initially polarised along the $x$-axis rotates in the $xy$-plane by
$\varphi=\tau\omega_{L}$. However, a superposition of such rotations is not a
new rotation in the $xy$-plane, and at $t=T$ the spin acquires also a non-zero
$z$-component. For a weakly perturbing Larmor clock one finds
${\langle}\sigma_{y}(T){\rangle}\approx\text{Re}[\tau_{w}]$ and
${\langle}\sigma_{z}(T){\rangle}\approx\text{Im}[\tau_{w}]$, with $\tau_{w}$
defined in Eq.(22).

Equation (13) describes a multi-slit problem where a particle can reach the
“point on the screen”, $|\phi{\rangle}$, by passing through a continuum of
“slits”, labelled by $\tau$, as shown schematically in Fig.3a. As in the
original double-slit case, one has a choice between keeping the transition
intact, and not knowing the duration $\tau$, or measuring $\tau$ at the cost
of destroying the interference, and with it the studied transition [21]-[23].
The only exception is the classical case, where
$A(\phi\leftarrow\tau\leftarrow\psi)$ rapidly oscillates everywhere but in the
vicinity of the classical value $\tau=\tau_{cl}$, and $\omega_{L}\tau_{cl}$
defines a unique angle by which the spin rotates. In a classically forbidden
tunnelling transition no unique duration can be selected. To make the matter
worse, the $A(\phi\leftarrow\tau\leftarrow\psi)$ is not itself small, and the
very small tunnelling amplitude in the l.h.s. of Eq.(13) is a result of a very
precise cancellation. Any attempt to perturb this delicate balance is,
therefore, likely to destroy tunnelling.
The second method to measure the delay, experienced by a particle in a
potential, is to compare the particles’s final position with that of its free
moving counterpart [24] (see Fig.4).

[angle=0,width=6.0 cm, height= 2.5cm]FIG4.pdf

Figure 4: The distance between the centre of mass of a tunnelling wave packet
and that of its freely propagating counterpart. An attempt to deduce from it
the duration spent in the barrier meets with the same difficulty as the
application of a Larmor clock shown in Fig.3b.

For a classically forbidden transition, it meets with the same difficulty. The
state of a particle with a momentum $p$ and energy $E$, transmitted across a
finite range barrier, or well, is given by $T(p)\exp(ipx)$, where $T(p)$ is
the barrier’s transmission amplitude. Using a Fourier transform
$\xi(x^{\prime})=(2\pi)^{-1}\int_{-\infty}^{\infty}T(p^{\prime})\exp(ip^{\prime}x^{\prime})dp^{\prime}$,
one can write the transmitted state as a superposition of plane waves, each
displaced in space by a distance $x^{\prime}$ [25], [26]

$\displaystyle T(p)\exp(ipx)=\int
dx^{\prime}\xi(x^{\prime})\exp[ip(x-x^{\prime})].$ (14)

A similar expression exists for a transmitted wave packet with a mean momentum
$p_{0}$,

$\displaystyle\psi^{T}(x,t)=\exp[ip_{0}x-iE(p_{0})t]\times$ (15)
$\displaystyle\int G_{0}(x-x^{\prime},t)\eta(x^{\prime},p_{0})dx^{\prime},$

where $\eta(x^{\prime},p_{0})\equiv\exp(-ip_{0}x^{\prime})\xi(x^{\prime})$,
and $G_{0}(x,t)$ is the envelope of the same initial wave packet, as it would
be in the absence of the potential,

$\displaystyle\psi_{0}(x,t)\equiv\exp[ip_{0}x-iE(p_{0})t]G_{0}(x,t).$ (16)

In general, a sum of spatial shifts is not a new shift, and no single shape is
selected from the collection of the envelopes $G(x-x^{\prime},t)$ in Eq.(15).
Again, the only exception is the classical limit, where an oscillatory
$\eta(x^{\prime},p_{0})$ develops a stationary region around a classical value
$x^{\prime}=x^{\prime}_{cl}(p_{0})$, and one recovers the classical result
$\psi^{T}(x,t)\approx\psi_{0}(x-x^{\prime}_{cl},t)$. For a barrier not
supporting bound states, $\xi(x^{\prime})\equiv 0$ for $x^{\prime}>0$ [25], so
that none of the envelopes in (15) are advanced, relative to the freely
propagating $G_{0}(x,t)$. In a tunnelling regime, the spatial delay
experienced by the particle is lost to interference [26], just like the
duration $\tau$ of the previous example. The reader may reasonably ask if,
according to the Uncertainty Principle, a tunnelling time cannot exist, how
could McColl conclude [24] that tunnelling is a delay-free process? And what
was measured in the experiment reported in [10]?

##  6 Complex times and weak ”measurements”

To answer these questions, we consider again the double-slit case (1)-(4), but
this time with the probe $\overline{D}$ replaced by a von Neumann pointer [27]
with position $x$, prepared in a state $|G_{0}{\rangle}$,
${\langle}x|G_{0}{\rangle}\sim\exp(-x^{2}/\Delta x^{2})$. The pointer is set
up to measure an operator
$\hat{B}=|up{\rangle}B_{1}{\langle}up|+|down{\rangle}B_{2}{\langle}down|$ with
eigenvalues $B_{1}$ and $B_{2}$. Now the state of the pointer, provided the
spin is found in a state $|ok{\rangle}$, is given by

$\displaystyle G(x,t_{2})=G_{0}(x-B_{1})A_{1}+G_{0}(x-B_{2})A_{2},$ (17)

where we use a shorthand $A_{1}$ and $A_{2}$ for $A^{S}(ok\leftarrow
up\leftarrow s_{0})$ and $A^{S}(ok\leftarrow down\leftarrow s_{0})$,
respectively. The probability for the observed system to arrive in
$|ok{\rangle}$ is

$\displaystyle P(ok)=\int|G(x,t_{2})|^{2}dx=|A_{1}|^{2}+|A_{2}|^{2}$ (18)
$\displaystyle+2\text{Re}[A_{2}^{*}A_{1})]\times\int G(x-B_{1})G(x-B_{2})dx.$

The pointer carries a record of the spin’s condition at $t_{1}$, and in the
accurate limit $\Delta x\to 0$ it is always possible to find out whether it
was $up$ or $down$. In the opposite limit, $\Delta x\to\infty$, a pointer’s
reading $x$ cannot be used to distinguish between the paths, and
$P(ok)=|A_{1}+A_{2}|^{2}$. The possible pointer readings are distributed
between $-\infty$ and $\infty$, and the value of $\hat{B}$, measured in this
way, is well and truly indeterminate.
One can also use Eq.(18) to calculate the average pointer reading
${\langle}x{\rangle}$, conditional on the system arriving at $|ok{\rangle}$.
For $\Delta x\to\infty$ one obtains [28]

$\displaystyle{\langle}x{\rangle}\xrightarrow[\Delta
x\to\infty]{}\text{Re}\left[\frac{B_{1}A_{1}+B_{2}A_{2}}{A_{1}+A_{2}}\right].$
(19)

At first glance, Eq.(19) appears to contradict the Uncertainty Principle,
because a definite value, ${\langle}x{\rangle}$, has been obtained in a
situation where everything was meant to be indeterminate. This is, however,
not so, since the initial and final states $|s_{0}{\rangle}$ and
$|ok{\rangle}$ can always be chosen so as to give the r.h.s. of Eq.(19) any
desired value - large, small, positive, negative, or zero [28]. The
Uncertainty Principle still applies, albeit at a different level. Clearly, one
thing that can be learnt from Eq.(19) is that adding the system’s amplitudes,
multiplied by $B_{1}$ and $B_{2}$, dividing the result by their sum, and
taking the real part of the fraction, gives a particular number. It is far
less clear, since Feynman’s rules [11] give no clue in this regard, whether
Eq.(19) can have any other significance.
The non-perturbing “weak” limit $\Delta x\to\infty$ was first studied in [29],
where the quantity in brackets in Eq.(19) was called the “weak value of an
operator $\hat{B}$”, equally written as

$\displaystyle
B_{w}\equiv\frac{{\langle}ok|\hat{B}|s_{0}{\rangle}}{{\langle}ok|s_{0}{\rangle}}.$
(20)

The authors of [29] made two claims which, while no doubt helping subsequent
popularity of the subject [30], have led to a fair amount of confusion,
including that surrounding the tunnelling time problem discussed here.
Firstly, it was claimed that Eq.(20) “defined a new kind of quantum variable”,
whereas, as we have seen, they were describing a particular combination of the
familiar probability amplitudes. Secondly, they found it surprising that a
weak value of a spin-$1/2$ component could take a value of $100$. According to
the Uncertainty Principle, it would be surprising if it could not.
It is easy to see what all this means for the quest to find and measure “the
tunnelling time”. Neglecting the spreading of the wave packet,
$G_{0}(x,t)\approx G(x-v_{0}t)$, and comparing Eq.(15) with Eq.(17), we note
that we are dealing with an inaccurate measurement of the quantity which we
earlier described as the spatial delay, or shift, $x^{\prime}$, with which the
transmitted particle with a momentum $p_{0}$ leaves the barrier. The
particle’s own position $x$ plays the role of the pointer, and by sending
$\Delta x\to\infty$ we can make the measurement weak. In this limit, the
distance between the centre of mass of the transmitted wave packet and its
freely propagating counterpart is given by an analogue of Eq.(19) [26]

$\displaystyle{\langle}x{\rangle}^{T}-{\langle}x{\rangle}_{free}\xrightarrow[\Delta
x\to\infty]{}\text{Re}[x^{\prime}_{w}]\equiv\quad\quad\quad$ (21)
$\displaystyle\text{Re}\left[\frac{\int
x^{\prime}\eta(x^{\prime},p_{0})dx^{\prime}}{\int\eta(x^{\prime},p_{0})dx^{\prime}}\right]=-\partial_{p}\Phi(p_{0}),$

where $\Phi(p)$ is the phase of the transmission amplitude,
$T(p)=|T(p)|\exp[i\Phi(p)]$. This can be verified experimentally [31], but
what can be learnt from such an experiment? The answer is the same as before:
a barrier can be characterised by distribution of virtual shifts it imposes
upon the transmitted particle. The measured distance (21) is a weighted sum of
the corresponding amplitudes.
An attempt to give this quantity a deeper meaning immediately meets with
difficulties. For a broad rectangular barrier of a height $V$, width $d$, and
a non-relativist particle of a mass $\mu$, one finds
$T(p,V)\sim\exp(-ipd)\exp(-d\sqrt{2\mu V-p^{2}})$. According to Eq.(21), the
(small) tunnelled pulse is advanced by roughly the barrier’s width $d$, no
matter how broad the barrier is. This Hartmann effect [32] has more dramatic
consequences if one tries to deduce from Eq.(21) the “time $\mathcal{T}$ the
particle has spent in the barrier.” The result $\mathcal{T}\approx 0$ seems to
point towards a conflict between quantum mechanics and special relativity [9].
The problem, however, has no relativistic implications. In Eq. (15) the
envelopes $G(x-x^{\prime},t)$, from whose front tails the transmitted wave
packet is built, are all delayed even relative to free propagation, and
certainly so relative to the motion at the speed of light.
Time measurements by means of a weakly perturbing Larmor clock [10],[19]
suffer from the same deficiency [21] \- [23] . They inevitably involve a
“complex time”, ${\tau_{w}}$ [23],

$\displaystyle\tau_{w}\equiv\frac{\int_{0}^{T}\tau
A(\phi\leftarrow\tau\leftarrow\psi)d\tau}{\int_{0}^{T}A(\phi\leftarrow\tau\leftarrow\psi)d\tau}=\text{Re}[\tau_{w}]+i\text{Im}[\tau_{w}],$
(22)

another “mean value”, calculated with an alternating complex value
distribution, which cannot be interpreted as a meaningful duration spent in a
given region of space. Indeed, by choosing $|\phi{\rangle}$ so that
${\langle}\phi|\hat{U}(T)|\psi{\rangle}\to 0$, one can always ensure that
$\text{Re}[{\tau_{w}}]$, $\text{Im}[\tau_{w}]$ , and $|\tau_{w}|$ exceed the
duration of motion, $T$, which rather proves the point. A measured value of
$\text{Re}[\tau_{w}]$ represents a relation between the amplitudes
$A(\phi\leftarrow\tau\leftarrow\psi)$ in Eq.(13), while the Uncertainty
Principle ensures that the duration spent in the region where the spin’s
rotates, remains indeterminate, just like the slit chosen in the double-slit
experiment. We are back to Feynman’s “only mystery quantum mechanics”.

##  7 Conclusions and discussion

In summary, one finds Feynman Uncertainty Principle at the centre of many
recent developments in elementary quantum mechanics. The principle, we recall,
implies that in a an interference phenomenon the system’s past remains
indeterminate, and an inquiry inevitably destroys the phenomenon. In [13]
Feynman’s wrote “They can give you a wider class of experiments than just the
two slit interference experiment. But that is just repeating the same thing to
drive it in.” Well, they did, and often without mentioning the “thing” in
question. Today “Feynman’s blind alley” [13] comes adorned with many
extraordinary, yet not particularly useful claims. Most of them arise either
from mixing incompatible scenarios (we promised not to use the words
“contextual paradox”), or from giving probability amplitudes a meaning they
were not suppose to have in the first place [12].
Thus, accepting the general principles of quantum mechanics, as given in [11],
one notes that the seeming conflict between the statements (9)-(12) does not
prove that quantum theory is “inconsistent” but only that an interference
pattern cannot co-exist with the knowledge of the slit used by an electron.
Neither does it imply that each Observer is entitled to his or her own facts.
The probabilities (9)-(12) refer to mutually exclusive situations, when some
of the virtual scenarios can be distinguished, and some others cannot.
The so-called “weak measurements”, used in [2] -[4],[29], quantify particular
relations between Feynman’s amplitudes [28], but provide no insight into the
amplitudes’ meaning or physical significance beyond what was said in [11]. A
quantum particle is not in two places at the same time - it is either in one
place, or it is impossible to say where it is [12]. Similarly, using the “weak
values” as the evidence of the particle’s presence does not prove that photons
can be found in a place they never entered [5], but only that two non-zero
amplitudes may sum up to zero [33].
For the same reason, tunnelling which results from interference between
different delays, whichever way one looks at it, cannot be described by a
single meaningful duration, and poses no threat to special relativity. Finding
such a delay would invalidate the UP, and uproot quantum mechanics “protected”
by the principle [11]. This is why the claim to have measured the traversal
time and “resolved the controversy regarding how long a tunnelling particle
spends in the barrier region” made in [10] by Ramos et al is grossly
misleading, even if the experiment itself is technically perfect.
In summary, it may not be a bad idea to check whether a proposed experiment,
no matter how spectacular, could be just another illustration of Feynman’s
“only mystery” [11]. If it is quantum mechanics could only be accused of being
quantum mechanics as we know it. If it is not, something new may be learnt as
a result. Yet by ignoring the Uncertainty Principle altogether one risks
arriving at a conclusion, which is either wrong, or after scrutiny, will prove
to be the “repetition of the same thing” [13], in other words, a platitude.

###### Acknowledgements.

Financial support of MCIU, through the grant
PGC2018-101355-B-100(MCIU/AEI/FEDER,UE) and the Basque Government Grant No
IT986-16, is acknowledged by DS.

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