# Analogue of the Cauchy–Schwarz Inequality for Determinants:
A Simple Proof

Avram Sidi
Computer Science Department
Technion - Israel Institute of Technology
Haifa 32000, Israel
e-mail<EMAIL_ADDRESS>
https://asidi.cswp.cs.technion.ac.il

###### Abstract

In this note, we present a simple proof of an analogue of the Cauchy–Schwarz
inequality relevant to products of determinants. Specifically, we show that

$|\det(A^{*}MB)|^{2}\leq\det(A^{*}MA)\cdot\det(B^{*}MB),\quad
A,B\in\mbox{$\mathbb{C}$}^{m\times n},$

where $M\in\mbox{$\mathbb{C}$}^{m\times m}$ is hermitian positive definite.
Here $m$ and $n$ are arbitrary. In case $m\leq n$, equality holds trivially.
Equality holds when $m>n$ and $\text{rank}(A)=\text{rank}(B)=n$ if and only if
the columns of $A$ and the columns of $B$ span the same subspace of
$\mbox{$\mathbb{C}$}^{m}$.

Mathematics Subject Classification 2020: 15A15

Keywords and expressions: Cauchy–Schwarz inequality, determinants.

The Cauchy–Schwarz inequality for vectors in $\mbox{$\mathbb{C}$}^{n}$ states
that, if $x,y\in\mbox{$\mathbb{C}$}^{n}$, and $(a,b)$ is the inner product in
$\mbox{$\mathbb{C}$}^{n}$, then

$|(x,y)|^{2}\leq(x,x)\cdot(y,y),$

with equality if and only if $x$ and $y$ are linearly dependent. There are
many analogues of this theorem in different settings, and these can be found
in many books and papers on linear algebra and related subjects. In this work,
we prove an analogue of this inequality that concerns determinants. Our result
here also follows from Marcus and Moore [2]. (See the remarks at the end of
this note.)

We begin with the following lemma:

######  Lemma 1.

Let $U$ and $V$ be two rectangular unitary matrices in
$\mbox{$\mathbb{C}$}^{m\times n}$ with $m>n$, in the sense

$U^{*}U=I_{n},\quad V^{*}V=I_{n}.$

Then

$|\det(U^{*}V)|\leq 1.$ (1)

Equality holds if and only if the columns of $U$ and the columns of $V$ span
the same subspace of $\mbox{$\mathbb{C}$}^{m}$.

###### Proof.

The matrices $U$ and $V$ have the following columnwise partitionings:

$U=[u_{1}|u_{2}|\cdots|u_{n}],\quad u_{i}^{*}u_{j}=\delta_{i,j};\quad
V=[v_{1}|v_{2}|\cdots|v_{n}],\quad v_{i}^{*}v_{j}=\delta_{i,j}.$

Then $W=U^{*}V\in\mbox{$\mathbb{C}$}^{n\times n}$ and the $(i,j)$ element of
$W$ is $u_{i}^{*}v_{j}$. Therefore, $U^{*}V$ has the columnwise partitioning

$U^{*}V=W=[w_{1}|w_{2}|\cdots|w_{n}];\quad
w_{j}=[u_{1}^{*}v_{j},u_{2}^{*}v_{j},\ldots,u_{n}^{*}v_{j}]^{T},\quad
j=1,2,\dots,n,$ (2)

and

$\|w_{j}\|^{2}=\sum_{i=1}^{n}|u_{i}^{*}v_{j}|^{2},\quad j=1,2,\ldots,n.$ (3)

Thus, by Hadamard’s inequality,222Hadamard’s inequality: If $H$ is an $s\times
s$ matrix with columnwise partitioning $H=[h_{1}|h_{2}|\cdots|h_{s}]$, then
$|\det
H|\leq\prod^{s}_{j=1}\|h_{j}\|,\quad\|h_{j}\|=\sqrt{h_{j}^{*}h_{j}},\quad
j=1,\ldots,s.$ A proof of Hadamard’s inequality, can be found in Horn and
Johnson [1, p. 477]. there holds

$|\det(U^{*}V)|=|\det W|\leq\prod^{n}_{j=1}\|w_{j}\|.$ (4)

Let us now add to the set of the $n$ (orthonormal) vectors
$u_{1},u_{2},\ldots,u_{n}$ the $m-n$ (orthonormal) vectors
$u_{n+1},u_{n+2},\ldots,u_{m}$, such that $u_{1},\ldots,u_{m}$ is an
orthonormal basis for $\mbox{$\mathbb{C}$}^{m}$, hence for every vector
$x\in\mbox{$\mathbb{C}$}^{m}$, we have

$x=\sum^{m}_{i=1}(u_{i}^{*}x)u_{i}\quad\text{and}\quad\|x\|^{2}=\sum^{m}_{i=1}|u_{i}^{*}x|^{2}.$

Next, let us define

$S_{U}=\text{span}\\{u_{1},u_{2},\ldots,u_{n}\\}\quad\text{and}\quad
S_{U}^{\perp}=\text{span}\\{u_{n+1},u_{n+2},\ldots,u_{m}\\}.$

Here $S_{U}$ is the column space of $U$ and $S^{\perp}_{U}$ is the orthogonal
complement of $S_{U}$, and every vector $x\in\mbox{$\mathbb{C}$}^{m}$ is of
the form

$x=\hat{x}+\tilde{x};\quad\hat{x}=\sum^{n}_{i=1}(u_{i}^{*}x)u_{i}\in
S_{U},\quad\tilde{x}=\sum^{m}_{i=n+1}(u_{i}^{*}x)u_{i}\in
S^{\perp}_{U};\quad\hat{x}^{*}\tilde{x}=0,$

therefore,

$\|x\|^{2}=\|\hat{x}\|^{2}+\|\tilde{x}\|^{2};\quad\|\hat{x}\|^{2}=\sum^{n}_{i=1}|u_{i}^{*}x|^{2},\quad\|\tilde{x}\|^{2}=\sum^{m}_{i=n+1}|u_{i}^{*}x|^{2}.$

Then, for $v_{j}$, the $j^{\text{th}}$ column of the matrix $V$, we have
$v_{j}=\hat{v}_{j}+\tilde{v}_{j}$, where

$\|\hat{v}_{j}\|^{2}=\sum^{n}_{i=1}|u_{i}^{*}v_{j}|^{2}=\|w_{j}\|^{2}\quad\text{by
\eqref{eqbn}},\quad\|\tilde{v}_{j}\|^{2}=\sum^{m}_{i=n+1}|u_{i}^{*}v_{j}|^{2}.$

Therefore,

$1=\|v_{j}\|^{2}=\|w_{j}\|^{2}+\|\tilde{v}_{j}\|^{2}\quad\Rightarrow\quad\|w_{j}\|\leq\|v_{j}\|=1;\quad
j=1,2,\ldots,n.$ (5)

This forces $\prod^{n}_{j=1}\|w_{j}\|\leq 1$ in (4), thus proving (1).

Let us now denote the column space of $V$ by $S_{V}$. That is,

$S_{V}=\text{span}\\{v_{1},v_{2},\ldots,v_{n}\\}.$

and let us assume that $S_{V}\neq S_{U}$. Then, at least one of the vectors
$v_{1},v_{2},\ldots,v_{n}$, say $v_{p}$, does not belong to $S_{U}$, and this
implies that

$v_{p}=\hat{v}_{p}+\tilde{v}_{p},\quad\tilde{v}_{p}\neq 0.$

As a result,

$1=\|v_{p}\|^{2}=\|w_{p}\|^{2}+\|\tilde{v}_{p}\|^{2}>\|w_{p}\|^{2}\quad\Rightarrow\quad\|w_{p}\|<\|v_{p}\|=1.$
(6)

As a result of (5) and (6), we have $\prod^{n}_{j=1}\|w_{j}\|<1$ in (4), which
forces strict inequality in (1).

We now show that equality holds in (1) when $S_{U}=S_{V}$. In this case, each
$v_{i}$ is a linear combination of the $u_{j}$. That is,

$V=U\Sigma,\quad\text{for some $\Sigma\in\mbox{$\mathbb{C}$}^{n\times n}$}.$

Consequently,

$I_{n}=V^{*}V=(U\Sigma)^{*}(U\Sigma)=\Sigma^{*}(U^{*}U)\Sigma=\Sigma^{*}\Sigma\quad\Rightarrow\quad\Sigma^{*}\Sigma=I_{n},$

that is, $\Sigma$ is a unitary matrix in $\mbox{$\mathbb{C}$}^{n\times n}$.
Consequently,

$U^{*}V=U^{*}(U\Sigma)=(U^{*}U)\Sigma=\Sigma\quad\Rightarrow\quad\det(U^{*}V)=\det\Sigma,$

from which,

$|\det(U^{*}V)|=|\det\Sigma|=1.$

This completes the proof. ∎

######  Theorem 2.

Let $A,B\in\mbox{$\mathbb{C}$}^{m\times n}$, with $m,n$ arbitrary. Then

$|\det(A^{*}B)|^{2}\leq\det(A^{*}A)\cdot\det(B^{*}B).$ (7)

  1. 1.

If $m<n$, equality holds in (7), both sides vanishing there.

  2. 2.

If $m=n$, equality holds in (7).

  3. 3.

(a) If $m>n$ and $\text{\em rank}(A)<n$ or $\text{\em rank}(B)<n$, equality
holds in (7), both sides vanishing there.

(b) If $m>n$ and $\text{\em rank}(A)=\text{\em rank}(B)=n$, equality holds in
(7) if and only if the columns of $A$ and the columns of $B$ span the same
subspace of $\mbox{$\mathbb{C}$}^{m}$.

###### Proof.

We start by noting that all three matrices $A^{*}A$, $B^{*}B$, and $A^{*}B$
are $n\times n$.

  1. 1.

If $m<n$, each one of the matrices $A^{*}A$, $B^{*}B$, and $A^{*}B$ is of rank
at most $m$, hence is singular. Therefore, (7) holds, both sides vanishing
there.

  2. 2.

If $m=n$, then $A$ and $B$ are square. Therefore,

$\det(A^{*}A)=(\det A^{*})(\det A),\quad\det(B^{*}B)=(\det B^{*})(\det B)$

and

$\det(A^{*}B)=(\det A^{*})(\det B).$

The result in (7) now follows with equality there by invoking $\det
C^{*}=\overline{\det C}$ for every square matrix $C$.

  3. 3.

(a) If $m>n$, and either $\text{rank}(A)<n$ (hence $A^{*}A$ is singular) or
$\text{rank}(B)<n$ (hence $B^{*}B$ is singular), we have that $A^{*}B$ is
singular as well. Therefore, (7) holds with equality, both sides vanishing
there.

(b) If $m>n$ and $\text{rank}(A)=\text{rank}(B)=n$, we proceed as follows:
Consider the QR factorizations of $A$ and $B$, namely,

$A=Q_{A}R_{A},\quad B=Q_{B}R_{B};\quad
Q_{A},Q_{B}\in\mbox{$\mathbb{C}$}^{m\times n},\quad
R_{A},R_{B}\in\mbox{$\mathbb{C}$}^{n\times n},$

where $Q_{A}$ and $Q_{B}$ are unitary in the sense that

$Q_{A}^{*}Q_{A}=I_{n},\quad Q_{B}^{*}Q_{B}=I_{n},$

and $R_{A}$ and $R_{B}$ are upper triangular square matrices with nonzero
diagonal elements. Now, it is easy to verify that

$A^{*}A=R_{A}^{*}R_{A},\quad B^{*}B=R_{B}^{*}R_{B},\quad
A^{*}B=R_{A}^{*}(Q_{A}^{*}Q_{B})R_{B}.$

Note that $Q_{A}^{*}Q_{B}\in\mbox{$\mathbb{C}$}^{n\times n}$, just as
$R_{A},R_{B}$. Therefore, since $R_{A}$ and $R_{B}$ are upper triangular
matrices with nonzero diagonal elements,

$\det(A^{*}A)=(\det R_{A}^{*})(\det R_{A})=|\det R_{A}|^{2},$

and

$\det(B^{*}B)=(\det R_{B}^{*})(\det R_{B})=|\det R_{B}|^{2}.$

Next,

$\det(A^{*}B)=(\det(Q_{A}^{*}Q_{B}))(\det R_{A}^{*})(\det R_{B}).$

Therefore,

$|\det(A^{*}B)|=|\det(Q_{A}^{*}Q_{B})|\,|\det R_{A}|\,|\det R_{B}|,$

which implies that

$|\det(A^{*}B)|^{2}=|\det(Q_{A}^{*}Q_{B})|^{2}\det(A^{*}A)\det(B^{*}B).$

The rest of the proof can now be achieved (i) by realizing that the subspaces
spanned by $A$ and $Q_{A}$ are the same, and so are the subspaces spanned by
$B$ and $Q_{B}$, and (ii) by applying Lemma 1 to $|\det(Q_{A}^{*}Q_{B})|$
since $m>n$. We leave the details to the reader.

This completes the proof. ∎

By applying Theorem 2 to the matrices $\tilde{A}=M^{1/2}A$ and
$\tilde{B}=M^{1/2}B$, where $M\in\mbox{$\mathbb{C}$}^{m\times m}$ is hermitian
positive definite, we obtain the following general form of Theorem 2.

######  Theorem 3.

Let $A,B\in\mbox{$\mathbb{C}$}^{m\times n}$, with $m,n$ arbitrary, and let
$M\in\mbox{$\mathbb{C}$}^{m\times m}$ be hermitian positive definite. Then

$|\det(A^{*}MB)|^{2}\leq\det(A^{*}MA)\cdot\det(B^{*}MB).$ (8)

  1. 1.

If $m<n$, equality holds in (8), both sides vanishing there.

  2. 2.

If $m=n$, equality holds in (8).

  3. 3.

(a) If $m>n$ and $\text{\em rank}(A)<n$ or $\text{\em rank}(B)<n$, equality
holds in (8), both sides vanishing there.

(b) If $m>n$ and $\text{\em rank}(A)=\text{\em rank}(B)=n$, equality holds in
(8) if and only if the columns of $A$ and the columns of $B$ span the same
subspace of $\mbox{$\mathbb{C}$}^{m}$.

Remarks.

  1. 1.

The proof of the Cauchy–Schwarz inequality in $\mbox{$\mathbb{C}$}^{n}$ makes
use of the fact that the inner product $(x,y)$ in $\mbox{$\mathbb{C}$}^{n}$ is
bilinear in $x$ and $y$. Because $X^{*}MY$ is bilinear in $X$ and $Y$, one
might think that the proof of the Cauchy–Schwarz inequality in
$\mbox{$\mathbb{C}$}^{n}$ can be applied to $\det(A^{*}MB)$, $\det(A^{*}MA)$,
and $\det(B^{*}MB)$ to obtain (8). This is not possible, however, since
$\det(A^{*}MB)$ does not have the bilinearity property; for example,
$\det((A_{1}+A_{2})^{*}MB)$ is not necessarily equal to
$\det(A_{1}^{*}MB)+\det(A_{2}^{*}MB).$

  2. 2.

The problem treated here is a special case of a general problem treated by
Marcus and Moore in [2, Eq. (1) on p. 111 and Theorem on p. 115]. These
authors use advanced techniques to obtain the relations which must exist
between the $m\times m$ matrices $M_{1},M_{2},M_{3},M_{4}$ so that the
relation

$\det(A^{*}M_{1}B)\det(B^{*}M_{2}A)\leq\det(A^{*}M_{3}A)\det(B^{*}M_{4}B)$

holds for all $m\times n$ matrices $A$ and $B$ when $m>n$.

Because we are restricting our problem to the special case in which
$M_{1}=M_{2}=M_{3}=M_{4}=M$, $M$ being an $m\times m$ positive definite
hermitian matrix, we are able to carry out our analysis by employing rather
elementary techniques of linear algebra that are easily accessible.

## References

  * [1] R.A. Horn and C.R. Johnson.  Matrix Analysis.  Cambridge University Press, Cambridge, 1985.
  * [2] M. Marcus and K. Moore.  A determinant formulation of the Cauchy-Schwarz inequality.  Linear Algebra Appl., 36:111–127, 1981.