#  On some properties of three different types of triangular blocked tensors
††thanks: This work was supported by the NSF of China (Grant Nos. 11231004,
11271288 and 11571123) and the NSF of Guangdong Provincial (Grant No.
2015A030313377).

Jiayu Shaoa,***Email address<EMAIL_ADDRESS>Lihua Youb,†††Corresponding
author<EMAIL_ADDRESS>
(a Department of Mathematics, Tongji University, Shanghai, 200092, P.R. China
b School of Mathematical Sciences, South China Normal University, Guangzhou,
510631, P.R. China
)

###### Abstract

We define three types of upper (and lower) triangular blocked tensors, which
are all generalizations of the triangular blocked matrices. We study some
basic properties and characterizations of these three types of triangular
blocked tensors. We obtain the formulas for the determinants, characteristic
polynomials and spectra of the first and second type triangular blocked
tensors, and give an example to show that these formulas no longer hold for
the third type triangular blocked tensors. We prove that the product of any
two $(n_{1},\cdots,n_{r})$-upper (or lower) triangular blocked tensors of the
first or second or third type is still an $(n_{1},\cdots,n_{r})$-upper (or
lower) triangular blocked tensor of the same type. We also prove that, if an
$(n_{1},\cdots,n_{r})$-upper triangular blocked tensor of the first or second
or third type has a left $k$-inverse, then its unique left $k$-inverse is
still an $(n_{1},\cdots,n_{r})$-upper triangular blocked tensor of the same
type. Also if it has a right $k$-inverse, then all of its right $k$-inverses
are still $(n_{1},\cdots,n_{r})$-upper triangular blocked tensors of the same
type. By showing that the left $k$-inverse (if any) of a weakly irreducible
nonsingular $M$-tensor is a positive tensor, we show that the left $k$-inverse
(if any) of a first or second or third type canonical
$(n_{1},\cdots,n_{r})$-upper triangular blocked nonsingular $M$-tensor is an
$(n_{1},\cdots,n_{r})$-upper triangular blocked tensor of the same type all of
whose diagonal blocks are positive tensors. We also show that every order $m$
dimension $n$ tensor is permutation similar to some third type normal upper
triangular blocked tensor (all of whose diagonal blocks are irreducible). We
give an example to show that this is not true for the first type canonical
upper triangular blocked tensor.

AMS classification: 15A42, 15A18, 15A69

Keywords: tensor, triangular blocked tensor, product, inverse, determinant,
$M$-tensor

##  1 Introduction

In recent years, the study of tensors and the spectra of tensors (and
hypergraphs) with their various applications has attracted extensive attention
and interest, since the work of L.Qi ([14]) and L.H.Lim ([11]) in 2005.

As was in [14], an order $m$ dimension $n$ tensor
$\mathbb{A}=(a_{i_{1}i_{2}\cdots i_{m}})_{1\leq i_{j}\leq n\ (j=1,\cdots,m)}$
over the complex field $\mathbb{C}$ is a multidimensional array with all
entries $a_{i_{1}i_{2}\cdots i_{m}}\in\mathbb{C}\ \
(i_{1},\cdots,i_{m}\in[n]=\\{1,\cdots,n\\})$.

In this paper, we define and study three different types of the general
$(n_{1},\cdots,n_{r})$-upper (and lower) triangular blocked tensors.

It is well known that the triangular blocked matrices are very important and
useful in the study and applications of matrices. For tensors, Hu et al [8]
gave a determinant formula for the special case $r=2$ of some type of the
$(n_{1},\cdots,n_{r})$-triangular blocked tensors. Also, Shao et al [16]
defined a type of lower triangular blocked tensors which is essentially
equivalent to the second type upper triangular blocked tensors defined in this
paper (see Theorem 2.5 of this paper for the proof). Shao et al [16] also
studied some other basic properties of such type triangular blocked tensors.

Recently, Hu, Huang and Qi in [9] introduced and studied the “nonnegative
tensor partition” which is also essentially equivalent to the second type
triangular blocked tensors defined in this paper (up to a permutation
similarity). They obtained that every order $m$ dimension $n$ tensor is
permutation similar to such a type of upper triangular blocked tensor each of
whose diagonal blocks are weakly irreducible (also see Proposition 1 of [10]).
Hu and Qi [10] also used this “nonnegative tensor partition” to study some
spectral properties of nonnegative tensors, and obtain a necessary and
sufficient condition for a nonnegative tensor to have a positive eigenvector.

In this paper, we first give the definitions of three different types of the
general $(n_{1},\cdots,n_{r})$-upper (and lower) triangular blocked tensors,
which are all the natural generalizations of the $(n_{1},\cdots,n_{r})$-upper
(and lower) triangular blocked matrices. Then we study some properties of
these three types of triangular blocked tensors.

In some sense, the first and second type upper triangular blocked tensors
defined in this paper are generalizations of the weakly reducible tensors (see
Definition 1.1 below) which corresponds to the case $r=2$ (with two diagonal
blocks) in Definitions 2.1 and 2.2 up to a permutation similarity (also see
Remarks 2.1 and 2.2 in §2), while the third type upper triangular blocked
tensors defined in this paper is a generalization of the reducible tensors
(see Definition 1.1 below) which corresponds to the case $r=2$ in Definitions
2.3 up to a permutation similarity (also see Remark 2.4 in §2).

We first study some basic properties and characterizations of these three
types of triangular blocked tensors in §2. Then in §3, we obtain the formulas
for the determinants, characteristic polynomials and spectra of the first two
types of triangular blocked tensors, and give an example to show that these
formulas no longer hold for the third type triangular blocked tensors. We
prove in §4 that the product of any two $(n_{1},\cdots,n_{r})$ upper (or
lower) triangular blocked tensors of the first or second or third type is
still an $(n_{1},\cdots,n_{r})$ upper (or lower) triangular blocked tensor of
the same type.

We also prove in §5 that, if an $(n_{1},\cdots,n_{r})$ upper (or lower)
triangular blocked tensor of the first or second or third type has a left
$k$-inverse, then its unique left $k$-inverse is still an
$(n_{1},\cdots,n_{r})$ upper (or lower) triangular blocked tensor of all the
three types. Also if it has a right $k$-inverse, then all of its right
$k$-inverses are still $(n_{1},\cdots,n_{r})$ upper (or lower) triangular
blocked tensors of all the three types. Furthermore, by showing (in Theorem
5.3) that the left $k$-inverse (if any) of a weakly irreducible nonsingular
$M$-tensor is a positive tensor, we show that the left $k$-inverse (if any) of
a first or second or third type canonical $(n_{1},\cdots,n_{r})$-upper (or
lower) triangular blocked nonsingular $M$-tensor is an
$(n_{1},\cdots,n_{r})$-upper (or lower) triangular blocked tensor of all the
three types all of whose diagonal blocks are positive tensors.

In §6, we show that every order $m$ dimension $n$ tensor is permutation
similar to some third type normal upper triangular blocked tensor (all of
whose diagonal blocks are irreducible). On the other hand, we also give an
example to show that not every tensor can be permutational similar to some
first type normal upper triangular blocked tensor (all of whose diagonal
blocks are weakly irreducible).

Now we introduce some basic concepts of tensors which are relevant to the main
results of this paper.

###### Definition 1.1

([4, 7, 17]) Let $\mathbb{A}$ be an order $m$ dimension $n$ tensor.

(1) If there exists a proper subset $I$ of the set $[n]$ such that

$a_{i_{1}i_{2}\cdots i_{m}}=0\quad(\forall\ i_{1}\in I,\ \mbox{and all of
the}\ i_{2},\cdots,i_{m}\notin I),$ (1.1)

then $\mathbb{A}$ is called reducible (or sometimes $I$-reducible). If
$\mathbb{A}$ is not reducible, then $\mathbb{A}$ is called irreducible.

(2) If there exists a proper subset $I$ of the set $[n]$ such that

$a_{i_{1}i_{2}\cdots i_{m}}=0\quad(\forall\ i_{1}\in I,\ \mbox{and at least
one of the}\ i_{2},\cdots,i_{m}\notin I),$ (1.2)

then $\mathbb{A}$ is called weakly reducible (or sometimes $I$-weakly
reducible). If $\mathbb{A}$ is not weakly reducible, then $\mathbb{A}$ is
called weakly irreducible.

The following is an equivalent definition of the determinant of an order $m$
dimension $n$ tensor with $m\geq 2$ (see [8] and [16]).

###### Definition 1.2

([8, 16]) Let $\mathbb{A}$ be an order $m$ dimension $n$ tensor with $m\geq
2$. Then $\det(\mathbb{A})$ is the unique polynomial on the entries of
$\mathbb{A}$ satisfying the following three conditions:

(1) $\det(\mathbb{A})=0$ if and only if the system of homogeneous equations
$\mathbb{A}x=0$ has a nonzero solution.

(2) $\det(\mathbb{I})=1$, where $\mathbb{I}$ is the unit tensor.

(3) $\det(\mathbb{A})$ is an irreducible polynomial on the entries of
$\mathbb{A}$, when the entries $a_{i_{1}\cdots i_{m}}\ (1\leq
i_{1},\cdots,i_{m}\leq n)$ of $\mathbb{A}$ are all viewed as independent
distinct variables.

By using the definition of determinants, we can define the characteristic
polynomial $\phi_{\mathbb{A}}(\lambda)$ of a tensor $\mathbb{A}$ to be the
determinant $\det(\lambda\mathbb{I}-\mathbb{A})$, where $\mathbb{I}$ is the
unit tensor. The spectrum of the tensor $\mathbb{A}$ is defined to be the
multi-set of the roots of the characteristic polynomial of $\mathbb{A}$.

The following definitions about the product of tensors and the permutation
similarity of tensors can be found in [15].

###### Definition 1.3

([15]) Let $\mathbb{A}$ (and $\mathbb{B}$) be an order $m\geq 2$ (and order
$k\geq 1$), dimension $n$ tensor, respectively. Define the product
$\mathbb{A}\mathbb{B}$ to be the following tensor $\mathbb{C}$ of order
$(m-1)(k-1)+1$ and dimension $n$:

$c_{i\alpha_{1}\cdots\alpha_{m-1}}=\sum_{i_{2},\cdots,i_{m}=1}^{n}a_{ii_{2}\cdots
i_{m}}b_{i_{2}\alpha_{1}}\cdots b_{i_{m}\alpha_{m-1}}\quad(i\in[n],\
\alpha_{1},\cdots,\alpha_{m-1}\in[n]^{k-1}).$ (1.3)

###### Definition 1.4

([15]) Let $\mathbb{A}$ and $\mathbb{B}$ be two order $m$ dimension $n$
tensors. If there exists a permutation matrix $P$ such that
$\mathbb{B}=P\mathbb{A}P^{T}$, then we say that $\mathbb{A}$ and $\mathbb{B}$
are permutational similar.

It was proved in [15] that similar (and thus permutational similar) tensors
have the same determinants, same characteristic polynomials and the same
spectra.

Let $I$ be a nonempty subset of the set $[n]$. Then the (principal) subtensor
$\mathbb{A}[I]$ of $\mathbb{A}$ is the subtensor consisting of those entries
$a_{i_{1}i_{2}\cdots i_{m}}$ of $\mathbb{A}$ all of whose subscripts
$i_{1},i_{2},\cdots,i_{m}$ are in $I$.

##  2 The definitions of the three types of triangular blocked tensors and
their basic properties

We first recall the definition of the triangular blocked matrices.

Let $A$ be a matrix of order $n$. Let $n_{1},\cdots,n_{r}\ (r\geq 2)$ be
positive integers with $n_{1}+\cdots+n_{r}=n$. Write $S_{0}=0$ and

$S_{j}=n_{1}+\cdots+n_{j},\qquad
I_{j}=\\{S_{j-1}+1,\cdots,S_{j}\\}\quad(j=1,\cdots,r).$

If for any $j=2,\cdots,r$, we have

$a_{ii_{2}}=0\quad(\forall i\in I_{j},\ \mbox{and}\ i_{2}\leq S_{j-1}),$

then we say that $A$ is an $(n_{1},\cdots,n_{r})$-upper triangular blocked
matrix.

$\left(\begin{array}[]{ccccc}A_{1}&\cdots&\ast&\cdots&\ast\\\
\vdots&\ddots&\vdots&\cdots&\ast\\\ O&\cdots&A_{j}&\cdots&\ast\\\
\vdots&\cdots&\vdots&\ddots&\vdots\\\ O&\cdots&O&\cdots&A_{r}\\\
\end{array}\right)$

In the following, we want to generalize this definition from matrices to
tensors.

###  2.1 The definitions of the three types of triangular blocked tensors

In this subsection, we give the definitions of the three types of triangular
blocked tensors. Here the definition of our second type upper triangular
blocked tensors can be proved to be equivalent to the Definition 4.3 of [16]
on a type of lower triangular blocked tensors (see Theorem 2.5 of this paper),
and is also essentially equivalent to the “nonnegative tensor partition”
defined and studied by Hu, Huang and Qi in [9], and by Hu and Qi in [10] (up
to a permutation similarity).

###### Definition 2.1

Let $\mathbb{A}$ be an order $m$ and dimension $n$ tensor,
$n_{1},\cdots,n_{r}\ (r\geq 2)$ be positive integers with
$n_{1}+\cdots+n_{r}=n$. Write $S_{0}=0$ and

$S_{j}=n_{1}+\cdots+n_{j},\qquad
I_{j}=\\{S_{j-1}+1,\cdots,S_{j}\\}\quad(j=1,\cdots,r).$

(1) If for any $j=2,\cdots,r$, we have

$a_{ii_{2}\cdots i_{m}}=0\quad(\forall i\in I_{j},\ \mbox{and}\
\min\\{i_{2},\cdots,i_{m}\\}\leq S_{j-1}),$ (2.1)

then we say that $\mathbb{A}$ is a first type $(n_{1},\cdots,n_{r})$-upper
triangular blocked tensor, or $(n_{1},\cdots,n_{r})$-1stUTB tensor (or simply
$(n_{1},\cdots,n_{r})$-UTB tensor) with diagonal blocks
$\mathbb{A}_{1},\cdots,\mathbb{A}_{r}$, where
$\mathbb{A}_{j}=\mathbb{A}[I_{j}]$ is the $j$-th diagonal block of this UTB
tensor.

(Here the condition $\min\\{i_{2},\cdots,i_{m}\\}\leq S_{j-1}$ means that at
least one index of $i_{2},\cdots,i_{m}$ belongs to $I_{1}\cup\cdots\cup
I_{j-1}$.)

(2) If for any $j=1,\cdots,r-1$, we have

$a_{ii_{2}\cdots i_{m}}=0\quad(\forall i\in I_{j},\ \mbox{and}\
\max\\{i_{2},\cdots,i_{m}\\}\geq S_{j}+1),$ (2.2)

then we say that $\mathbb{A}$ is a first type $(n_{1},\cdots,n_{r})$-lower
triangular blocked tensor (or simply $(n_{1},\cdots,n_{r})$-LTB tensor) with
diagonal blocks $\mathbb{A}_{1},\cdots,\mathbb{A}_{r}$.

(3) If for any $j=1,\cdots,r$, we have

$a_{ii_{2}\cdots i_{m}}=0\quad(\forall i\in I_{j},\ \mbox{and at least one of
}\ i_{2},\cdots,i_{m}\notin I_{j}),$ (2.3)

then we say that $\mathbb{A}$ is an $(n_{1},\cdots,n_{r})$-diagonal blocked
tensor with diagonal blocks $\mathbb{A}_{1},\cdots,\mathbb{A}_{r}$.

There of course might be other possible ways to define the triangular blocked
tensors. For example, the following (second type) definition is equivalent to
Definition 4.3 in [16] (see Theorem 2.5 below), and is also essentially
equivalent to the “nonnegative tensor partition” defined and studied by Hu,
Huang and Qi in [9], and by Hu and Qi in [10] (up to a permutation
similarity).

###### Definition 2.2

Let $\mathbb{A}$ be an order $m$ and dimension $n$ tensor,
$n_{1},\cdots,n_{r}\ (r\geq 2)$ be positive integers with
$n_{1}+\cdots+n_{r}=n$. Write $S_{0}=0$ and

$S_{j}=n_{1}+\cdots+n_{j},\qquad
I_{j}=\\{S_{j-1}+1,\cdots,S_{j}\\}\quad(j=1,\cdots,r).$

(1) If for any $j=2,\cdots,r$, we have

$a_{ii_{2}\cdots i_{m}}=0\quad(\forall i\in I_{j},\
\min\\{i_{2},\cdots,i_{m}\\}\leq S_{j-1}\ \mbox{and}\
\max\\{i_{2},\cdots,i_{m}\\}\leq S_{j}),$ (2.4)

then $\mathbb{A}$ is called a second type $(n_{1},\cdots,n_{r})$-upper
triangular blocked tensor (or 2ndUTB tensor).

(Here the condition $\min\\{i_{2},\cdots,i_{m}\\}\leq S_{j-1}$ means that at
least one index of $i_{2},\cdots,i_{m}$ belongs to $I_{1}\cup\cdots\cup
I_{j-1}$, while the condition $\max\\{i_{2},\cdots,i_{m}\\}\leq S_{j}$ means
that all indices of $i_{2},\cdots,i_{m}$ belong to $I_{1}\cup\cdots\cup
I_{j}$.)

(2) If for any $j=1,\cdots,r-1$, we have

$a_{ii_{2}\cdots i_{m}}=0\quad(\forall i\in I_{j},\
\max\\{i_{2},\cdots,i_{m}\\}\geq S_{j}+1\ \mbox{and}\
\min\\{i_{2},\cdots,i_{m}\\}\geq S_{j-1}+1),$ (2.5)

then $\mathbb{A}$ is called a second type $(n_{1},\cdots,n_{r})$-lower
triangular blocked tensor (or 2ndLTB tensor).

###### Remark 2.1

From Definitions 2.1 and 2.2 we can see that, in the case $r=2$, and thus
$j=2$ is the only case for $2\leq j\leq r$, we see that the condition
$\max\\{i_{2},\cdots,i_{m}\\}\leq S_{j}=S_{r}=n$ holds automatically. So in
this case $r=2$, the $(n_{1},n_{2})$-UTB tensors and $(n_{1},n_{2})$-2ndUTB
tensors are the same.

###### Remark 2.2

Also it is easy to see that, in the case $r=2$, if $\mathbb{A}$ is an
$(n_{1},n_{2})$-UTB (and 2ndUTB) tensor, then $\mathbb{A}$ is $I_{2}$-weakly
reducible. On the other hand, a tensor $\mathbb{A}$ of dimension $n$ is weakly
reducible if and only if there exists some integer $k$ with $1\leq k\leq n-1$
such that $\mathbb{A}$ is permutation similar to some $(k,n-k)$-UTB (and
2ndUTB) tensor.

###### Example 2.1

For the special case $r=2$, $n_{1}=k$ and $n_{2}=n-k\ (1\leq k\leq n-1)$, we
have that $\mathbb{A}$ is a $(k,n-k)$-UTB (and 2ndUTB) tensor if and only if
the following condition holds:

$a_{ii_{2}\cdots i_{m}}=0\quad(\forall i>k,\ \mbox{and}\
\min\\{i_{2},\cdots,i_{m}\\}\leq k).$ (2.6)

Proof. Since $r=2$, we only need to check that (2.1) holds for the case $j=2$.
Now in this case we have $i\in I_{2}\Longleftrightarrow i>k$, and
$\min\\{i_{2},\cdots,i_{m}\\}<S_{j-1}+1=k+1\Longleftrightarrow\min\\{i_{2},\cdots,i_{m}\\}\leq
k$. So in this case (2.1) is equivalent to (2.6). $\square$

Now we define the third type triangular blocked tensors.

###### Definition 2.3

Let $\mathbb{A}$ be an order $m$ and dimension $n$ tensor,
$n_{1},\cdots,n_{r}\ (r\geq 2)$ be positive integers with
$n_{1}+\cdots+n_{r}=n$. Write $S_{0}=0$ and

$S_{j}=n_{1}+\cdots+n_{j},\qquad
I_{j}=\\{S_{j-1}+1,\cdots,S_{j}\\}\quad(j=1,\cdots,r).$

(1) If for any $j=2,\cdots,r$, we have

$a_{ii_{2}\cdots i_{m}}=0\quad(\forall i\in I_{j},\ \mbox{and}\
\max\\{i_{2},\cdots,i_{m}\\}\leq S_{j-1}),$ (2.7)

then we say that $\mathbb{A}$ is a third type $(n_{1},\cdots,n_{r})$-upper
triangular blocked tensor (or 3rdUTB tensor).

(Here the condition $\max\\{i_{2},\cdots,i_{m}\\}\leq S_{j-1}$ means that all
indices of $i_{2},\cdots,i_{m}$ belong to $I_{1}\cup\cdots\cup I_{j-1}$.)

(2) If for any $j=1,\cdots,r-1$, we have

$a_{ii_{2}\cdots i_{m}}=0\quad(\forall i\in I_{j},\ \mbox{and}\
\min\\{i_{2},\cdots,i_{m}\\}\geq S_{j}+1),$ (2.8)

then we say that $\mathbb{A}$ is a third type $(n_{1},\cdots,n_{r})$-lower
triangular blocked tensor (or 3rdLTB tensor).

###### Remark 2.3

It is easy to see from Definition 2.3 that, in the case $r=2$, if $\mathbb{A}$
is an $(n_{1},n_{2})$-3rdUTB tensor, then $\mathbb{A}$ is $I_{2}$-reducible.
On the other hand, a tensor $\mathbb{A}$ of dimension $n$ is reducible if and
only if there exists some integer $k$ with $1\leq k\leq n-1$ such that
$\mathbb{A}$ is permutation similar to some $(k,n-k)$-3rdUTB tensor.

Now we give some more remarks on the definitions of the above three types of
upper triangular blocked tensors.

###### Remark 2.4

(1) It is easy to see from the definitions that an $(n_{1},\cdots,n_{r})$-UTB
tensor is always an $(n_{1},\cdots,n_{r})$\- 2ndUTB tensor, and an
$(n_{1},\cdots,n_{r})$-2ndUTB tensor is always an $(n_{1},\cdots,n_{r})$\-
3rdUTB tensor.

Also it is easy to see that, in this case $m=2$, all the three types of
$(n_{1},\cdots,n_{r})$-UTB tensors are the same.

(2) In case of $m=2$, namely if the tensor $\mathbb{A}$ is a matrix of order
$n$, then our definitions of all the three types are equivalent to the
definition of upper (or lower) triangular blocked matrices. So these
definitions are all generalizations of triangular blocked matrices.

(3) If $n_{1}=\cdots=n_{r}=1$, then an $(n_{1},\cdots,n_{r})$-UTB (or LTB)
tensor is called an upper (or lower) triangular tensor. We also have the
similar remarks for the second and third types $(n_{1},\cdots,n_{r})$-UTB (or
LTB) tensors.

(4) Up to some permutation similarity, Definition 2.2 is essentially
equivalent to the (nonnegative) tensor partition defined in [10] (Proposition
1). Also, [10] mentioned that Proposition 1 in [10] is the main result of [9]
.

(5) It is easy to see from the definitions that, an $(n_{1},\cdots,n_{r})$-UTB
(or 2ndUTB) tensor is permutation similar to an $(n_{r},\cdots,n_{1})$-LTB (or
2nd LTB) tensor via the permutation $\sigma:[n]\rightarrow[n]$ with
$\sigma(i)=n+1-i$.

(6) It is easy to see from the definitions that $\mathbb{A}$ being a first or
second or third type $(n_{1},\cdots,n_{r})$-UTB tensor is independent of the
diagonal blocks $\mathbb{A}_{1},\cdots,\mathbb{A}_{r}$.

The following proposition is an easy consequence of Definition 2.1.

###### Proposition 2.1

$\mathbb{A}$ is an $(n_{1},\cdots,n_{r})$-diagonal blocked tensor if and only
if $\mathbb{A}$ is both an $(n_{1},\cdots,n_{r})$-UTB tensor and an
$(n_{1},\cdots,n_{r})$-LTB tensor.

The following example shows that Proposition 2.1 would no longer hold for the
second and third type triangular blocked tensors.

###### Example 2.2

Let $\mathbb{A}$ be an $(n_{1},\cdots,n_{r})$-diagonal blocked tensor of
dimension $n$ and order 3. Take

$2\leq j\leq r-1,\quad i\in I_{j},\quad p\leq S_{j-1}\quad\mbox{and}\quad
q>S_{j}.$

Let $\mathbb{B}$ be the tensor of dimension $n$ and order 3 with $b_{ipq}=1$
and all the other entries 0. Then the tensor $\mathbb{A}+\mathbb{B}$ is both
an $(n_{1},\cdots,n_{r})$-2ndUTB (and thus 3rdUTB) tensor and an
$(n_{1},\cdots,n_{r})$-2ndLTB (and thus 3rdLTB) tensor, but it is not an
$(n_{1},\cdots,n_{r})$-diagonal blocked tensor.

###### Example 2.3

If $G$ is a $k$-uniform hypergraph with $r$ connected components
$G_{1},\cdots,G_{r}$. Then the adjacency tensor $\mathbb{A}(G)$ of $G$ is a
diagonal blocked tensor with diagonal blocks
$\mathbb{A}(G_{1}),\cdots,\mathbb{A}(G_{r})$ (each of which is weakly
irreducible).

###  2.2 A basic property of the third type triangular blocked tensors

###### Theorem 2.1

Let the meaning of the notations be as in Definition 2.1 $r\geq 2$, $1\leq
t\leq r-1$, write $k=S_{t}$ and $I=[k]$. Then $\mathbb{A}$ is an
$(n_{1},\cdots,n_{r})$-3rdUTB tensor with diagonal blocks
$\mathbb{A}_{1},\cdots,\mathbb{A}_{r}$ if and only if the following three
conditions are all satisfied:

(1) $\mathbb{A}$ is an $(k,n-k)$-3rdUTB tensor with diagonal blocks
$\mathbb{A}[I]$ and $\mathbb{A}[\overline{I}]$.

(2) $\mathbb{A}[I]$ is an $(n_{1},\cdots,n_{t})$-3rdUTB tensor with diagonal
blocks $\mathbb{A}_{1},\cdots,\mathbb{A}_{t}$.

(3) $\mathbb{A}[\overline{I}]$ is an $(n_{t+1},\cdots,n_{r})$-3rdUTB tensor
with diagonal blocks $\mathbb{A}_{t+1},\cdots,\mathbb{A}_{r}$.

Proof. Firstly we write $\mathbb{B}=\mathbb{A}[\overline{I}]$. Thus we have

$b_{ii_{2}\cdots
i_{m}}=a_{(i+k)(i_{2}+k)\cdots(i_{m}+k)}\quad(i,i_{2},\cdots,i_{m}\in[n-k]).$

For $j=1,\cdots,r-t$, we further write $n^{\prime}_{j}=n_{t+j}$, and
consequently write

$S^{\prime}_{j}=n^{\prime}_{1}+\cdots+n^{\prime}_{j},\quad
I^{\prime}_{j}=\\{S^{\prime}_{j-1}+1,\cdots,S^{\prime}_{j}\\}\quad(j=1,\cdots,r-t).$

Then we have $S^{\prime}_{j}=S_{t+j}-S_{t}=S_{t+j}-k$.

Necessity.

(1). By Definition 2.3 we only need to verify the following

$a_{ii_{2}\cdots i_{m}}=0\quad(\forall i>k,\ \mbox{and}\
\max\\{i_{2},\cdots,i_{m}\\}\leq k).$

Now assume that $i\in I_{j}$. Then $i>k=S_{t}\Longrightarrow j>t$, or $j-1\geq
t$. So $\max\\{i_{2},\cdots,i_{m}\\}\leq k=S_{t}\leq S_{j-1}$. By the
hypothesis that $\mathbb{A}$ is an $(n_{1},\cdots,n_{r})$-3rdUTB tensor we
have $a_{ii_{2}\cdots i_{m}}=0$. So $\mathbb{A}$ is an $(k,n-k)$-3rdUTB tensor
with diagonal blocks $\mathbb{A}[I]$ and $\mathbb{A}[\overline{I}]$.

(2). Since the condition in Eq.(2.6) holds for all $j=2,\cdots,r$ for the
tensor $\mathbb{A}$, it also holds for all $j=2,\cdots,t$ for the tensor
$\mathbb{A}[I]$. So $\mathbb{A}[I]$ is an $(n_{1},\cdots,n_{t})$-3rdUTB tensor
with diagonal blocks $\mathbb{A}_{1},\cdots,\mathbb{A}_{t}$.

(3). By Definition 2.3 we need to show that for any $j=2,\cdots,n-t$, we have

$b_{ii_{2}\cdots i_{m}}=0\quad(\forall i\in I^{\prime}_{j},\ \mbox{and}\
\max\\{i_{2},\cdots,i_{m}\\}\leq S^{\prime}_{j-1}).$ (2.9)

Now

$i\in I^{\prime}_{j}\Longrightarrow S_{t+j-1}-k+1=S^{\prime}_{j-1}+1\leq i\leq
S_{j}^{\prime}=S_{t+j}-k$

$\Longrightarrow S_{t+j-1}+1\leq i+k\leq S_{t+j}$

$\Longrightarrow i+k\in I_{t+j}$.

And

$\max\\{i_{2},\cdots,i_{m}\\}\leq
S^{\prime}_{j-1}\Longrightarrow\max\\{i_{2},\cdots,i_{m}\\}\leq S_{t+j-1}-k$

$\Longrightarrow\max\\{i_{2}+k,\cdots,i_{m}+k\\}\leq S_{t+j-1}$.

So by the hypothesis that $\mathbb{A}$ is an $(n_{1},\cdots,n_{r})$-3rdUTB
tensor we have

$b_{ii_{2}\cdots i_{m}}=a_{(i+k)(i_{2}+k)\cdots(i_{m}+k)}=0.$

So $\mathbb{B}=\mathbb{A}[\overline{I}]$ is an $(n_{t+1},\cdots,n_{r})$-3rdUTB
tensor with diagonal blocks $\mathbb{A}_{t+1},\cdots,\mathbb{A}_{r}$ (where
$(n_{t+1},\cdots,n_{r})=(n_{1}^{\prime},\cdots,n^{\prime}_{r-t})$).

Sufficiency.

By Definition 2.3 we need to show that (2.7) holds for $j=2,\cdots,r$. Suppose
that $2\leq j\leq r$, $i\in I_{j}$, $\max\\{i_{2},\cdots,i_{m}\\}\leq
S_{j-1}$. We consider the following two cases.

Case 1: $j\leq t$.

Then by $\max\\{i_{2},\cdots,i_{m}\\}\leq S_{j-1}\leq S_{t}$ we see that
$a_{ii_{2}\cdots i_{m}}$ is an entry of $\mathbb{A}[I]$. So by condition (2)
we have $a_{ii_{2}\cdots i_{m}}=0$.

Case 2: $j\geq t+1$.

Then we consider the following two subcases.

Subcase 2.1: $\max\\{i_{2},\cdots,i_{m}\\}\leq S_{t}=k$.

Then by $i\in I_{j}\Longrightarrow i>S_{j-1}\geq S_{t}=k$, we have
$a_{ii_{2}\cdots i_{m}}=0$ by condition (1).

Subcase 2.2: $\max\\{i_{2},\cdots,i_{m}\\}>k$.

Now we still have

$\max\\{i_{2},\cdots,i_{m}\\}\leq
S_{j-1}=S^{\prime}_{j-t-1}+k\Longrightarrow\max\\{i_{2}-k,\cdots,i_{m}-k\\}\leq
S^{\prime}_{j-t-1}.$

Also

$i\in I_{j}\Longrightarrow S_{j-1}<i\leq S_{j}\Longrightarrow
S^{\prime}_{j-t-1}<i-k\leq S^{\prime}_{j-t}\Longrightarrow i-k\in
I^{\prime}_{j-t}.$

So by condition (3) and using Definition 2.3 for $\mathbb{B}$, we have
$b_{(i-k)(i_{2}-k)\cdots(i_{m}-k)}=0$, which is equivalent to $a_{ii_{2}\cdots
i_{m}}=0$. $\square$

In the following we will see that, for the first type triangular blocked
tensors, only the necessity part of Theorem 2.1 holds (see Example 2.4 and
Lemma 2.1 below), and the sufficiency part only holds for the case $t=1$ (see
Theorem 2.2). We will also see that, for the second type triangular blocked
tensors, only the sufficiency part of Theorem 2.1 holds (see Example 2.4 and
Lemma 2.2 below), and the necessity part only holds for the case $t=r-1$ (see
Theorem 2.4).

###### Example 2.4

Take $n=3$, $m=3$ and $\mathbb{A}$ to be a (0,1) tensor of order 3 and
dimension 3 with

$a_{213}=1,$

and all the other entries of $\mathbb{A}$ are zero. Taking
$n_{1}=n_{2}=n_{3}=1$. Then we have

(1). $\mathbb{A}$ is a (1,1,1)-2ndUTB tensor, but is not a (1,2)-2ndUTB
tensor. Because if $\mathbb{A}$ is a (1,2)-2ndUTB tensor, then it should
satisfy the following condition:

$a_{ii_{2}i_{3}}=0\quad(\forall i\geq 2,\ \min\\{i_{2},i_{3}\\}\leq 1\
\mbox{and}\ \max\\{i_{2},i_{3}\\}\leq 3).$

But $a_{213}=1$ means that the above condition does not hold. This shows that
the necessity part (1) of Theorem 2.1 does not hold for the second type UTB
tensor (by taking $t=1$, and thus $k=1$ in Theorem 2.1).

(2). By taking $t=2$, and thus $k=2$ in Theorem 2.1, we can check that
$\mathbb{A}$ satisfies all the three conditions in Theorem 2.1, but
$\mathbb{A}$ is not a (1,1,1)-UTB tensor. Because if $\mathbb{A}$ is a
(1,1,1)-UTB tensor, then it should satisfy the following condition (taking
$j=2$ in (2.1)):

$a_{ii_{2}i_{3}}=0\quad(\forall i\in I_{2}=\\{2\\},\ \min\\{i_{2},i_{3}\\}\leq
1).$

But $a_{213}=1$ means that the above condition does not hold. This shows that
the sufficiency part of Theorem 2.1 does not hold for the first type UTB
tensor.

###  2.3 Some basic properties and characterizations of the first type
triangular blocked tensors

This subsection mainly contains two results. Firstly we show in Lemma 2.1 that
the necessity part of Theorem 2.1 also holds for the first type triangular
blocked tensors. Secondly we use Lemma 2.1 to prove Theorem 2.2 which gives a
necessary and sufficient condition for the first type
$(n_{1},\cdots,n_{r})$-UTB tensors and will be used later in Section 4.

###### Lemma 2.1

Let the meaning of the notations be as in Definition 2.1, $r\geq 2$, $1\leq
t\leq r-1$, write $k=S_{t}$ and $I=[k]$. If $\mathbb{A}$ is an
$(n_{1},\cdots,n_{r})$-UTB tensor with diagonal blocks
$\mathbb{A}_{1},\cdots,\mathbb{A}_{r}$, then we have:

(1) $\mathbb{A}$ is an $(k,n-k)$-UTB tensor with diagonal blocks
$\mathbb{A}[I]$ and $\mathbb{A}[\overline{I}]$.

(2) $\mathbb{A}[I]$ is an $(n_{1},\cdots,n_{t})$-UTB tensor with diagonal
blocks $\mathbb{A}_{1},\cdots,\mathbb{A}_{t}$.

(3) $\mathbb{A}[\overline{I}]$ is an $(n_{t+1},\cdots,n_{r})$-UTB tensor with
diagonal blocks $\mathbb{A}_{t+1},\cdots,\mathbb{A}_{r}$.

Since the proof of Lemma 2.1 is similar to the necessity part of Theorem 2.1,
we choose to omit its proof.

Using Lemma 2.1, we can obtain the following characterization for the first
type upper triangular blocked tensors. This means that, for the first type
upper triangular blocked tensors, the sufficiency part of Theorem 2.1 still
holds for the case $t=1$. This theorem will be used later in the proof of
Theorem 4.1.

###### Theorem 2.2

Let the meaning of the notations be as in Definition 2.1, $r\geq 2$. Then
$\mathbb{A}$ is an $(n_{1},\cdots,n_{r})$-UTB tensor with diagonal blocks
$\mathbb{A}_{1},\cdots,\mathbb{A}_{r}$ if and only if $\mathbb{A}$ satisfies
the following two conditions:

(1) $\mathbb{A}$ is an $(n_{1},n-n_{1})$-UTB tensor with diagonal blocks
$\mathbb{A}_{1}$ and $\mathbb{A}[\overline{I_{1}}]$.

(2) $\mathbb{A}[\overline{I_{1}}]$ is an $(n_{2},\cdots,n_{r})$-UTB tensor
with diagonal blocks $\mathbb{A}_{2},\cdots,\mathbb{A}_{r}$.

Proof. The necessity part follows from Lemma 2.1. Now we prove the sufficiency
part. Similarly as in Lemma 2.1, we write
$\mathbb{B}=\mathbb{A}[\overline{I_{1}}]$. Thus we have:

$b_{ii_{2}\cdots
i_{m}}=a_{(i+n_{1})(i_{2}+n_{1})\cdots(i_{m}+n_{1})}\quad(i,i_{2},\cdots,i_{m}\in[n-n_{1}]).$

For $j=1,\cdots,r-1$, we further write $n^{\prime}_{j}=n_{j+1}$, and
consequently write

$S^{\prime}_{j}=n^{\prime}_{1}+\cdots+n^{\prime}_{j},\quad
I^{\prime}_{j}=\\{S^{\prime}_{j-1}+1,\cdots,S^{\prime}_{j}\\}\quad(j=1,\cdots,r-1).$

Then we have $S^{\prime}_{j}=S_{j+1}-n_{1}$.

By Definition 2.1 we need to show that (2.1) holds for $j=2,\cdots,r$. Suppose
that $2\leq j\leq r$, $i\in I_{j}$ and $\min\\{i_{2},\cdots,i_{m}\\}\leq
S_{j-1}$ (say, $i_{2}=\min\\{i_{2},\cdots,i_{m}\\}$). Then we consider the
following two cases:

Case 1: $i_{2}\leq n_{1}$.

Then by $i\in I_{j}\Longrightarrow i>S_{j-1}\geq S_{1}=n_{1}$, we have
$a_{ii_{2}\cdots i_{m}}=0$ by Example 2.1 and condition (1).

Case 2: $i_{2}>n_{1}$.

Then by hypothesis we have $i_{2}\leq
S_{j-1}=S^{\prime}_{j-2}+n_{1}\Longrightarrow i_{2}-n_{1}\leq
S^{\prime}_{j-2}$. Also

$i\in I_{j}\Longrightarrow S_{j-1}+1\leq i\leq S_{j}\Longrightarrow
S^{\prime}_{j-2}+1\leq i-n_{1}\leq S^{\prime}_{j-1}\Longrightarrow i-n_{1}\in
I^{\prime}_{j-1}.$

So by condition (2) we have $b_{(i-n_{1})(i_{2}-n_{1})\cdots(i_{m}-n_{1})}=0$,
which is equivalent to $a_{ii_{2}\cdots i_{m}}=0$. $\square$

Using Theorem 2.2, we can obtain the following equivalent definition for the
$(n_{1},\cdots,n_{r})$-UTB tensor (we omit its proof since this result will
not be used in the remaining of this paper.

###### Theorem 2.3

Let the meaning of the notations be as in Definition 2.1, $r\geq 2$. Then
$\mathbb{A}$ is an $(n_{1},\cdots,n_{r})$-UTB tensor with diagonal blocks
$\mathbb{A}_{1},\cdots,\mathbb{A}_{r}$ if and only if for each
$i=1,\cdots,r-1$, $\mathbb{A}[I_{i}\cup\cdots\cup I_{r}]$ is an
$(n_{i},n_{i+1}+\cdots+n_{r})$-UTB tensor with diagonal blocks
$\mathbb{A}_{i}$ and $\mathbb{A}[I_{i+1}\cup\cdots\cup I_{r}]$.

###  2.4 Some basic properties and characterizations of the second type
triangular blocked tensors

This subsection mainly contains three results. Firstly we show in Lemma 2.2
that the sufficiency part of Theorem 2.1 also holds for the second type
triangular blocked tensors. Secondly we use Lemma 2.2 to prove Theorem 2.4
which gives a necessary and sufficient condition for the second type
$(n_{1},\cdots,n_{r})$-UTB tensors and will be used later in Section 3, 4 and
6. Thirdly we show in Theorem 2.5 that the definition of the second type UTB
tensors in this paper is essentially equivalent to a definition in Definition
4.3 of [16].

###### Lemma 2.2

Let the meaning of the notations be as in Definition 2.1, $r\geq 2$, $1\leq
t\leq r-1$, write $k=S_{t}$ and $I=[k]$. If $\mathbb{A}$ satisfies the
following three conditions:

(1) $\mathbb{A}$ is an $(k,n-k)$-2ndUTB tensor with diagonal blocks
$\mathbb{A}[I]$ and $\mathbb{A}[\overline{I}]$.

(2) $\mathbb{A}[I]$ is an $(n_{1},\cdots,n_{t})$-2ndUTB tensor with diagonal
blocks $\mathbb{A}_{1},\cdots,\mathbb{A}_{t}$.

(3) $\mathbb{A}[\overline{I}]$ is an $(n_{t+1},\cdots,n_{r})$-2ndUTB tensor
with diagonal blocks $\mathbb{A}_{t+1},\cdots,\mathbb{A}_{r}$.

Then $\mathbb{A}$ is an $(n_{1},\cdots,n_{r})$-2ndUTB tensor with diagonal
blocks $\mathbb{A}_{1},\cdots,\mathbb{A}_{r}$.

Proof. Similarly as in Lemma 2.1, we write
$\mathbb{B}=\mathbb{A}[\overline{I}]$. Thus we have

$b_{ii_{2}\cdots
i_{m}}=a_{(i+k)(i_{2}+k)\cdots(i_{m}+k)}\quad(i,i_{2},\cdots,i_{m}\in[n-k]).$

For $j=1,\cdots,r-t$, we further write $n^{\prime}_{j}=n_{t+j}$, and
consequently write

$S^{\prime}_{j}=n^{\prime}_{1}+\cdots+n^{\prime}_{j},\quad
I^{\prime}_{j}=\\{S^{\prime}_{j-1}+1,\cdots,S^{\prime}_{j}\\}\quad(j=1,\cdots,r-t).$

Then we have $S^{\prime}_{j}=S_{t+j}-S_{t}=S_{t+j}-k$.

By Definition 2.2 we need to show that (2.4) holds for $j=2,\cdots,r$. Suppose
that $2\leq j\leq r$, $i\in I_{j}$, $\min\\{i_{2},\cdots,i_{m}\\}\leq S_{j-1}$
(say, $i_{2}=\min\\{i_{2},\cdots,i_{m}\\}$), and
$\max\\{i_{2},\cdots,i_{m}\\}\leq S_{j}$. We consider the following two cases.

Case 1: $j\leq t$.

Then by $\max\\{i_{2},\cdots,i_{m}\\}\leq S_{j}\leq S_{t}$ we see that
$a_{ii_{2}\cdots i_{m}}$ is an entry of $\mathbb{A}[I]$. So by condition (2)
we have $a_{ii_{2}\cdots i_{m}}=0$.

Case 2: $j\geq t+1$.

Then for $i_{2}=\min\\{i_{2},\cdots,i_{m}\\}\leq S_{j-1}$ and
$\max\\{i_{2},\cdots,i_{m}\\}\leq S_{j}$, we consider the following two
subcases.

Subcase 2.1: $i_{2}\leq S_{t}=k$.

Then by $i\in I_{j}\Longrightarrow i>S_{j-1}\geq S_{t}=k$, we have
$a_{ii_{2}\cdots i_{m}}=0$ by Example 2.1nd condition (1).

Subcase 2.2: $i_{2}>k$.

But we still have $i_{2}\leq S_{j-1}=S^{\prime}_{j-t-1}+k\Longrightarrow
i_{2}-k\leq S^{\prime}_{j-t-1}$. Also

$i\in I_{j}\Longrightarrow S_{j-1}<i\leq S_{j}\Longrightarrow
S^{\prime}_{j-t-1}<i-k\leq S^{\prime}_{j-t}\Longrightarrow i-k\in
I^{\prime}_{j-t}.$

Furthermore, we have $\max\\{i_{2},\cdots,i_{m}\\}\leq
S_{j}\Longrightarrow\max\\{i_{2}-k,\cdots,i_{m}-k\\}\leq S^{\prime}_{j-t}$. So
by condition (3) and using Definition 2.2 for $\mathbb{B}$, we have
$b_{(i-k)(i_{2}-k)\cdots(i_{m}-k)}=0$, which is equivalent to $a_{ii_{2}\cdots
i_{m}}=0$. $\square$

Using Lemma 2.2, we can obtain the following characterization for the second
type upper triangular blocked tensors. This means that, for the second type
upper triangular blocked tensors, the necessity part of Theorem 2.1 still
holds for the case $t=r-1$. Also, the following Theorem 2.4 will be used later
in the proofs of Theorems 3.3, 4.2 and 6.3.

###### Theorem 2.4

Let the meaning of the notations be as in Definition 2.1, $r\geq 2$. Then
$\mathbb{A}$ is an $(n_{1},\cdots,n_{r})$-2ndUTB tensor with diagonal blocks
$\mathbb{A}_{1},\cdots,\mathbb{A}_{r}$ if and only if $\mathbb{A}$ satisfies
the following two conditions:

(1) $\mathbb{A}$ is an $(n-n_{r},n_{r})$-2ndUTB tensor with diagonal blocks
$\mathbb{A}[I_{1}\cup\cdots\cup I_{r-1}]$ and $\mathbb{A}_{r}$.

(2) $\mathbb{A}[I_{1}\cup\cdots\cup I_{r-1}]$ is an
$(n_{1},\cdots,n_{r-1})$-2ndUTB tensor with diagonal blocks
$\mathbb{A}_{1},\cdots,\mathbb{A}_{r-1}$.

Proof. The sufficiency part follows from Lemma 2.2. Now we prove the necessity
part.

(1). Write $k=n-n_{r}=n_{1}+\cdots+n_{r-1}=S_{r-1}$. By Example 2.1 we only
need to verify the following

$a_{ii_{2}\cdots i_{m}}=0\quad(\forall i>k,\ \mbox{and}\
\min\\{i_{2},\cdots,i_{m}\\}\leq k).$

Now assume that $i\in I_{j}$. Then $S_{j}\geq i>k=S_{r-1}\Longrightarrow
j>r-1\Longrightarrow j=r$. In this case, $\max\\{i_{2},\cdots,i_{m}\\}\leq
S_{j}=S_{r}=n$ holds automatically. So by the hypothesis that $\mathbb{A}$ is
an $(n_{1},\cdots,n_{r})$-2ndUTB tensor and Definition 2.2 and the fact that
$k=S_{r-1}$ we have $a_{ii_{2}\cdots i_{m}}=0$.

(2). Since the condition in Eq.(2.4) holds for all $j=2,\cdots,r$ for the
tensor $\mathbb{A}$, it also holds for all $j=2,\cdots,r-1$ for the tensor
$\mathbb{A}[I_{1}\cup\cdots\cup I_{r-1}]$. So the conclusion follows from
Definition 2.2. $\square$

The following theorem shows that our definition for the
$(n_{1},\cdots,n_{r})$-2ndUTB tensors is essentially equivalent to Definition
4.3 in [16]. We would like to point out that this theorem will not be used
later in this paper.

###### Theorem 2.5

Let the meaning of the notations be as in Definition 2.1, $r\geq 2$. Then
$\mathbb{A}$ is an $(n_{1},\cdots,n_{r})$-2ndUTB tensor with diagonal blocks
$\mathbb{A}_{1},\cdots,\mathbb{A}_{r}$ if and only if for each $i=2,\cdots,r$,
$\mathbb{A}[I_{1}\cup\cdots\cup I_{i}]$ is an
$((n_{1}+\cdots+n_{i-1}),n_{i})$-2ndUTB tensor with diagonal blocks
$\mathbb{A}[I_{1}\cup\cdots\cup I_{i-1}]$ and $\mathbb{A}_{i}$.

Proof. Necessity. The case $i=r$ follows from the necessity part (1) of
Theorem 2.4. Now we assume $1\leq i\leq r-1$. By the necessity part (2) of
Theorem 2.4, $\mathbb{A}[I_{1}\cup\cdots\cup I_{r-1}]$ is an
$(n_{1},\cdots,n_{r-1})$-2ndUTB tensor with diagonal blocks
$\mathbb{A}_{1},\cdots,\mathbb{A}_{r-1}$. Thus by induction on $r$ for the
tensor $\mathbb{A}[I_{1}\cup\cdots\cup I_{r-1}]$, we obtain the desired
result.

Sufficiency. We use induction on $r$. The case $r=2$ is trivial. Now we assume
that $r\geq 3$. By induction and the case $i=2,\cdots,r-1$ of the hypothesis,
$\mathbb{A}[I_{1}\cup\cdots\cup I_{r-1}]$ is an
$(n_{1},\cdots,n_{r-1})$-2ndUTB tensor with diagonal blocks
$\mathbb{A}_{1},\cdots,\mathbb{A}_{r-1}$. By the case $i=r$ of the hypothesis,
and using the sufficiency part of the Theorem 2.4, we conclude that
$\mathbb{A}$ is an $(n_{1},\cdots,n_{r})$-2ndUTB tensor with diagonal blocks
$\mathbb{A}_{1},\cdots,\mathbb{A}_{r}$. $\square$

Comparing Theorem 2.5 with Definition 4.3 and Remark 4.1 in [16] we can see
that, if we transfer Definition 4.3 of [16] from the form of lower triangular
blocked tensors to its equivalent form of upper triangular blocked tensors,
then that type of $(n_{1},\cdots,n_{r})$-lower triangular blocked tensors is
actually equivalent to the second type $(n_{1},\cdots,n_{r})$-upper triangular
blocked tensors defined in this paper.

##  3 The determinants, characteristic polynomials and spectra of the
triangular blocked tensors

In this section, we will prove some formulas for the determinants and
characteristic polynomials of the first and second type UTB tensors, and give
an example to show that these formulas do not hold for the third type UTB
tensors.

The following result for the special case $r=2$ was proved by Hu et al in [8].
For the convenience and self-containedness, here we give an outline of a
simplified proof.

###### Theorem 3.1

([8]) Let $\mathbb{A}$ be an order $m$ dimension $n$ and $(k,n-k)$-UTB (also
2ndUTB by Remark 2.1) tensor with diagonal blocks $\mathbb{A}_{1}$ and
$\mathbb{A}_{2}$. Then

$\det\mathbb{A}=(\det\mathbb{A}_{1})^{(m-1)^{n-k}}(\det\mathbb{A}_{2})^{(m-1)^{k}}$
(3.1)

Proof. Step 1: To show that
$\det\mathbb{A}=0\Longrightarrow(\det\mathbb{A}_{1})(\det\mathbb{A}_{2})=0$.
Suppose not, then

$(\det\mathbb{A}_{1})(\det\mathbb{A}_{2})\neq
0\Longrightarrow\left\\{\begin{array}[]{c}\det\mathbb{A}_{2}\neq
0\Longrightarrow\mathbb{A}_{2}z=0\ \ \mbox{has only zero solution}\\\
\det\mathbb{A}_{1}\neq 0\Longrightarrow\mathbb{A}_{1}y=0\ \ \mbox{has only
zero solution}\\\ \end{array}\right.$

$\stackrel{{\scriptstyle(\ast)}}{{\Longrightarrow}}\mathbb{A}x=\mathbb{A}\left(\begin{array}[]{c}y\\\
z\\\ \end{array}\right)=0$ has only zero solution

$\Longrightarrow\det\mathbb{A}\neq 0$.

where the implication $(\ast)$ holds because: $\mathbb{A}$ is an $(k,n-k)$-UTB
tensor with diagonal blocks $\mathbb{A}_{1}$ and $\mathbb{A}_{2}$ implies that

$\mathbb{A}\left(\begin{array}[]{c}y\\\ z\\\
\end{array}\right)=\left(\begin{array}[]{c}*\\\ \mathbb{A}_{2}z\\\
\end{array}\right)\quad\mbox{and}\quad\mathbb{A}\left(\begin{array}[]{c}y\\\
0\\\ \end{array}\right)=\left(\begin{array}[]{c}\mathbb{A}_{1}y\\\ 0\\\
\end{array}\right).$

Step 2: By using Hilbert’s Nullstellensatz ([5]) and the irreducibility of the
polynomials $\det\mathbb{A}_{1}$ and $\det\mathbb{A}_{2}$ ([8]), we have Every
irreducible factor of $\det\mathbb{A}$ is an irreducible factor of
$(\det\mathbb{A}_{1})(\det\mathbb{A}_{2})$.

Thus there exist constants $c,r_{1},r_{2}$ (independent of the tensor
$\mathbb{A}$) such that

$\det\mathbb{A}=c\cdot(\det\mathbb{A}_{1})^{r_{1}}(\det\mathbb{A}_{2})^{r_{2}}.$
(3.2)

Step 3: Take $\mathbb{A}=\mathbb{I}$ to be the identity tensor, then both
$\mathbb{A}_{1}$ and $\mathbb{A}_{2}$ are also the identity tensors, thus we
obtain $c=1$ from (3.2).

Step 4 (To further determine $r_{1},r_{2}$): Take $\mathbb{A}$ to be the
diagonal tensor with the first $k$ diagonal entries $a$, and all the rest
diagonal entries $b$, then we have:

$\det\mathbb{A}=a^{k(m-1)^{n-1}}b^{(n-k)(m-1)^{n-1}},$ (3.3)

and

$\det\mathbb{A}_{1}=a^{k(m-1)^{k-1}},\qquad
det\mathbb{A}_{2}=b^{(n-k)(m-1)^{n-k-1}}.$ (3.4)

Substituting (3.3), (3.4) and $c=1$ into (3.2), we obtain

$a^{k(m-1)^{n-1}}b^{(n-k)(m-1)^{n-1}}=a^{r_{1}k(m-1)^{k-1}}b^{r_{2}(n-k)(m-1)^{n-k-1}}.$
(3.5)

Comparing the exponents of $a$ and $b$ of the both sides of (3.5), we obtain
$r_{1}=(m-1)^{n-k},\ r_{2}=(m-1)^{k}.$ $\square$

The following example shows that Theorem 3.1 does not hold for the third type
$(k,n-k)$-UTB tensors.

###### Example 3.1

Take a $(0,1)$ tensor $\mathbb{A}$ of order $m=3$ and dimension $n=2$ with

$a_{111}=a_{122}=1,\ a_{112}=a_{121}=0\quad and\quad a_{211}=a_{222}=0,\
a_{212}=a_{221}=1.$

Then $\mathbb{A}$ is a (1,1)-3rdUTB tensor (since $a_{211}=0$) with the two
diagonal blocks $\mathbb{A}_{1}=a_{111}=1$ and $\mathbb{A}_{2}=a_{222}=0$. In
this case, the right hand side of (3.1) is zero since $\det\mathbb{A}_{2}=0$.

On the other hand, we now show that $\det\mathbb{A}\neq 0$. By the definition
we can verify that $\mathbb{A}x=0$ is the following system of equations:

$\left\\{\begin{array}[]{c}x_{1}^{2}+x_{2}^{2}=0;\\\ 2x_{1}x_{2}=0,\\\
\end{array}\right.$

which has only a zero solution. Thus we have $\det\mathbb{A}\neq 0$ by the
definition of the tensor determinants, and so (3.1) does not hold for this
$(1,1)$-3rdUTB tensor $\mathbb{A}$.

Now we show that Theorem 3.1 also holds for lower triangular blocked tensors.

###### Theorem 3.2

Theorem 3.1 also holds for lower triangular blocked tensors.

Proof. Write $k^{\prime}=n-k$. Then by (5) of Remark 2.4 we have that:

$\mathbb{A}$ is an order $m$ dimension $n$ and $(k,n-k)$-LTB tensor with
diagonal blocks $\mathbb{A}_{1}$ and $\mathbb{A}_{2}$

$\Longrightarrow\mathbb{A}$ is permutation similar to a
$(k^{\prime},n-k^{\prime})$-UTB tensor $\mathbb{B}$ with diagonal blocks
$\mathbb{A}_{2}$ and $\mathbb{A}_{1}$.

Thus by [15] (similar tensors have the same determinants) and using Theorem
3.1 for $\mathbb{B}$ we have:

$\det\mathbb{A}=\det\mathbb{B}=(\det\mathbb{A}_{2})^{(m-1)^{n-k^{\prime}}}(\det\mathbb{A}_{1})^{(m-1)^{k^{\prime}}}=(\det\mathbb{A}_{1})^{(m-1)^{n-k}}(\det\mathbb{A}_{2})^{(m-1)^{k}}.$

$\square$

By using Theorem 3.1 and Lemma 2.1, we can use induction to obtain the formula
for determinants of general $(n_{1},\cdots,n_{r})$-UTB (and 2ndUTB) tensors as
follows. (The case of the second type also appeared in Shao et al [16].)

###### Theorem 3.3

Let $\mathbb{A}$ be an order $m$ dimension $n$ and $(n_{1},\cdots,n_{r})$-UTB
(or 2ndUTB) tensor with diagonal blocks
$\mathbb{A}_{1},\cdots,\mathbb{A}_{r}$. Then we have:

$\det\mathbb{A}=\prod_{i=1}^{r}(\det\mathbb{A}_{i})^{(m-1)^{n-n_{i}}}.$ (3.6)

Proof. We use induction on $r$. If $r=2$, the result is just Theorem 3.1. Now
we assume $r\geq 3$. Write $\mathbb{B}=\mathbb{A}[I_{1}\cup\cdots\cup
I_{r-1}]$, then by Lemma 2.1 and Theorem 2.4 we know that $\mathbb{A}$ is an
$(n-n_{r},n_{r})$-UTB (or 2ndUTB) tensor with diagonal blocks $\mathbb{B}$ and
$\mathbb{A}_{r}$. Thus by Theorem 3.1 we have

$\det\mathbb{A}=(\det\mathbb{B})^{(m-1)^{n_{r}}}(\det\mathbb{A}_{r})^{(m-1)^{n-n_{r}}}.$
(3.7)

Also by Lemma 2.1 and Theorem 2.4 we know that $\mathbb{B}$ is an
$(n_{1},\cdots,n_{r-1})$-UTB (or 2ndUTB) tensor with diagonal blocks
$\mathbb{A}_{1},\cdots,\mathbb{A}_{r-1}$. So by induction we have

$\det\mathbb{B}=\prod_{i=1}^{r-1}(\det\mathbb{A}_{i})^{(m-1)^{n-n_{r}-n_{i}}}.$
(3.8)

Substituting (3.8) into (3.7), we obtain (3.6). $\square$

###### Theorem 3.4

Let $\mathbb{A}$ be an order $m$ dimension $n$ and $(n_{1},\cdots,n_{r})$-UTB
(or 2ndUTB) tensor with diagonal blocks
$\mathbb{A}_{1},\cdots,\mathbb{A}_{r}$. Then we have the following relations
for the characteristic polynomial and spectrum of $\mathbb{A}$:

$\phi_{\mathbb{A}}(\lambda)=\prod_{i=1}^{r}(\phi_{\mathbb{A}_{i}}(\lambda))^{(m-1)^{n-n_{i}}},$
(3.9)

and

$Spec(\mathbb{A})=\bigcup_{i=1}^{r}(Spec(\mathbb{A}_{i}))^{(m-1)^{n-n_{i}}},$
(3.10)

where the notation $S^{k}$ of a multi-set $S$ denotes the repetition of $k$
times of the set $S$, and

$\rho(\mathbb{A})=\max\\{\rho(\mathbb{A}_{i})\ |\ i=1,\cdots,r\\}.$ (3.11)

###### Corollary 3.1

Let $G$ be a $k$-uniform hypergraph with the $r$ connected components
$G_{1},\cdots,G_{r}$. Then we have

$Spec(\mathbb{A}(G))=\bigcup_{i=1}^{r}(Spec(\mathbb{A}(G_{i})))^{(m-1)^{n-n_{i}}},$
(3.12)

and

$\rho(\mathbb{A}(G))=\max\\{\rho(\mathbb{A}(G_{i}))\ |\ i=1,\cdots,r\\}.$
(3.13)

Proof. By Example 2.3, the adjacency tensor $\mathbb{A}(G)$ of $G$ is a
diagonal blocked tensor with diagonal blocks
$\mathbb{A}(G_{1}),\cdots,\mathbb{A}(G_{r})$. $\square$

##  4 The products of the triangular blocked tensors

In this section we will show that the product of any two
$(n_{1},\cdots,n_{r})$ upper (or lower) triangular blocked tensors of the
first or second or third type is still an $(n_{1},\cdots,n_{r})$ upper (or
lower) triangular blocked tensor of the same type. We first prove the case
$r=2$ in Lemma 4.1 (for the first and second types) and Lemma 4.2 (for the
third type).

The first result (but not the second result about the two diagonal blocks of
$\mathbb{A}\mathbb{B}$) of the following lemma can be found in [16]
(Proposition 4.1). Recall that in Remark 2.1, we have mentioned that in the
case $r=2$, the $(n_{1},n_{2})$-UTB tensors and $(n_{1},n_{2})$-2ndUTB tensors
are the same.

###### Lemma 4.1

Let $\mathbb{A}$ and $\mathbb{B}$ be order $m$ and order $k$, dimension $n$
and $(p,n-p)$-UTB (also 2ndUTB by Remark 2.1) tensor with diagonal blocks
$\mathbb{A}_{1}$, $\mathbb{A}_{2}$ and $\mathbb{B}_{1}$, $\mathbb{B}_{2}$,
respectively. Then their product $\mathbb{A}\mathbb{B}$ is also a
$(p,n-p)$-UTB tensor with the two diagonal blocks
$\mathbb{A}_{1}\mathbb{B}_{1}$ and $\mathbb{A}_{2}\mathbb{B}_{2}$.

Proof. Write $\mathbb{C}=\mathbb{A}\mathbb{B}$. Then by the definition of the
tensor product we have (see (1.3)):

$c_{i\alpha_{1}\cdots\alpha_{m-1}}=\sum_{i_{2},\cdots,i_{m}=1}^{n}a_{ii_{2}\cdots
i_{m}}b_{i_{2}\alpha_{1}}\cdots b_{i_{m}\alpha_{m-1}}\qquad(i\in[n],\
\alpha_{1},\cdots,\alpha_{m-1}\in[n]^{k-1}).$ (4.1)

The proof of the first result that $\mathbb{C}$ is a $(p,n-p)$-UTB tensor can
be found in [16] (Proposition 4.1).

Now we show that the diagonal blocks $\mathbb{C}_{1}$ and $\mathbb{C}_{2}$ of
$\mathbb{A}\mathbb{B}$ are $\mathbb{A}_{1}\mathbb{B}_{1}$ and
$\mathbb{A}_{2}\mathbb{B}_{2}$. For the first diagonal block $\mathbb{C}_{1}$,
let $i$ and all the subscripts in $\alpha_{1},\cdots,\alpha_{m-1}$ are all
$\leq p$. In this case, if all $b_{i_{j}\alpha_{j-1}}\neq 0$, then we must
have all $i_{j}\leq p\ (j=2,\cdots,m)$ since $\mathbb{B}$ is $(p,n-p)$-UTB.
Thus we have

$c_{i\alpha_{1}\cdots\alpha_{m-1}}=\sum_{i_{2},\cdots,i_{m}=1}^{n}a_{ii_{2}\cdots
i_{m}}b_{i_{2}\alpha_{1}}\cdots
b_{i_{m}\alpha_{m-1}}=\sum_{i_{2},\cdots,i_{m}=1}^{p}a_{ii_{2}\cdots
i_{m}}b_{i_{2}\alpha_{1}}\cdots
b_{i_{m}\alpha_{m-1}}=(\mathbb{A}_{1}\mathbb{B}_{1})_{i\alpha_{1}\cdots\alpha_{m-1}},$

so we obtain $\mathbb{C}_{1}=\mathbb{A}_{1}\mathbb{B}_{1}$.

Now we consider the second diagonal block $\mathbb{C}_{2}$. Let $i$ and all
the subscripts in $\alpha_{1},\cdots,\alpha_{m-1}$ are all $>p$. In this case,
if $\min\\{i_{2},\cdots,i_{m}\\}\leq p$, then we will have $a_{ii_{2}\cdots
i_{m}}=0$ since $\mathbb{A}$ is $(p,n-p)$-UTB. Thus we have

$c_{i\alpha_{1}\cdots\alpha_{m-1}}=\sum_{i_{2},\cdots,i_{m}=1}^{n}a_{ii_{2}\cdots
i_{m}}b_{i_{2}\alpha_{1}}\cdots
b_{i_{m}\alpha_{m-1}}=\sum_{i_{2},\cdots,i_{m}=p+1}^{n}a_{ii_{2}\cdots
i_{m}}b_{i_{2}\alpha_{1}}\cdots b_{i_{m}\alpha_{m-1}}.$

Let $j=i-p$, $j_{t}=i_{t}-p$, and $\beta_{j}$ be the sequence of the
subscripts obtained by subtracting $p$ from all the subscripts in
$\alpha_{j}$. Then we have

$\displaystyle(\mathbb{A}_{2}\mathbb{B}_{2})_{j\beta_{1}\cdots\beta_{m-1}}$
$\displaystyle=\sum_{j_{2},\cdots,j_{m}=1}^{n-p}(\mathbb{A}_{2})_{jj_{2}\cdots
j_{m}}(\mathbb{B}_{2})_{j_{2}\beta_{1}}\cdots(\mathbb{B}_{2})_{j_{m}\beta_{m-1}}$
$\displaystyle=\sum_{i_{2},\cdots,i_{m}=p+1}^{n}a_{ii_{2}\cdots
i_{m}}b_{i_{2}\alpha_{1}}\cdots
b_{i_{m}\alpha_{m-1}}=c_{i\alpha_{1}\cdots\alpha_{m-1}}.$

So we obtain $\mathbb{C}_{2}=\mathbb{A}_{2}\mathbb{B}_{2}$. $\square$

Now we show that Lemma 4.1 is also true for the 3rdUTB tensors.

###### Lemma 4.2

Lemma 4.1 is also true for the 3rdUTB tensors.

Proof. The proof is similar to the proof of Lemma 4.1. We only give the proof
starting from the following step.

We first prove that $\mathbb{C}$ is a $(p,n-p)$-3rdUTB tensor. By Definition
2.3 we need to show that $\mathbb{C}$ satisfies the following condition:

$c_{i\alpha_{1}\cdots\alpha_{m-1}}=0\quad(\forall i>p,\ \mbox{and all indices
in $\alpha_{1},\cdots,\alpha_{m-1}$ are $\leq p$}).$ (4.2)

Now we consider the following two cases to show that every term in the
summation of (4.1) is zero.

Case 1: $\max\\{i_{2},\cdots,i_{m}\\}\leq p$.

Then $a_{ii_{2}\cdots i_{m}}=0$ since $i>p$ and $\mathbb{A}$ is
$(p,n-p)$-3rdUTB.

Case 2: $i_{j}>p$ for some $j\in\\{2,\cdots,m\\}$.

Then $b_{i_{j}\alpha_{j-1}}=0$, since all indices in $\alpha_{j-1}$ are $\leq
p$, and $\mathbb{B}$ is $(p,n-p)$-3rdUTB.

Combining these two cases, we see that every term in the summation of (4.1) is
zero. So (4.2) holds and thus $\mathbb{C}$ is also a $(p,n-p)$-3rdUTB tensor.

The proof of the second conclusion is similar to that of Lemma 4.1. $\square$

The following theorem gives the results on the products of the first type or
third type UTB tensors.

###### Theorem 4.1

Let $\mathbb{A}$ and $\mathbb{B}$ be order $m$ and order $k$, dimension $n$
and $(n_{1},\cdots,n_{r})$-UTB (or 3rdUTB) tensor with diagonal blocks
$\mathbb{A}_{1},\cdots,\mathbb{A}_{r}$ and
$\mathbb{B}_{1},\cdots,\mathbb{B}_{r}$, respectively. Then their product
$\mathbb{A}\mathbb{B}$ is also an $(n_{1},\cdots,n_{r})$-UTB (or 3rdUTB)
tensor with diagonal blocks
$\mathbb{A}_{1}\mathbb{B}_{1},\cdots,\mathbb{A}_{r}\mathbb{B}_{r}$.

Proof. The proof for the first type and third type UTB tensors are the same.
So here we only give the proof for the first type UTB tensors.

We use Theorem 2.2, Lemma 4.1 and induction on $r$. If $r=2$, this is just
Lemma 4.1. So we assume $r\geq 3$. Write
$\mathbb{C}=\mathbb{A}[I_{2}\cup\cdots\cup I_{r}]$ and
$\mathbb{D}=\mathbb{B}[I_{2}\cup\cdots\cup I_{r}]$. Then by Theorem 2.2 we
know that:

(i) $\mathbb{A}$ is an $(n_{1},n-n_{1})$-UTB tensor with diagonal blocks
$\mathbb{A}_{1}$ and $\mathbb{C}$.

(ii) $\mathbb{B}$ is an $(n_{1},n-n_{1})$-UTB tensor with diagonal blocks
$\mathbb{B}_{1}$ and $\mathbb{D}$.

(iii) $\mathbb{C}$ and $\mathbb{D}$ are $(n_{2},\cdots,n_{r})$-UTB tensors
with diagonal blocks $\mathbb{A}_{2},\cdots,\mathbb{A}_{r}$ and
$\mathbb{B}_{2},\cdots,\mathbb{B}_{r}$, respectively.

Thus by (i), (ii) and Lemma 4.1 we know that $\mathbb{A}\mathbb{B}$ is an
$(n_{1},n-n_{1})$-UTB tensor with diagonal blocks
$\mathbb{A}_{1}\mathbb{B}_{1}$ and $\mathbb{C}\mathbb{D}$.

By using (iii) and induction on $r$, we also know that $\mathbb{C}\mathbb{D}$
is an $(n_{2},\cdots,n_{r})$-UTB tensor with diagonal blocks
$\mathbb{A}_{2}\mathbb{B}_{2},\cdots,\mathbb{A}_{r}\mathbb{B}_{r}$.

Combining these two results with Theorem 2.2, we conclude that
$\mathbb{A}\mathbb{B}$ is also an $(n_{1},\cdots,n_{r})$-UTB tensor with
diagonal blocks
$\mathbb{A}_{1}\mathbb{B}_{1},\cdots,\mathbb{A}_{r}\mathbb{B}_{r}$. $\square$

Theorem 4.1 for the first and third types used the induction from top to
bottom. But the following Theorem 4.2 for the second type needs to use the
induction from bottom to top.

###### Theorem 4.2

Let $\mathbb{A}$ and $\mathbb{B}$ be order $m$ and order $k$, dimension $n$
and $(n_{1},\cdots,n_{r})$-2ndUTB tensor with diagonal blocks
$\mathbb{A}_{1},\cdots,\mathbb{A}_{r}$ and
$\mathbb{B}_{1},\cdots,\mathbb{B}_{r}$, respectively. Then their product
$\mathbb{A}\mathbb{B}$ is also an $(n_{1},\cdots,n_{r})$-2ndUTB tensor with
diagonal blocks
$\mathbb{A}_{1}\mathbb{B}_{1},\cdots,\mathbb{A}_{r}\mathbb{B}_{r}$.

Proof. We use Theorem 2.4, Lemma 4.1 and induction on $r$. If $r=2$, this is
just Lemma 4.1. So we assume $r\geq 3$. Write
$\mathbb{C}=\mathbb{A}[I_{1}\cup\cdots\cup I_{r-1}]$ and
$\mathbb{D}=\mathbb{B}[I_{1}\cup\cdots\cup I_{r-1}]$. Then by Theorem 2.4 we
know that:

(i) $\mathbb{A}$ is an $(n-n_{r},n_{r})$-2ndUTB tensor with diagonal blocks
$\mathbb{C}$ and $\mathbb{A}_{r}$.

(ii) $\mathbb{B}$ is an $(n-n_{r},n_{r})$-2ndUTB tensor with diagonal blocks
$\mathbb{D}$ and $\mathbb{B}_{r}$.

(iii) $\mathbb{C}$ and $\mathbb{D}$ are $(n_{1},\cdots,n_{r-1})$-2ndUTB
tensors with diagonal blocks $\mathbb{A}_{1},\cdots,\mathbb{A}_{r-1}$ and
$\mathbb{B}_{1},\cdots,\mathbb{B}_{r-1}$, respectively.

Thus by (i), (ii) and Lemma 4.1 we know that $\mathbb{A}\mathbb{B}$ is an
$(n-n_{r},n_{r})$-2ndUTB tensor with diagonal blocks $\mathbb{C}\mathbb{D}$
and $\mathbb{A}_{r}\mathbb{B}_{r}$.

By using (iii) and induction on $r$, we also know that $\mathbb{C}\mathbb{D}$
is an $(n_{1},\cdots,n_{r-1})$-2ndUTB tensor with diagonal blocks
$\mathbb{A}_{1}\mathbb{B}_{1},\cdots,\mathbb{A}_{r-1}\mathbb{B}_{r-1}$.

Combining these two results with Theorem 2.4, we conclude that
$\mathbb{A}\mathbb{B}$ is also an $(n_{1},\cdots,n_{r})$-2ndUTB tensor with
diagonal blocks
$\mathbb{A}_{1}\mathbb{B}_{1},\cdots,\mathbb{A}_{r}\mathbb{B}_{r}$. $\square$

##  5 The inverses of the triangular blocked tensors

In [3], Bu et al. defined the left and right inverses of a tensor as
following.

###### Definition 5.1

Let $\mathbb{A}$ and $\mathbb{B}$ be tensors of dimension $n$ with order $m$
and $k$, respectively. If $\mathbb{A}\mathbb{B}=\mathbb{I}$, then $\mathbb{A}$
is called a left $m$-inverse of $\mathbb{B}$, and $\mathbb{B}$ is called a
right $k$-inverse of $\mathbb{A}$.

In [3], Bu et al. obtained some results on left 2-inverses and right
2-inverses of tensors.

In [12], W.Liu and W.Li further studied the left and right $k$-inverses of
tensors. They proved that a tensor $\mathbb{A}$ has a left (or right)
$k$-inverse if and only if $\mathbb{A}$ has a left (or right) $2$-inverse.
They also proved the uniqueness of the left $k$-inverse, and obtained the
expressions of the left and right $k$-inverses of $\mathbb{A}$.

In this section, we study the left and right $k$-inverses of an
$(n_{1},\cdots,n_{r})$-UTB tensors of all the three types. We will show that,
if a first or second or third type $(n_{1},\cdots,n_{r})$-UTB tensor
$\mathbb{A}$ has a left (or right) $k$-inverse, then its unique left
$k$-inverse is still an $(n_{1},\cdots,n_{r})$-UTB tensor of all the three
types, and all of its right $k$-inverses are still $(n_{1},\cdots,n_{r})$-UTB
tensors of all the three types. Furthermore, by showing (in Theorem
LABEL:The54) that the left $k$-inverse (if any) of a weakly irreducible
nonsingular $M$-tensor is a positive tensor, we show that the left $k$-inverse
(if any) of a first or second or third type normal $(n_{1},\cdots,n_{r})$-UTB
nonsingular $M$-tensor is an $(n_{1},\cdots,n_{r})$-UTB tensor of all the
three types all of whose diagonal blocks are positive tensors. (The meaning of
“normal” can be found in Definitions 6.1 and 6.2 of §6.)

###  5.1 The left inverses of the triangular blocked tensors

The $i$-th row of a tensor $\mathbb{A}$ of order $m$ and dimension $n$,
denoted by $R_{i}(\mathbb{A})$, is the row-subtensor of order $m-1$ and
dimension $n$ of $\mathbb{A}$ with the entries

$(R_{i}(\mathbb{A}))_{i_{2}\cdots i_{m}}=a_{ii_{2}\cdots i_{m}}.$

###### Definition 5.2

A tensor $\mathbb{A}$ of order $m$ and dimension $n$ is called row-subtensor
diagonal, or simply row diagonal, if all of its row-subtensors
$R_{1}(\mathbb{A}),\cdots,R_{n}(\mathbb{A})$ are diagonal tensors. Namely, if
we have

$a_{ii_{2}\cdots i_{m}}=0\quad(\mbox{if}\ i_{2},\cdots,i_{m}\ \mbox{are not
all equal}).$

###### Definition 5.3

([13]) Let $\mathbb{A}$ be an order $m$ and dimension $n$ tensor. Then the
majorization matrix $M(\mathbb{A})$ of $\mathbb{A}$ is the matrix of order $n$
with the entries

$(M(\mathbb{A}))_{ij}=a_{ij\cdots j}\quad(\forall i,j=1,\cdots,n).$

###### Proposition 5.1

Let $\mathbb{A}$ be an order $m$ and dimension $n$ tensor. Then $\mathbb{A}$
is row diagonal if and only if there exists a matrix $P$ of order $n$ such
that $\mathbb{A}=P\mathbb{I}_{m}$, and in this case, $P=M(\mathbb{A})$.

Proof. Sufficiency. If $\mathbb{A}=P\mathbb{I}_{m}$, then by the definition of
tensor product we have

$a_{ii_{2}\cdots i_{m}}=\sum_{i_{1}=1}^{n}p_{ii_{1}}\delta_{i_{1}i_{2}\cdots
i_{m}}=0\quad(\mbox{if}\ i_{2},\cdots,i_{m}\ \mbox{are not all equal}),$

and

$a_{ij\cdots j}=\sum_{i_{1}=1}^{n}p_{ii_{1}}\delta_{i_{1}j\cdots
j}=p_{ij}\quad(\forall i,j=1,\cdots,n).$ (5.1)

Thus $\mathbb{A}$ is row diagonal, and $P=M(\mathbb{A})$.

Necessity. If $\mathbb{A}$ is row diagonal. Write $P=M(\mathbb{A})$ and
$\mathbb{B}=P\mathbb{I}_{m}$. Then by the sufficiency part we know that
$\mathbb{B}$ is row diagonal with $M(\mathbb{B})=P=M(\mathbb{A})$. Thus we
have $\mathbb{A}=\mathbb{B}=P\mathbb{I}_{m}=M(\mathbb{A})\mathbb{I}_{m}$.
$\square$

The following result can be found in [12] by W.Li et al. in a slightly
different version.

###### Lemma 5.1

([12], Theorem 3.1 and Corollary 3.3) Let $k\geq 2$ and $\mathbb{A}$ be an
order $m$ and dimension $n$ tensor. Then the following three conditions are
equivalent:

(1) $\mathbb{A}=P\mathbb{I}_{m}$ for some nonsingular matrix $P$.

(2) $\mathbb{A}$ has a left 2-inverse.

(3) $\mathbb{A}$ has a left $k$-inverse.

And in this case, the unique left 2-inverse of $\mathbb{A}$ is $P^{-1}$, and
the unique left $k$-inverse of $\mathbb{A}$ is $\mathbb{I}_{k}P^{-1}$.

(Notice that by Proposition 5.1 we know that condition (1) is also equivalent
to the following condition:

(4) $\mathbb{A}$ is row diagonal, and its majorization matrix $M(\mathbb{A})$
is nonsingular.)

Proof. See [12] Theorem 3.1 and Corollary 3.3. $\square$

###### Lemma 5.2

Let $P$ be a matrix of order $n$, $\mathbb{A}=P\mathbb{I}_{m}$ be an order $m$
dimension $n$, and a first or second or third type $(n_{1},\cdots,n_{r})$-UTB
tensor. Then $P$ is an $(n_{1},\cdots,n_{r})$-UTB matrix.

Proof. Since $\mathbb{A}$ is a first or second or third type
$(n_{1},\cdots,n_{r})$-UTB tensor, we have $a_{it\cdots t}=0$ for any
$j=2,\cdots,r$, $i\in I_{j}$ and $t\leq S_{j-1}$. Now
$\mathbb{A}=P\mathbb{I}_{m}$, so by (5.3) we have $p_{it}=a_{it\cdots t}=0$
for any $j=2,\cdots,r$, $i\in I_{j}$ and $t\leq S_{j-1}$, so by definition $P$
is an $(n_{1},\cdots,n_{r})$-UTB matrix. $\square$

###### Theorem 5.1

Let $k\geq 2$, $\mathbb{A}$ be an order $m$ dimension $n$, and the first or
second or third type $(n_{1},\cdots,n_{r})$-UTB tensor with diagonal blocks
$\mathbb{A}_{1},\cdots,\mathbb{A}_{r}$. If $\mathbb{A}$ has a left
$k$-inverse, then its (unique) left $k$-inverse is an
$(n_{1},\cdots,n_{r})$-UTB tensor of all the three types, with the diagonal
blocks $\mathbb{B}_{1},\cdots,\mathbb{B}_{r}$, where $\mathbb{B}_{i}$ is the
(unique) left $k$-inverse of $\mathbb{A}_{i}\ (i=1,\cdots,r)$.

Proof. By Lemma 5.1 we see that $\mathbb{A}=P\mathbb{I}_{m}$ for some
nonsingular matrix $P$, since $\mathbb{A}$ has a left $k$-inverse. By Lemma
5.2 we also know that $P$ is an $(n_{1},\cdots,n_{r})$-UTB matrix. Thus
$P^{-1}$ is also an $(n_{1},\cdots,n_{r})$-UTB matrix. By Lemma 5.1 again we
further know that its (unique) left $k$-inverse is
$\mathbb{B}=\mathbb{I}_{k}P^{-1}$. Now both $\mathbb{I}_{k}$ and $P^{-1}$ are
$(n_{1},\cdots,n_{r})$-UTB tensors of all the three types, so by Theorem 4.1,
the product $\mathbb{B}=\mathbb{I}_{k}P^{-1}$ is an $(n_{1},\cdots,n_{r})$-UTB
tensor of all the three types.

Now $\mathbb{B}$ is the left $k$-inverse of $\mathbb{A}$. So by using Theorem
4.1 again, $\mathbb{I}_{(k-1)(m-1)+1}=\mathbb{B}\mathbb{A}$ is an
$(n_{1},\cdots,n_{r})$-UTB tensor, with the diagonal blocks
$\mathbb{B}_{1}\mathbb{A}_{1},\cdots,\mathbb{B}_{r}\mathbb{A}_{r}$. But these
blocks must all be identity tensors (as some diagonal blocks of
$\mathbb{I}_{(k-1)(m-1)+1}$), so each $\mathbb{B}_{i}$ is the (unique) left
$k$-inverse of $\mathbb{A}_{i}\ (i=1,\cdots,r)$. $\square$

###  5.2 The right inverses of the triangular blocked tensors

Now we discuss the right inverses of the three types of UTB tensors. The
following result was obtained by Liu and Li in [12].

###### Lemma 5.3

([12], Theorem 3.4) Let $k\geq 2$ and $\mathbb{A}$ be an order $m$ dimension
$n$ tensor. Then the following three conditions are equivalent:

(1) There exists an nonsingular matrix $Q$ of order $n$, such that
$\mathbb{A}=\mathbb{I}_{m}Q$.

(2) $\mathbb{A}$ has a right 2-inverse.

(3) $\mathbb{A}$ has a right $k$-inverse.

Proof. (2)$\Longleftrightarrow$(3) was already proved in [12] (Theorem 3.4).
For the convenience of the readers, here we use the results on left inverses
to give a direct proof of (1)$\Longleftrightarrow$(3).

(1)$\Longrightarrow$(3): Obviously $Q^{-1}\mathbb{I}_{k}$ is a right
$k$-inverse of $\mathbb{A}=\mathbb{I}_{m}Q$.

(3)$\Longrightarrow$(1): Let $\mathbb{B}$ be a right $k$-inverse of
$\mathbb{A}$. Then $\mathbb{A}$ is the (unique) left $m$-inverse of
$\mathbb{B}$. So by Lemma 5.1 we have $\mathbb{B}=P\mathbb{I}_{k}$, and so
$\mathbb{A}=\mathbb{I}_{m}P^{-1}$ for some nonsingular matrix $P$. $\square$

The following Lemma 5.4 will be used in the proof of Theorem 5.2.

###### Lemma 5.4

Let $Q$ be a matrix of order $n$, $\mathbb{A}=\mathbb{I}_{m}Q$ be an order $m$
dimension $n$, and a first or second or third type $(n_{1},\cdots,n_{r})$-UTB
tensor. Then $Q$ is an $(n_{1},\cdots,n_{r})$-UTB matrix.

Proof. Since $\mathbb{A}$ is a first or second or third type
$(n_{1},\cdots,n_{r})$-UTB tensor, we have $a_{it\cdots t}=0$ for any
$j=2,\cdots,r$, $i\in I_{j}$ and $t\leq S_{j-1}$.

On the other hand, by the condition $\mathbb{A}=\mathbb{I}_{m}Q$ and the
definition of the tensor product we also have:

$a_{it\cdots t}=\sum_{j_{2},\cdots,j_{m}=1}^{n}\delta_{ij_{2}\cdots
j_{m}}q_{j_{2}t}\cdots q_{j_{m}t}=q_{it}^{m-1}.$

Thus we have $q_{it}^{m-1}=a_{it\cdots t}=0$ for all $j=2,\cdots,r$, $i\in
I_{j}$ and $t\leq S_{j-1}$. So by definition $Q$ is an
$(n_{1},\cdots,n_{r})$-UTB matrix. $\square$

The following theorem and its proof hold for all three types of UTB tensors
(since Lemma 5.4 holds for all three types of UTB tensors).

###### Theorem 5.2

Let $k\geq 2$, and $\mathbb{A}$ be an order $m$ dimension $n$, the first or
second or third type $(n_{1},\cdots,n_{r})$-UTB tensor with diagonal blocks
$\mathbb{A}_{1},\cdots,\mathbb{A}_{r}$, and $\mathbb{A}$ has a right
$k$-inverse. Then any right $k$-inverse $\mathbb{B}$ of $\mathbb{A}$ is an
$(n_{1},\cdots,n_{r})$-UTB tensor of all the three types, with the diagonal
blocks $\mathbb{B}_{1},\cdots,\mathbb{B}_{r}$, where $\mathbb{B}_{i}$ is a
right $k$-inverse of $\mathbb{A}_{i}\ (i=1,\cdots,r)$.

Proof. Since $\mathbb{B}$ is a right $k$-inverse $\mathbb{A}$, $\mathbb{B}$
has a left $m$-inverse $\mathbb{A}$. So by Lemma 5.1 we know that
$\mathbb{B}=P\mathbb{I}_{k}$ for some nonsingular matrix $P$, and in this case
we have $\mathbb{A}=\mathbb{I}_{m}P^{-1}$.

By hypothesis $\mathbb{A}$ is a first or second or third type
$(n_{1},\cdots,n_{r})$-UTB tensor, so by Lemma 5.4 we know that $P^{-1}$ (and
thus $P$) is an $(n_{1},\cdots,n_{r})$-UTB matrix.

Now both $P$ and $\mathbb{I}_{k}$ are the $(n_{1},\cdots,n_{r})$-UTB tensors
of all the three types, so by Theorem 4.1 we know that their product
$\mathbb{B}$ is also an $(n_{1},\cdots,n_{r})$-UTB tensor of all the three
types.

Also $\mathbb{B}$ is a right $k$-inverse of $\mathbb{A}$ with the diagonal
blocks $\mathbb{B}_{1},\cdots,\mathbb{B}_{r}$. So by using Theorem 4.1 again,
$\mathbb{I}_{(k-1)(m-1)+1}=\mathbb{A}\mathbb{B}$ is an
$(n_{1},\cdots,n_{r})$-UTB tensor, with the diagonal blocks
$\mathbb{A}_{1}\mathbb{B}_{1},\cdots,\mathbb{A}_{r}\mathbb{B}_{r}$. But these
blocks must all be identity tensors (as some diagonal blocks of
$\mathbb{I}_{(k-1)(m-1)+1}$), so each $\mathbb{B}_{i}$ is a right $k$-inverse
of $\mathbb{A}_{i}\ (i=1,\cdots,r)$. $\square$

###  5.3 The left $k$-inverses of the normal $(n_{1},\cdots,n_{r})$-UTB
nonsingular $M$-tensors

In [6], Ding, Qi and Wei generalized the concept of $M$-matrices to
$M$-tensors.

A real tensor $\mathbb{A}$ is called a $Z$-tensor, if all of its off-diagonal
entries are non-positive. Equivalently, $\mathbb{A}$ is a $Z$-tensor if and
only if $\mathbb{A}$ can be expressed as $\mathbb{A}=s\mathbb{I}-\mathbb{B}$,
where $\mathbb{B}$ is a nonnegative tensor, and $\mathbb{I}$ is the unit
tensor.

A $Z$-tensor $\mathbb{A}=s\mathbb{I}-\mathbb{B}$ is called an $M$-tensor, if
$s\geq\rho(\mathbb{B})$; And is called an nonsingular $M$-tensor, if
$s>\rho(\mathbb{B})$.

It is well-known in matrix theory ([1]) that if $A$ is an irreducible
nonsingular $M$-matrix, then $A^{-1}$ is a positive matrix.

An $(n_{1},\cdots,n_{r})$-UTB tensor $\mathbb{A}$ of the first or second type
is called the first or second type normal, if all its diagonal blocks
$\mathbb{A}_{1},\cdots,\mathbb{A}_{r}$ are weakly irreducible. An
$(n_{1},\cdots,n_{r})$-UTB tensor $\mathbb{A}$ of the third type is called the
third type normal, if all its diagonal blocks
$\mathbb{A}_{1},\cdots,\mathbb{A}_{r}$ are irreducible (also see Definitions
6.1 and 6.2 in §6).

In this subsection, we will mainly prove the following Theorem 5.4.

Theorem 5.4 If $\mathbb{A}$ is an order $m$ dimension $n$ and the first or
second or third type normal $(n_{1},\cdots,n_{r})$-UTB nonsingular $M$-tensor,
and $\mathbb{A}$ has a left $k$-inverse $\mathbb{B}$. Then $\mathbb{B}$ is an
$(n_{1},\cdots,n_{r})$-UTB tensor of all the three types all of whose diagonal
blocks $\mathbb{B}_{1},\cdots,\mathbb{B}_{r}$ are positive tensors.

In order to prove Theorem 5.4, we first need to prove some preliminary results
(The following Lemma 5.5 and Theorem 5.3).

###### Lemma 5.5

If an order $m$ dimension $n$ row diagonal tensor
$\mathbb{A}=P\mathbb{I}_{m}$, where $P$ is a matrix of order $n$. Then we
have:

(1)
$\det(\lambda\mathbb{I}_{m}-\mathbb{A})=(\det(\lambda\mathbb{I}_{2}-P))^{(m-1)^{n-1}}$.

(2) $\mathbb{A}$ is a $Z$-tensor if and only if $P$ is a $Z$-matrix.

(3) $\mathbb{A}$ is weakly irreducible if and only if $P$ is irreducible.

(4) $\mathbb{A}$ is an $M$-tensor if and only if $P$ is an $M$-matrix.

(5) $\mathbb{A}$ is a nonsingular $M$-tensor if and only if $P$ is a
nonsingular $M$-matrix.

Proof. (1). We have

$\lambda\mathbb{I}_{m}-\mathbb{A}=\lambda\mathbb{I}_{2}\mathbb{I}_{m}-P\mathbb{I}_{m}=(\lambda\mathbb{I}_{2}-P)\mathbb{I}_{m}.$

Thus by the formula of the determinants of the product of tensors ([16]) we
have

$\det(\lambda\mathbb{I}_{m}-\mathbb{A})=(\det(\lambda\mathbb{I}_{2}-P))^{(m-1)^{n-1}}(\det\mathbb{I}_{m})=(\det(\lambda\mathbb{I}_{2}-P))^{(m-1)^{n-1}}.$

(2). Since $\mathbb{A}$ is row diagonal, we have $a_{ii_{2}\cdots i_{m}}=0$ if
$i_{2},\cdots,i_{m}$ are not all equal. Thus we have:

$\mathbb{A}\ \mbox{is a $Z$-tensor}\Longleftrightarrow a_{ij\cdots j}\leq 0\
(\forall i\neq j)\Longleftrightarrow p_{ij}\leq 0\ (\forall i\neq
j)\Longleftrightarrow P\ \mbox{ is a $Z$-matrix}.$

(3). By Proposition 5.1 we know that $P=M(\mathbb{A})$ is the majorization
matrix of $\mathbb{A}$.

On the other hand, let $G(|\mathbb{A}|)$ be the “representation matrix” of
$|\mathbb{A}|$ defined as ([17]):

$(G(|\mathbb{A}|))_{ij}=\sum_{\\{i_{2},\cdots,i_{m}\\}\ni j}|a_{ii_{2}\cdots
i_{m}}|.$ (5.2)

Then by the hypothesis that $\mathbb{A}$ is row diagonal we can verify that

$(G(|\mathbb{A}|))_{ij}=\sum_{\\{i_{2},\cdots,i_{m}\\}\ni j}|a_{ii_{2}\cdots
i_{m}}|=|a_{ij\cdots j}|=(|M(\mathbb{A})|)_{ij}.$ (5.3)

Thus we have $G(|\mathbb{A}|)=|M(\mathbb{A})|$. So by [17] (Definitions 2.1
and 2.5) we have

$\mathbb{A}\ \mbox{is weakly irreducible}\Longleftrightarrow|\mathbb{A}|\
\mbox{is weakly irreducible}\Longleftrightarrow G(|\mathbb{A}|)\ \mbox{is
irreducible}$

$\Longleftrightarrow|M(\mathbb{A})|$ is irreducible $\Longleftrightarrow
M(\mathbb{A})$ is irreducible

$\Longleftrightarrow P$ is irreducible.

(4). By (1) we have that

$\mbox{The real parts of all eigenvalues of $\mathbb{A}$ are}\ \geq
0\Longleftrightarrow\mbox{The real parts of all eigenvalues of $P$ are}\ \geq
0.$

So we have

$\displaystyle\mathbb{A}\ \mbox{is an $M$-tensor}$
$\displaystyle\Longleftrightarrow$ $\displaystyle\mathbb{A}\ \mbox{is a
$Z$-tensor and the real parts of all eigenvalues of $\mathbb{A}$ are}\ \geq 0$
$\displaystyle\Longleftrightarrow$ $\displaystyle P\ \mbox{is a $Z$-matrix and
the real parts of all eigenvalues of $P$ are}\ \geq 0$
$\displaystyle\Longleftrightarrow$ $\displaystyle P\ \mbox{is an $M$-matrix}.$

(5). Replace all “$\geq 0$” by “$>0$”, and “$M$-tensor” by “nonsingular
$M$-tensor” in the proof of (4). $\square$

From Lemma 5.5 we can obtain the following result which is a generalization of
the corresponding result for irreducible nonsingular $M$-matrices.

###### Theorem 5.3

Let $\mathbb{A}$ be an order $m$ dimension $n$ weakly irreducible nonsingular
$M$-tensor, and $\mathbb{A}$ has a left $k$-inverse $\mathbb{B}$. Then
$\mathbb{B}>0$.

Proof. Since $\mathbb{A}$ has a left $k$-inverse, we have that
$\mathbb{A}=P\mathbb{I}_{m}$ (for some matrix $P$) is row diagonal by Lemma
5.1. Thus by (3) and (5) of Lemma 5.5, $P$ is an irreducible nonsingular
$M$-matrix. By the result of $M$-matrices ([1]), we have $P^{-1}>0$.

On the other hand, we know that the left $k$-inverse
$\mathbb{B}=\mathbb{I}_{k}P^{-1}$. Write $Q=P^{-1}$, then by the definition of
the tensor product we have

$b_{ii_{2}\cdots i_{k}}=\sum_{j_{2},\cdots,j_{k}=1}^{n}\delta_{ij_{2}\cdots
j_{k}}q_{j_{2}i_{2}}\cdots q_{j_{k}i_{k}}=q_{ii_{2}}\cdots q_{ii_{k}}>0.$

Thus we have $\mathbb{B}=\mathbb{I}_{k}P^{-1}>0$. $\square$

Now we are ready to prove Theorem 5.4.

Proof of Theorem 5.4: Let the diagonal blocks of $\mathbb{A}$ be
$\mathbb{A}_{1},\cdots,\mathbb{A}_{r}$ (all of which are weakly irreducible by
hypothesis).

Firstly, by Theorem 5.1 we know that $\mathbb{A}$ has a left $k$-inverse
implying that each $\mathbb{A}_{i}$ also has a left $k$-inverse.

Secondly, we show that $\mathbb{A}$ is a nonsingular $M$-tensor implying that
each $\mathbb{A}_{i}$ is also a nonsingular $M$-tensor:

Write $\mathbb{A}=s\mathbb{I}_{m}-\mathbb{C}$, where $\mathbb{C}\geq 0$ and
$s>\rho(\mathbb{C})$. Then $\mathbb{C}=s\mathbb{I}_{m}-\mathbb{A}$ is an
$(n_{1},\cdots,n_{r})$-UTB tensor with the diagonal blocks, say
$\mathbb{C}_{1},\cdots,\mathbb{C}_{r}$. Thus we have

$\mathbb{A}_{i}=s\mathbb{I}_{m}(n_{i})-\mathbb{C}_{i}\qquad(i=1,\cdots,r),$

where $\mathbb{I}_{m}(n_{i})$ is the unit tensor of order $m$ and dimension
$n_{i}$.

Now $s>\rho(\mathbb{C})=\max_{1\leq i\leq r}\rho(\mathbb{C}_{i})$ implying
that each $\mathbb{A}_{i}$ is also a nonsingular $M$-tensor.

Thus we see that this weakly irreducible $\mathbb{A}_{i}$ satisfies all the
conditions of Theorem 5.3. So by Theorem 5.3 we conclude that its left
$k$-inverse $\mathbb{B}_{i}>0\ (i=1,\cdots,r)$.

##  6 The reducible and weakly reducible normal form of tensors

In this section, we study the reducible and weakly reducible normal forms of
tensors. Since all the problems and results considered in this section depend
only on the zero-nonzero pattern of the tensors, we may assume without loss of
generality that all the tensors considered in this section are nonnegative
tensors.

First we recall some definitions and results in the matrix case.

An $(n_{1},\cdots,n_{r})$-UTB matrix $A$ is called normal (or canonical), if
all of whose diagonal blocks are irreducible. This concept can be generalized
to tensors as follows.

###### Definition 6.1

An $(n_{1},\cdots,n_{r})$-UTB (or 2ndUTB) tensor $\mathbb{A}$ is called a
first type (or second type) (weakly reducible) normal tensor, if all of whose
diagonal blocks are weakly irreducible.

###### Definition 6.2

An $(n_{1},\cdots,n_{r})$-3rdUTB tensor $\mathbb{A}$ is called a third type
(reducible) normal tensor, if all of whose diagonal blocks are irreducible.

In the matrix case, we have the following two well-known results about the
existence and uniqueness of the normal form of a matrix of order $n$.

###### Theorem 6.1

([2]) Every matrix $A$ of order $n$ is permutational similar to some normal
upper (or lower) triangular blocked matrix.

###### Theorem 6.2

([2]) Let $A$ be an $(n_{1},\cdots,n_{r})$ normal upper triangular blocked
matrix of order $n$ with diagonal blocks $A_{1},\cdots,A_{r}$, and $B$ be an
$(m_{1},\cdots,m_{t})$ normal upper triangular blocked matrix of order $n$
with diagonal blocks $B_{1},\cdots,B_{t}$, respectively. If $A$ and $B$ are
permutational similar, then $r=t$, and there exist some permutation $\sigma$:
$[r]\rightarrow[r]$, such that $A_{i}$ and $B_{\sigma(i)}$ are permutational
similar for $i=1,\cdots,r$.

Shao et al in [16] and Hu et al in [9] obtained that every order $m$ dimension
$n$ tensor is permutation similar to some normal 2ndUTB tensor (also see
Proposition 1 of [10]), thus generalized Theorem 6.1 from matrices to the
second type normal upper triangular blocked tensors. In this section, we will
show in Theorem 6.3 that Theorem 6.1 can also be generalized to the third type
normal upper triangular blocked tensors. We also give an example to show that
not every tensor can be permutational similar to some first type normal upper
triangular blocked tensor.

First we have the following lemma.

###### Lemma 6.1

(1) If an order $m$ dimension $n$ tensor $\mathbb{A}$ is $I$-weakly reducible
with $|I|=n-k\ (1\leq k\leq n-1)$. Then $\mathbb{A}$ is permutation similar to
some $(k,n-k)$-UTB (and thus 2ndUTB by Remark 2.1) tensor.

(2) If $\mathbb{A}$ is $I$-reducible with $|I|=n-k\ (1\leq k\leq n-1)$. Then
$\mathbb{A}$ is permutation similar to some $(k,n-k)$-3rdUTB tensor.

Proof. Take the permutation $\sigma:[n]\rightarrow[n]$, such that
$\sigma(I)=\\{k+1,\cdots,n\\}$. Take the permutation matrix $P=P_{\sigma}$.
Then the tensor $P\mathbb{A}P^{T}$ is a $(k,n-k)$-UTB (or 3rdUTB) tensor which
is permutation similar to $\mathbb{A}$. $\square$

###### Theorem 6.3

Every order $m$ dimension $n$ tensor $\mathbb{A}$ is permutation similar to
some normal 3rdUTB tensor.

Proof. We use induction on $n$. If $\mathbb{A}$ is irreducible, then
$\mathbb{A}$ itself is already in the required form. If $\mathbb{A}$ is
reducible, then by Lemma 6.1 there exists some $k$ with $1\leq k\leq n-1$ such
that $\mathbb{A}$ is permutation similar to some $(k,n-k)$-3rdUTB tensor
$\mathbb{B}$ with two diagonal blocks $\mathbb{B}_{1}$ and $\mathbb{B}_{2}$.

(The following blocked forms for the matrix cases illustrate the ideas of the
proof.)

$\mathbb{A}\sim\mathbb{B}=\left(\begin{array}[]{cc}B_{1}&\ast\\\ O&B_{2}\\\
\end{array}\right)$

By induction we know that $\mathbb{B}_{1}$ and $\mathbb{B}_{2}$ are
respectively permutation similar to some normal 3rdUTB tensors $\mathbb{C}$
and $\mathbb{D}$ all of whose diagonal blocks
$\mathbb{C}_{1},\cdots,\mathbb{C}_{p}$ and
$\mathbb{D}_{1},\cdots,\mathbb{D}_{q}$ are irreducible.

$\mathbb{B}_{1}\sim\mathbb{C}=\left(\begin{array}[]{ccc}\mathbb{C}_{1}&\ldots&\ast\\\
\vdots&\ddots&\vdots\\\ O&\ldots&\mathbb{C}_{p}\\\ \end{array}\right),\ \
\mathbb{B}_{2}\sim\mathbb{D}=\left(\begin{array}[]{ccc}\mathbb{D}_{1}&\ldots&\ast\\\
\vdots&\ddots&\vdots\\\ O&\ldots&\mathbb{D}_{q}\\\
\end{array}\right)\Longrightarrow\mathbb{B}\sim\left(\begin{array}[]{cc}\mathbb{C}&\ast\\\
O&\mathbb{D}\\\ \end{array}\right):=\mathbb{A}^{*},$

where $\mathbb{A}^{*}$ defined above is a $(k,n-k)$-3rdUTB tensor with two
diagonal blocks $\mathbb{C}$ and $\mathbb{D}$. By Theorem 2.1 we know that the
tensor $\mathbb{A}^{*}$ is a 3rdUTB tensor with the diagonal blocks
$\mathbb{C}_{1},\cdots,\mathbb{C}_{p}$ and
$\mathbb{D}_{1},\cdots,\mathbb{D}_{q}$ (they are all irreducible), so
$\mathbb{A}^{*}$ is a normal 3rdUTB tensor. But we also have
$\mathbb{A}\sim\mathbb{B}\sim\mathbb{A}^{*}$, thus we obtained the desired
result. $\square$

Using the proof similar to that of Theorem 6.3, we can also prove the
following result for the second type weakly reducible normal form which was
also obtained by Shao et al in [16] and Hu et al in [9].

###### Theorem 6.4

Every order $m$ dimension $n$ tensor $\mathbb{A}$ is permutation similar to
some normal 2ndUTB tensor.

The following example shows that not every tensor can be permutational similar
to some first type normal upper triangular blocked tensor.

###### Example 6.1

Let $\mathbb{A}$ be a $(0,1)$ tensor of order $m=3$ and dimension $n=4$ with
the entries

$a_{ijk}=\left\\{\begin{array}[]{cc}1,&\mbox{ if }1\leq i\leq 3,\
4\in\\{j,k\\};\\\ 0,&\mbox{otherwise.}\\\ \end{array}\right.$

Then $\mathbb{A}$ is not permutational similar to any first type normal upper
triangular blocked tensor.

Proof. Firstly, $\mathbb{A}$ is weakly reducible since the fourth row of
$\mathbb{A}$ is a zero row. Now if $\mathbb{A}$ is $\sigma$-permutational
similar to some first type normal $(n_{1},\cdots,n_{r})$-upper triangular
blocked tensor $\mathbb{B}$ with diagonal blocks
$\mathbb{B}_{1},\cdots,\mathbb{B}_{r}$, then $r\geq 2$. By Lemma 2.1 we know
that $\mathbb{B}$ satisfies the following two conditions (write
$I_{1}=[n_{1}]$):

(i) $\mathbb{B}[I_{1}]=\mathbb{B}_{1}$ is weakly irreducible.

(ii) $\mathbb{B}$ is $\overline{I_{1}}$-weakly reducible.

Now take $I=\sigma(\overline{I_{1}})$, then we see that $I$ is a proper subset
of $[n]$ satisfying the following two conditions:

(1) $\mathbb{A}[\overline{I}]$ is weakly irreducible.

(2) $\mathbb{A}$ is $I$-weakly reducible.

Now we consider the following two cases.

Case 1: $4\notin I$.

Take any $i\in I$. Then $i\neq 4$, so by definition we have $a_{i44}=1$,
contradicting (2) that $\mathbb{A}$ is $I$-weakly reducible.

Case 2: $4\in I$.

Subcase 2.1: $|\overline{I}|\geq 2$.

Then $4\in I\Longrightarrow\overline{I}\subseteq\\{1,2,3\\}$. Thus by
definition we have $\mathbb{A}[\overline{I}]=0$, contradicting (1) that
$\mathbb{A}[\overline{I}]$ is weakly irreducible (since $|\overline{I}|\geq
2$).

Subcase 2.2: $|\overline{I}|=1$.

Without loss of generality we may assume that $\overline{I}=\\{1\\}$. Then by
definition we have $a_{214}=1$, contradicting (2) that $\mathbb{A}$ is
$I$-weakly reducible. $\square$

###### Remark 6.1

Notice that we have not generalized the uniqueness of the normal form for
matrices in Theorem 6.2 to all the three types normal UTB tensors. We think
that these possible generalizations could be the problems for further study.

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