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Open Mathematics

formerly Central European Journal of Mathematics

Editor-in-Chief: Gianazza, Ugo / Vespri, Vincenzo


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Volume 16, Issue 1

Issues

Volume 13 (2015)

A note on the three-way generalization of the Jordan canonical form

Lu-Bin Cui
  • Corresponding author
  • School of Mathematics and Information Sciences, Henan Normal University, XinXiang, Henan 453007, China
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/ Ming-Hui Li
Published Online: 2018-08-08 | DOI: https://doi.org/10.1515/math-2018-0078

Abstract

The limit point 𝓧 of an approximating rank-R sequence of a tensor Ƶ can be obtained by fitting a decomposition (S, T, U) ⋅ 𝓖 to Ƶ. The decomposition of the limit point 𝓧 = (S, T, U) ⋅ 𝓖 with 𝓖 = blockdiag(𝓖1, … , 𝓖m) can be seen as a three order generalization of the real Jordan canonical form. The main aim of this paper is to study under what conditions we can turn 𝓖j into canonical form if some of the upper triangular entries of the last three slices of 𝓖j are zeros. In addition, we show how to turn 𝓖j into canonical form under these conditions.

Keywords: Low-rank tensor approximations; Jordan canonical form; Tensor decomposition

MSC 2010: 15A18; 15A69

1 Introduction

A tensor can be regarded as a higher-order generalization of a matrix, which takes the form

A=(ai1,,im)ai1,,imR1i1,,imn

Such a multi-array 𝓐 is said to be an mth-order n-dimensional square real tensor with nm entries ai1,…,im. In this paper, we only consider the case m = 3 and real-valued three-way arrays.

Definition 1.1

([1]). Let 𝓐 be a mth-order n-dimensional tensor. The mode-k matrix (or k-th matrix unfolding) A(k) ∈ ℝn×nm−1 is a matrix containing the element ai1i2im.

When 𝓐 is a 3rd-order n-dimensional tensor, its mode-k matrices are:

A(1)=[A(:,1,1)A(:,n,1)A(:,1,2)A(:,n,2)A(:,n,n)];A(2)=[A(1,:,1)A(1,:,n)A(2,:,1)A(2,:,n)A(n,:,n)];A(3)=[A(1,1,:)A(n,1,:)A(1,2,:)A(n,2,:)A(n,n,:)].

Definition 1.2

([2]). The multilinear rank of an I × J × K array is defined as the triplet(mode-1 rank, mode-2 rank, mode-3 rank). The mode-k rank of a tensor 𝓐 is defined as the rank of mode-k matrix.

Obviously, a three order tensor has 3 mode-k ranks and the different mode-k ranks of tensor are not necessarily the same [3]. In addition, the rank and the mode-k rank of a same tensor are not necessarily equal even though all the mode-k ranks are equal.

Let

SR(I,J,K)={YRI×J×K|rank(Y)R}(1)

Fitting the CP decomposition [4] to Ƶ boils down to solving the following minimization problem:

MinimizeZYsubjecttoYSR(I,J,K)(2)

Hence, we are looking for a best rank-R approximation [5] to Ƶ. In [6], A.Stegeman has shown that such a best rank-R approximation may not exist due to the set SR(I, J, K) not being closed for R ≥ 2. In this case, we are trying to compute the approximation results in diverging rank-1 terms [7]. This phenomenon can be seen as a three-way generalization of approximate diagonalization of a nondiagonalizable matrix. In [6, 8], A.Stegeman has shown that, analogous to the matrix case, the limit point of the approximating rank-R sequence satisfies a three-way generalization of the real Jordan canonical form. [6, 9] show that the limit point 𝓧 is a boundary point of SR(I, J, K) and can be obtained by fitting a decomposition (S, T, U) ⋅ 𝓖 to Ƶ, with 𝓖 = blockdiag(𝓖1, … ,𝓖m) and core block 𝓖j of size dj × dj × dj and in sparse canonical form. The decomposition of 𝓧 has been introduced in [10, 11, 12], where the block terms are (Sj,Tj,Uj) ⋅ 𝓖j. Nondiverging rank-1 terms have an associated core block with dj = 1, and core blocks with dj ≥ 2 are the limit of a group of dj diverging rank-1 terms.

For groups of two, or three, or four diverging rank-1 terms, [6, 8] have shown limit point 𝓧 = j=1m 𝓧j and its decomposition 𝓧 = (S, T, U) ⋅ 𝓖 = j=1m (Sj,Tj,Uj) ⋅ 𝓖j have the following results.

Lemma 1.3

([13]). For a group of dj = 2 diverging rank-1 terms, the limit 𝓧j can be written as 𝓧j = (Sj, Tj, Uj) ⋅ 𝓖j with Sj, Tj, Uj of rank 2, and 2 × 2 × 2 array 𝓖j given by

[10010100].(3)

we have rank(𝓖j) = 3. Here, we denote the 2 × 2 × 2 array 𝓖j with 2 × 2 slices G1 and G2 as [G1 |G2]. (3) is referred to as the canonical form of a boundary array of S2(2, 2, 2).

Lemma 1.4

([6]). For a group of dj = 3 diverging rank-1 terms, and min(I, J, K) ≥ 3, almost all limits 𝓧j with multilinear rank(3, 3, 3) can be written as 𝓧j = (Sj, Tj, Uj) ⋅ 𝓖j with Sj, Tj, Uj of rank 3, and 3 × 3 × 3 array 𝓖j given by

[1000000101000000001000000],(4)

where * denotes a nonzero entry. We have rank(𝓧j) = rank(𝓖j) = 5.

Lemma 1.5

([8]). For a group of dj = 4 diverging rank-1 terms, and min(I, J, K) ≥ 4, almost all limits 𝓧j with multilinear rank(4, 4, 4) can be written as 𝓧j = (Sj, Tj, Uj) ⋅ 𝓖j with Sj, Tj, Uj of rank 4, and 4 × 4 × 4 array 𝓖j given by

[10000000000001010000000000000010000000000000001000000000000],(5)

where * denotes a nonzero entry. We have rank(𝓧j) = rank(𝓖j) ≥ 7.

Remark 1.6

The proof of Lemma 1.5 in [8] has shown that 𝓖j has multilinear rank (4, 4, 4).

However, the proof of Lemma 1.5 in [8] does not take account of the cases that some of the upper triangular entries of the last three slices of 𝓖j are zeros. To make up for this defect, we assume that some of the upper triangular entries of the last three slices of 𝓖j are zeros and study whether we can turn 𝓖j into the canonical form (5). Firstly, we consider the following two examples.

Example 1.7

Let 𝓖j be a 4 × 4 × 4 array that satisfies the conditions of Lemma 1.5, and the (1, 4) entries of the last three slices are equal to zeros, i.e.,

Gj=[1000312032903316001000325034110361700100033003600390001000300030003].

By subtracting 3 times the first slice of 𝓖j from slice 2,3,4, then we obtain an all-zero diagonal in slice 2,3,4. Similarly, by subtracting 2 times the second slice from the third slice and 3 times the second slice from the fourth slice, 𝓖j is of the form

Gj=[10000120005000100010000250001000200100003000000000001000000000000].

Next, by subtracting 2 times the third slice from the fourth slice, we can turn the (1,3) and (2,4) entries of slice four into zeros. By subtracting 5 times the third slice from the second slice, we can turn the (2, 4) entry of slice two into zero. Then we obtain the following form

Gj=[10000123000500000010000200001000000100003000000000001000000000000].

In each slice of the above 𝓖j, we add 23/2 times row 2 to row 1 and subtract 23/2 times column 1 from column 2. Then we obtain the following form

Gj=[1000010000523/20000010000200001000000100003000000000001000000000000].

It is apparent from this example that if some of the upper triangular entries of the last three slices of 𝓖j are zeros, we can’t turn 𝓖j into canonical form (5).

Example 1.8

Let 𝓖j be a 4 × 4 × 4 array that satisfies the conditions of Lemma 1.5, and the (1, 4) entries of the second and the third slices are equal to zeros, i.e.,

Gj=[1000312032903316501000325034110361700100033003600390001000300030003].

Using the same method as Example 1.7, we can obtain

Gj=[1000010000523/20005010000200001000000100003000000000001000000000000].

Now, the fourth slice only has its (1,4) entry nonzero, we normalize it to one. Then by subtracting the 23/2 times the fourth slice from the third slice, the (1,4) entry of the third slice can be turned into zero. Then we obtain

Gj=[1000010000500001010000200001000000100003000000000001000000000000].

According to this example, we see that if some of the upper triangular entries of the last three slices of 𝓖j are zeros, we can turn 𝓖j into canonical form (5).

A natural question is under what conditions we can turn 𝓖j into canonical form (5) if some of the upper triangular entries of the last three slices of 𝓖j are zeros. The answer to this question is the main contribution of this paper.

Remark 1.9

It is worth noting that if some of the upper triangular entries of the last three slices of 𝓖j are zeros, the entry * of canonical form (5) may be zero. Therefore, in this paper, we mainly consider the following canonical form

[10000e¯0000h¯00001010000f¯0000i¯00000010000g¯000000000001000000000000],(6)

where ē, , , , ī may be zero, and there must be at least one nonzero entry in every slice.

Now, we prove why we require that there must be at least one nonzero entry in every slice. According to the the definition of the mode-k matrix of a tensor, (5) is actually the mode-1 matrix of 𝓖j. Because the first slice of 𝓖j is an identity matrix, it follows that the mode-1 rank of 𝓖j is always 4, no matter what values ē, , , , ī take.

The mode-2 matrix of 𝓖j is

[10000000000000000e¯0010000000000000h¯00f¯0010000000000100i¯00g¯001000].

Since each row of this matrix has a nonzero entry 1, it follows that mode-2 rank of 𝓖j is always 4, no matter what values ē, , , , ī take.

The mode-3 matrix of 𝓖j is

[10000100001000010000e¯0000f¯0000g¯000000000h¯0000i¯000000000000001000].

For this matrix, if there exists a row that its elements are all zeros, then the mode-3 rank of 𝓖j does not equal to 4. This contradicts the fact that the mode-3 rank of 𝓖j is 4. Consequently, ē, , can’t be zero at the same time, and , ī can’t be zero at the same time. This implies that there must be at least one nonzero entry in every slice.

Definition 1.10

For a matrix A=(a11a12a13a14a21a22a23a24a31a32a33a34a41a42a43a44), we call the main diagonal elements of matrix A as the 1st diagonal elements , and a12, a23, a34 as the 2nd diagonal elements, and a13, a24 as the 3rd diagonal elements, and a14 as the 4th diagonal element. The kth diagonal element of A is called zero if the elements of the kth diagonal are all zeros. The kth diagonal elements of A is called nonzero if the kth diagonal of A has a nonzero element.

The remaining of this paper is organized as follows. In section 2, we study some properties related to 𝓖j. In section 3, we discuss under what conditions can we turn the 2nd diagonal elements of any two slices of the last three slices of 𝓖j into zeros. In section 4, based on the results of the third section, we analyze under what conditions can we turn the 3rd and 4th diagonal elements of the last three slices of 𝓖j into zeros. Finally, some concluding remarks are given in section 5.

2 Some properties related to 𝓖j

Before we discuss what conditions we need to turn 𝓖j into canonical form (6), we first give some properties related to 𝓖j.

Property 2.1

If 𝓖j satisfies the conditions of Lemma 1.5, then it can be turned into

[1000a2e2h2j2a3e3h3j3a4e4h4j401000b2f2i20b3f3i30b4f4i4001000c2g200c3g300c4g40001000d2000d3000d4],(7)

and we can obtain the following three conclusions: (1) ap = bp = cp = dp (p = 2, 3, 4) hold for almost all 𝓖j;

(2) The equations

e3f2=f3e2,f3g2=g3f2e4f2=f4e2,f4g2=g4f2e4f3=f4e3,f4g3=g4f3(8)

hold for almost all 𝓖j; (3) The equations

e3i2e2i3+h3g2g3h2=0e4i2e2i4+h4g2g4h2=0e4i3e3i4+h4g3g4h3=0(9)

holds for almost all 𝓖j.

Proof

The proof of Lemma 3.3 in [8] has shown that 𝓖j can be turned into (7) if it satisfies the conditions of Lemma 3.3.

The first conclusion has been proved in Lemma 3.3 of [8].

Now we show the proof of the second conclusion. Similarly to the proof of the vectors, (ep, fp, gp) are proportional for p = 2,3, 4 in Lemma 3.3 of the [8]. We write A(n) in terms of p = 3 and compute Y2(n) = A(n)C2(n)(A(n))−1, which yields matrix (A.10). The entries in this matrix equal those of Y2(n) in (A.2). It follows that e3f2 = f3e2,f3g2 = g3f2. When we write A(n) in terms of p = 4 and compute Y2(n) = A(n)C2(n)(A(n))−1, this yields a matrix similar with (A.10). The entries in this matrix equal those of Y3(n) in (A.2). It follows that e4f2 = f4e2,f4g2 = g4f2. Analogously, when writing A(n) in terms of p = 3 and compute Y4(n) = A(n)C4(n)(A(n))−1, we obtain that e4f3 = f4e3,f4g3 = g4f3.

Now we prove the third conclusion. Similarly to the proof of the vectors (h3α h2,i3α i2) and (h4β h2, i4β i2) are proportional for α = e3/e2,β = e4/e2 in Lemma 3.3 of the [8], we write A(n) in terms of p = 3 and compute Y2(n) = A(n)C2(n)(A(n))−1, which yields matrix (A.10). The entries in this matrix equal those of Y2(n) in (A.2). It follows that e3i2e2i3+h3g2g3h2 = 0. We write A(n) in terms of p = 4 and compute Y2(n) = A(n)C2(n)(A(n))−1, which yields a matrix similar with (A.10). The entries in this matrix equal those of Y3(n) in (A.2). It follows that e4i2e2i4+h4g2g4h2 = 0. Analogously, when writing A(n) in terms of p = 3 and compute Y4(n) = A(n)C4(n)(A(n))−1, we obtain that e4i3e3i4+h4g3g4h3 = 0.

Remark 2.2

According to the first conclusion of Property 2.1, if 𝓖j satisfies the conditions of Lemma 1.5, by subtracting ap times the first slice of 𝓖j from slice p (p = 2, 3, 4), then it can be further turned into

[10000e2h2j20e3h3j30e4h4j4010000f2i200f3i300f4i40010000g2000g3000g40001000000000000].

Remark 2.3

According to the second conclusion of the Property 2.1, if f2f3f4 ≠ 0, the vectors (ep, fp, gp) (p = 2, 3, 4) are proportionate. If f2 = f3 = f4 = 0, the vectors (ep, fp, gp) (p = 2, 3, 4) are disproportionate.

By the first conclusion of Property 2.1, we have turned the first diagonal elements of the last three slices of 𝓖j into zeros. In the next section, we mainly consider turning the 2nd diagonal elements of any two slices of the last three slices of 𝓖j into zeros. Noting that if f2f3f4 ≠ 0, the vectors (ep, fp, gp) (p = 2, 3, 4) are proportionate. This implies that we can directly turn the 2nd diagonal elements of any two slices of the last three slices of 𝓖j into zeros if f2f3f4 ≠ 0. However, in this paper we assume some of the upper triangular entries of the last three slices of 𝓖j are zeros. This means the vectors (ep, fp, gp) (p = 2, 3, 4) may not be proportionate. Consequently, we first discuss all the combinations of ep, fp, gp (p = 2, 3, 4) when some of them are equal to zeros. On the other hand, noting that ep, fp, gp (p = 2, 3, 4) satisfies equations (8), it is easy to get the following conclusion.

Property 2.4

There are 91 combinations of ep, fp, gp (p = 2, 3, 4) when some of them are equal to zeros and satisfy the equations (8).

Proof

We traverse ep, fp, gp (p = 2, 3, 4) based on the number of nonzero entries, and remove some of the combinations that do not meet the conditions (8).

If one of ep, fp, gp (p = 2, 3, 4) is not equal to zero, and the remaining eight of them are equal to zeros, this yields 9 combinations. For convenience, we only write nonzero entry in each combination, i.e., e2, e3, e4, f2, f3, f4, g2, g3, g4. For example, e2 represents e2 ≠ 0,e3 = e4 = f2 = f3 = f4 = g2 = g3 = g4 = 0.

If two of ep, fp, gp (p = 2, 3, 4) are not equal to zeros, and the remaining seven of them are equal to zeros, this yields C92 combinations. However, some combinations do not satisfy conditions (8). For example, combination e2≠ 0, f3≠ 0,e3 = e4 = f2 = f4 = g2 = g3 = g4 = 0 contradicts the condition f3e2 = e3f2 of (8). Thus, we remove this combination. Similarly, we remove other combinations that do not meet conditions (8). Finally, by traversing we obtain the following 24 combinations that satisfy conditions (8): e2e3, e2e4, e3e4, f2f3, f2f4, f3f4, g2g3, g2g4, g3g4, e2f2, e2g2, f2g2, e3f3, e3g3, f3g3, e4f4, e4g4, f4g4, e2g3, e2g4, e3g2, e3g4, e4g2, e4g3.

Analogously, if three of ep, fp, gp (p = 2, 3, 4) are not equal to zeros, and the remaining six of them are equal to zeros, there are 24 combinations that satisfy conditions (8), i.e., e2e3e4, f2f3f4, g2g3g4, e2f2g2, e3f3g3, e4f4g4, e2g2g3, e2g2g4, e2g3g4, e3g2g3, e3g2g4, e3g3g4, e4g2g3, e4g2g4, e4g3g4, e2e3g2, e2e3g3, e2e3g4, e2e4g2, e2e4g3, e2e4g4, e3e4g2, e3e4g3,e3e4g4.

If four of ep, fp, gp (p = 2, 3, 4) are not equal to zeros, and the remaining five of them are equal to zeros, there are 21 combinations that satisfy conditions (8), i.e., e2e3f2f3, e2e4f2f4, e3e4f3f4, e2e3g2g3, e2e4g2g4, e3e4g3g4, f2f3g2g3, f2f4g2g4, f3f4g3g4, e2e3g2g4, e2e3g3g4, e2e4g2g3, e2e4g3g4, e3e4g2g3, e3e4g2g4, e2g2g3g4, e3g2g3g4, e4g2g3g4, e2e3e4g2, e2e3e4g3, e2e3e4g4.

If five of ep, fp, gp (p = 2, 3, 4) are not equal to zeros, and the remaining four of them are equal to zeros, there are 6 combinations that satisfy conditions (8), i.e., e2e3g2g3g4, e2e4g2g3g4, e3e4g2g3g4, e2e3e4g2g3, e2e3e4g2g4, e2e3e4g3g4.

If six of ep, fp, gp (p = 2, 3, 4) are not equal to zeros, and the remaining three of them are equal to zeros, there are 6 combinations that satisfy conditions (8), i.e., e2e3e4f2f3f4, e2e3e4g2g3g4, f2f3f4g2g3g4, e2f2g2e3f3g3, e2f2g2e4f4g4, e3f3g3e4f4g4.

If seven (or eight) of ep, fp, gp (p = 2, 3, 4) are not equal to zeros, the remaining two (or one) of them are equal to zeros (or zero), there is no combination that satisfies conditions (8).

The last one combination is that ep, fp, gp (p = 2, 3, 4) are all nonzero, i.e., e2e3e4f2f3f4g2g3g4. □

In the next section, we will study in which combinations of the 91 combinations in Property 2.4 we turn the 2nd diagonal elements of any two slices of the last three slices into zeros.

Remark 2.5

Noting that if ep, fp, gp (p = 2, 3, 4) are all zeros, this combination also satisfies the conditions (8). However, another question arises: Under this combination, we can turn 𝓖j into canonical form? Our answer is negative. In fact, if ep, fp, gp (p = 2, 3, 4) are all zeros, then the 2nd diagonal elements of the last three slices of 𝓖j are all zeros. This means in the process of transforming 𝓖j into canonical form, the 2nd diagonal elements of the last three slices of 𝓖j are always zeros. However, in canonical form (6), there exist a slice whose 2nd diagonal elements are nonzero. Therefore, if ep, fp, gp (p = 2, 3, 4) are all zeros, we can’t turn 𝓖j into canonical form.

3 Make the 2nd diagonal of any two slices of the last three slices of 𝓖j zero

In this section we analyze under what conditions we can turn the 2nd diagonal elements of any two slices of the last three slices of 𝓖j into zeros.

It is worth noting that if 2nd diagonal elements of any two slices of the last three slices of 𝓖j can be turned into zeros, then we can continue to analyze the 3rd and 4th diagonal elements. If the 2nd diagonal elements of any two slices of the last three slices of 𝓖j cannot be turned into zeros, it is meaningless to continue to analyze the 3rd and 4th diagonal elements. Therefore, in this section, we only consider the 2nd diagonal elements of the last three slices of 𝓖j.

Another thing we should pay attention to is why we discuss the conditions that turn the 2nd diagonal elements of any two slices of the last three slices of 𝓖j into zeros, instead of the conditions that turn the 2nd diagonal elements of the 3rd and the 4th slices into zeros. In fact, if the 2nd diagonal elements of the second slice and the fourth slice (or the second slice and the third slice) are equal to zeros, we can exchange the second slice and the third slice ( or the second slice and the fourth slice). The specific operation will be discussed in detail in the next section and so is omitted here.

Now, based on the Property 2.4, we discuss under what conditions we can turn 2nd diagonal elements of any two slices of the last slices of 𝓖j into zeros. For convenience of the following discussion, we divide the 91 combinations of Property 2.4 into three categories. We regard each category as a set. Each element of the set represents a combination, which can be represented by the nonzero entries of ep, fp, gp (p = 2, 3, 4). For example, the element e2 of set T1 represents e2 ≠ 0, e3 = e4 = f2 = f3 = f4 = g2 = g3 = g4 = 0. The element e2g3 of set T2 represents e2 ≠ 0,g3 ≠ 0, e3 = e4 = f2 = f3 = f4 = g2 = g4 = 0.

T1={e2,e3,e4,g2,g3,g4,e2e3,e2e4,e3e4,g2g3,g2g4,g3g4,e2f2,e2g2,f2g2,e3f3,e3g3,f3g3,e4f4,e4g4,f4g4,e2e3e4,g2g3g4,e2f2g2,e3f3g3,e4f4g4,e2e3f2f3,e2e4f2f4,e3e4f3f4,f2f3g2g3,f2f4g2g4,f3f4g3g4,e2e3e4f2f3f4,f2f3f4g2g3g4,e2f2g2e3f3g3,e2f2g2e4f4g4,e3f3g3e4f4g4,e2e3e4f2f3f4g2g3g4}T2={e2g3,e2g4,e3g2,e3g4,e4g2,e4g3,e2g2g3,e2g2g4,e2g3g4,e3g2g3,e3g2g4,e3g3g4,e4g2g3,e4g2g4,e4g3g4,e2e3g2,e2e3g3,e2e3g4,e2e4g2,e2e4g3,e2e4g4,e3e4g2,e3e4g3,e3e4g4,e2e3g2g4,e2e3g3g4,e2e4g2g3,e2e4g3g4,e3e4g2g3,e3e4g2g4,e2g2g3g4,e3g2g3g4,e4g2g3g4,e2e3e4g2,e2e3e4g3,e2e3e4g4,e2e3g2g3g4,e2e4g2g3g4,e3e4g2g3g4,e2e3e4g2g3,e2e3e4g2g4,e2e3e4g3g4}T3={f2,f3,f4,f2f3,f2f4,f3f4,f2f3f4,e2e3g2g3,e2e4g2g4,e3e4g3g4,e2e3e4g2g3g4}

Now we show that under each combination of T1 we can turn 2nd diagonal elements of any two slices of the last three slices of 𝓖j into zeros. For each combination in T1, nonzero entries in ep, fp, gp (p = 2, 3, 4) are either in the same slice or in the same position of different slices.

In fact, if nonzero entries are in the same slice, there are just two slices of the last three slices of 𝓖j whose 2nd diagonal elements are zeros. For example, the element e3f3g3 of T1 represents e3 ≠ 0,f3 ≠ 0,g3 ≠ 0,e2 = e4 = f2 = f4 = g2 = g4 = 0. From Remark 2.2, 𝓖j is of the form

[100000h2j20e3h3j300h4j40100000i200f3i3000i400100000000g300000001000000000000].

From the above 𝓖j, we can see that 2nd diagonal elements of the second and fourth slices are zeros.

If nonzero entries lie in the same position of different slices, it will yield two possibilities. The first case is that there exist two slices whose 2nd diagonal elements are nonzero. For these two slices, by subtracting one slice from the other slice we can turn the 2nd diagonal elements of one of them into zeros. After this, we can obtain two slices of the last three slices of 𝓖j whose 2nd diagonal elements are equal to zeros. For example, the element e3e4f3f4 in T1 represents e3 ≠ 0,e4 ≠ 0,f3 ≠ 0,f4 ≠ 0,e2 = f2 = g2 = g3 = g4 = 0. From Remark 2.2, 𝓖j is of the form

[100000h2j20e3h3j30e4h4j40100000i200f3i300f4i400100000000000000001000000000000].

For this 𝓖j, through subtracting e4/e3 times the third slice from the fourth slice (or e3/e4 times the fourth slice from the third slice), we can turn e4 and f4 (or e3 and f3 ) into zeros because e4f3 = f4e3 holds for almost all 𝓖j in (8). Then we get the second and fourth slices (or the second and third slices ) whose 2nd diagonal elements are zeros. The second case is that there exist three slices whose 2nd diagonal elements are nonzero. For these three slices, through subtracting one slice from another two slices, then we can turn the 2nd diagonal elements of two of them into zeros. For example, the element e2e3e4f2f3f4 in T1 represents e2 ≠ 0, e3 ≠ 0, e4 ≠ 0, f2 ≠ 0, f3 ≠ 0, f4 ≠ 0, g2 = g3 = g4 = 0. From Remark 2.2, 𝓖j is of the form

[10000e2h2j20e3h3j30e4h4j4010000f2i200f3i300f4i400100000000000000001000000000000].

For this 𝓖j, through subtracting e3/e2 times the second slice from the third slice and e4/e2 times the second slice from the fourth slice (or e2/e3 times the third slice from the second slice and e4/e3 times the third slice from the fourth slice or e2/e4 times the fourth slice from the second slice and e3/e4 times the fourth slice from the third slice), we can turn e3, f3, e4, f4 ( or e2, f2, e4, f4 or e2, f2, e3, f3 ) into zeros because e3f2 = f3e2, e4f2 = f4e2, e4f3 = f4e3 holds for almost all 𝓖j in (8). Then we get the third and fourth (or the second and fourth or the second and third) slices whose 2nd diagonal elements are zeros.

Now we show that under each combination of T2, we can’t turn 2nd diagonal elements of any two slices of the last three slices of 𝓖j into zeros. For each combination in T2, we find that nonzero entries in ep, fp, gp (p = 2, 3, 4) are in different position of different slices, which results in at least two slices whose 2nd diagonal elements can’t be turned into zeros. For example, the element e2g3 in T2 represents e2 ≠ 0, g3 ≠ 0, e3 = e4 = f2 = f3 = f4 = g2 = g4 = 0. From Remark 2.2, 𝓖j is of the form

[10000e2h2j200h3j300h4j40100000i2000i3000i400100000000g300000001000000000000].

For this 𝓖j, if we can turn e2 into zero, then we get the second and fourth slices whose 2nd diagonal elements are all zeros; if we can turn g3 into zero, then we get the third and fourth slices whose 2nd diagonal elements are all zeros. On the other hand, e2 can be turned into zero only by the (1, 2) entry of the third slice or the (1, 2) entry of the fourth slice; g3 can be turned into zero only by the (3, 4) entry of the second slice or the (3, 4) entry of the fourth slice. However, (1, 2) entries of the third and fourth slices are all zeros; (3, 4) entries of the second and fourth slices are all zeros; so e2 and g3 can’t be turned into zeros. This implies that if nonzero entries are in different position of different slices, we can’t turn any two slices of the last three slices whose 2nd diagonal elements into zeros. Therefore, it makes no sense to continue to analyze the 3rd and 4th diagonal elements. Consequently, in the next section, we no longer consider all the combinations in T2.

For each combination of T3, we have not found the relationship between ep, fp, gp and hp, ip, jp for p = 2, 3, 4. Consequently, in the next section analysis we no longer consider all the combinations in T3.

4 Make the 3rd and 4th diagonal of the last three slices of 𝓖j zero

In this section we analyze under what conditions we can turn the 3rd and the 4th diagonal elements of the last three slices of 𝓖j into zeros.

Our main idea is as follows. For each combination in T1, combining with expression (9), we can obtain the relationship between ep, gp and hp, ip (p = 2, 3, 4). Through the relationship between them, we give conditions that can turn the 3rd and 4th diagonal elements of the last three slices of 𝓖j into zeros. Under these conditions, we can turn 𝓖j into canonical form (6).

Theorem 4.1

If one of ep, fp, gp (p = 2, 3, 4) is not equal to zero, the remaining eight of them are equal to zeros, there are 6 combinations in T1, i.e., ep, gp (p = 2, 3, 4). If ex ≠ 0, the other of ep, fp, gp (p = 2, 3, 4) are equal to zeros, under conditions ix = iy = iz = 0,hyjzjy hz, or if gx ≠ 0, the other of ep, fp, gp (p = 2, 3, 4) are equal to zeros, under conditions hx = hy = hz = 0, jyiziy jz, where x, y, z ∈ {2, 3, 4} and xyz, we can turn 𝓖j into canonical form (6).

Proof

Here, we only consider the combination that e2 ≠ 0, the remaining eight of ep, fp, gp (p = 2, 3, 4) are equal to zeros. The combinations that e3 ≠ 0 (or e4 ≠ 0 or g2 ≠ 0 or g3 ≠ 0 or g4 ≠ 0), the remaining eight of ep, fp, gp (p = 2, 3, 4) are equal to zeros can be proved in a similar way as the combination that e2 ≠ 0, the remaining eight of ep, fp, gp (p = 2, 3, 4) are equal to zeros.

If e2 ≠ 0, e3 = e4 = f2 = f3 = f4 = g2 = g3 = g4 = 0, according to (9), we have i3 = i4 = 0, then 𝓖j is of the form

[10000e2h2j200h3j300h4j40100000i20000000000100000000000000001000000000000].

Because the mode-3 rank of 𝓖j is 4, then h3j4j3h4. Next, we show why the mode-3 rank of 𝓖j is not equal to 4 if h3j4 = j3h4. Suppose h3 = 0, then j3h4 = 0. This yields three possibilities. The first case is j3 = 0,h4 ≠ 0, then all the entries of the third slice are equal to zeros. The second case is j3 ≠ 0,h4 = 0, if j4 = 0, then all the entries of the fourth slice are equal to zeros; if j4 ≠ 0, by subtracting j4/j3 times the third slice from the fourth slice (or j3/j4 times the fourth slice from the third slice), we can turn j4 (or j3) into zero, then all the entries of the fourth (or third) slice are equal to zeros. The third case is j3 = 0,h4 = 0, then all the entries of the third slice are equal to zeros. The situations where we suppose that j4 = 0 or j3 = 0 or h4 = 0 can be dealt with analogously. Suppose h3,j4,j3,h4 are nonzero, by subtracting h4/h3 times the third slice from the fourth slice (or h3/h4 times the fourth slice from the third slice), we can turn h4 and j4 (or h3 and j3) into zeros. Then all the entries of the third slice (or the fourth slice) are equal to zeros. From the above discussion we can see that, if h3j4 = j3h4, there always exists a slice whose entries are all equal to zeros. This implies that the mode-3 rank of 𝓖j is 3. Thus we draw the conclusion that h3j4j3h4.

Now we prove that i2 must be equal to zero, because i2 can be turned into zero only through (2, 3) or (3, 4) entry of the second slice or through (2, 4) entry of the third or (2, 4) entry of the fourth slice. However, (2, 3) and (3, 4) entries of the second slice, (2, 4) entry of the third and fourth slices are all equal to zeros. This means in the process of transforming 𝓖j into canonical form, i2 can’t be turned into zero. Therefore, we must have i2 equal to zero.

Now we prove how we turn 𝓖j into canonical form if e2 ≠ 0,e3 = e4 = f2 = f3 = f4 = g2 = g3 = g4 = 0, under conditions that i2 = i3 = i4 = 0 and h3j4j3h4. In fact, under these conditions, 𝓖j have the following form

[10000e2h2j200h3j300h4j4010000000000000000100000000000000001000000000000].

Firstly, we discuss how to standardize the last two slices. Because h3j4j3h4, suppose h3 = 0, then j3h4 ≠ 0. If j4 = 0, then the third slice only has its (1, 4) entry j3 nonzero, the fourth slice only has its (2, 3) entry h4 nonzero. We normalize them to one. By exchanging the third slice and the fourth slice, then we obtain the following form

[10000e2h2j200100001010000000000000000100000000000000001000000000000].(10)

If j4 ≠ 0, by subtracting j4/j3 times the third slice from the fourth slice, we can turn j4 into zero. After this, the third slice only has j3 is nonzero, the fourth slice only has h4 is nonzero. Similarly, we normalize them to one. By exchanging the third slice and the fourth slice, then we obtain (10). The situations where we suppose that j4 = 0 or j3 = 0 or h4 = 0 can be dealt with analogously.

Next, we discuss how to standardize the second slice. If h2 = j2 = 0, then we have transformed 𝓖j into canonical form. If h2,j2 are nonzero, we can turn them into zeros. The specific method is: h2 can be turned into zero by subtracting h2 times the third slice from the second slice, j2 can be turned into zero by subtracting j2 times the fourth slice from the second slice.

Consequently, if ex ≠ 0, the remaining eight of ep, fp, gp (p = 2, 3, 4) are equal to zeros, under conditions ix = iy = iz = 0 and hyjzjyhz, where x = 2, y = 3, z = 4 or x = 2, y = 4, z = 3, we can turn 𝓖j into canonical form. □

Remark 4.2

In fact, theorem 4.1 contains the following six cases.

Let x = 2, y = 3, z = 4 or x = 2, y = 4, z = 3.

  • Case 1

    If e2 ≠ 0, e3 = e4 = f2 = f3 = f4 = g2 = g3 = g4 = 0, under conditions i2 = i3 = i4 = 0 and h3j4j3 h4, we can turn 𝓖j into canonical form.

  • Case 2

    If g2 ≠ 0, e2 = e3 = e4 = f2 = f3 = f4 = g3 = g4 = 0, under conditions h2 = h3 = h4 = 0 and j3i4i3 j4, we can turn 𝓖j into canonical form.

    Let x = 3, y = 2, z = 4 or x = 3, y = 4, z = 2.

  • Case 3

    If e3 ≠ 0, e2 = e4 = f2 = f3 = f4 = g2 = g3 = g4 = 0, under conditions i2 = i3 = i4 = 0 and h2j4j2 h4, we can turn 𝓖j into canonical form.

  • Case 4

    If g3 ≠ 0, e2 = e3 = e4 = f2 = f3 = f4 = g2 = g4 = 0, under conditions h2 = h3 = h4 = 0 and j2i4i2 j4, we can turn 𝓖j into canonical form.

    Let x = 4, y = 2, z = 3 or x = 4, y = 3, z = 2.

  • Case 5

    If e4 ≠ 0, e2 = e3 = f2 = f3 = f4 = g2 = g3 = g4 = 0, under conditions i2 = i3 = i4 = 0 and h2j3j2 h3, we can turn 𝓖j into canonical form.

  • Case 6

    If g4 ≠ 0, e2 = e3 = e4 = f2 = f3 = f4 = g2 = g3 = 0, under conditions h2 = h3 = h4 = 0 and j2i3i2 j3, we can turn 𝓖j into canonical form.

    For the sake of simplicity, we write the above six cases as follows:

    ex0,theotherarezero,ix=iy=iz=0,hyjzjyhz;gx0,theotherarezero,hx=hy=hz=0,iyjzjyiz;

    where x, y, z ∈ {2, 3, 4} and xyz.

    The following theorem is also written in a similar way, and we don’t repeat it. For example, exey ≠ 0, ix = iy = iz = 0, yjzy hz, where x = 2, y = 3, z = 3, in Theorem 4.3 represents if e2 ≠ 0, e3 ≠ 0, e4 = f2 = f3 = f4 = g2 = g3 = g4 = 0, under conditions i2 = i3 = i4 = 0 and 3j43 h4, can we turn 𝓖j into canonical form.

Theorem 4.3

If two of ep, fp, gp (p = 2, 3, 4) are not equal to zeros, the remaining seven of them are equal to zeros, there are 15 combinations in T1, i.e. e2e3, e2e4, e3e4, g2g3, g2g4, g3g4, e2f2, e3f3, e4f4, f2g2, f3g3, f4g4, e2g2, e3g3, e4g4. For each combinations, we can turn 𝓖j into canonical form under the following conditions:

exey0,ix=iy=iz=0,h~yjzj~yhz;gxgy0,hx=hy=hz=0,i~yjzj~yiz;exfx0,iy=iz=0,hyjzjyhz;fxgx0,hy=hz=0,iyjzjyiz;exgx0,iy=hy=0,izhz0,jy0,hxiz=ixhz;(exgx0,iyhyizhz0,hxiy=ixhy,hyjzjyhz);wherex,y,z{2,3,4}andxyz.

Proof

Firstly, we discuss the combination that e2e3 ≠ 0, the remaining seven of ep, fp, gp (p = 2, 3, 4) are equal to zeros. The combinations e2e4, e3e4, g2g3, g2g4, g3g4 can be similar as in the discussion with e2e3.

If e2e3 ≠ 0, e4 = f2 = f3 = f4 = g2 = g3 = g4 = 0, according to (9), we can obtain i4 = 0 and e2i3 = e3i2. Because e2e3 ≠ 0, then i2 and i3 are zero or nonzero at the same time. This yields two possibilities.

The first case is i2 = i3 = 0. Then 𝓖j is of the form

[10000e2h2j20e3h3j300h4j4010000000000000000100000000000000001000000000000].

By subtracting e3/e2 times the second slice from the third slice, we can turn e3 into zero. Then we obtain the following form:

[10000e2h2j200h~3j~300h4j4010000000000000000100000000000000001000000000000],

where 3 = h3e3h2/e2,3 = j3e3j2/e2. According to Theorem 4.1, under condition 3j43h4, we can turn 𝓖j into canonical form. Analogously, by subtracting e2/e3 times the third slice from the second slice, we can turn e2 into zero, under condition 2j42h4, where 2 = h2e2h3/e3, 2 = j2e2j3/e3, we can also turn 𝓖j into canonical form.

The second case is i2i3 ≠ 0. By subtracting e3/e2 times the second slice from the third slice, we can turn e3, i3 into zeros according to e2i3 = e3i2. Then we obtain the following form:

[10000e2h2j200h~3j~300h4j40100000i20000000000100000000000000001000000000000],

where 3 = h3e3h2/e2, 3 = j3e3j2/e2. According to Theorem 4.1, only under conditions i2 = 0 and 3j43 h4, we can turn 𝓖j into canonical form. This contradicts the fact that i2i3 ≠ 0.

Thus we conclude that if exey ≠ 0, the remaining seven of ep, fp, gp (p = 2, 3, 4) are equal to zeros, under conditions ix = iy = iz = 0, yjzy hz, where x = 2, y = 3, z = 4 or x = 3, y = 2, z = 4, we can turn 𝓖j into canonical form (6).

Next, we discuss the combination that e2f2 ≠ 0, the remaining seven of ep, fp, gp (p = 2, 3, 4) are equal to zeros. The combinations f2g2, e3f3, f3g3, e4f4, f4g4 can be similar as in the discussion with e2f2.

If e2f2≠ 0, e3 = e4 = f3 = f4 = g2 = g3 = g4 = 0, according to (9), we can obtain i3 = i4 = 0. Because the mode-3 rank of 𝓖j is 4, then h3j4j3h4. The way to standardize the last two slices is the same as in Theorem 4.1, so we don’t repeat it here. Now, we show how to standardize the second slice. The way to turn h2 and j2 into zeros is the same as in Theorem 4.1. It remains to consider how to turn i2 into zero. In fact, if i2 = 0, then we have transformed 𝓖j into canonical form. If i2 ≠ 0, for every slice of 𝓖j, by subtracting i2/f2 times column 3 from column 4 and adding i2/f2 times row 4 to row 3 , we can turn i2 into zero. Then the (1, 4) entry of slice three is turned into − i2/f2. Next, by subtracting − i2/f2 times the fourth slice form the third slice, we can turn (1, 4) entry of slice three into zero.

Consequently, if exfx≠ 0, the remaining seven of ep, fp, gp(p = 2, 3, 4) are equal to zeros, under conditions iy = iz = 0 and hyjzjxhz, where x = 2, y = 3, z = 4 or x = 2, y = 4, z = 3, we can turn 𝓖j into canonical form.

Next, we discuss the combination that e2g2≠ 0, the remaining seven of ep, fp, gp (p = 2, 3, 4) are equal to zeros. The combinations e3g3, e4g4 can be similar as in the discussion with e2g2.

If e2g2≠ 0, e3 = e4 = f2 = f3 = f4 = g3 = g4 = 0, according to (9), we have e2i3 = h3g2, e2i4 = h4g2. Because e2g2≠ 0, so we must have i3 and h3 being zero or nonzero at the same time, i4 and h4 being zero or nonzero at the same time. This yields four possibilities.

The first case is i3 = h3 = 0, i4h4≠ 0. Then 𝓖j is of the form

[10000e2h2j2000j300h4j40100000i20000000i40010000g2000000000001000000000000].

Because the mode-3 rank of 𝓖j is 4, then j3 ≠ 0. Next, we normalize j3 to one, and it can be used to turn j2 and j4 into zeros if j2 and j4 are nonzero. After this, by exchanging the third slice and the fourth slice, then we can obtain the following form

[10000e2h2000h4000010100000i2000i400000010000g2000000000001000000000000].

If h2 and i2 are equal to zeros, then we have turned 𝓖j into canonical form. If one of h2, i2 is equal to zero, we can’t turn 𝓖j into canonical form. In fact, if h2 = 0,i2 ≠ 0, for every slice, by subtracting i2/g2 times row 3 from row 2 and adding i2/g2 times column 2 to column 3, we can turn i2 into zero. But meanwhile, h2 is turned into nonzero. Similarly, if h2 ≠ 0,i2 = 0, while turning h2 into zero, i2 is turned into nonzero. Similarly to the discussion of the above, if h2 and i2 are nonzero, after turn h2 into zero, i2 is turned into ĩ2 = i2+h2g2/e2. While turning ĩ2 into zero, h2 is turned into nonzero. If we add a restriction condition h2i4 = i2h4, by subtracting h4/h2 times the third slice from the second slice, we can turn h2 and i2 into zeros. Hence, we have turned 𝓖j into canonical form. From the above discussion we can see that, if i3 = h3 = 0,i4h4≠ 0, only under condition h2i4 = i2h4 we can turn 𝓖j into canonical form.

The second case is i3h3 ≠ 0,i4 = h4 = 0. This situation can be similar as in the discussion with i3 = h3 = 0,i4h4≠ 0.

The third case is i3 = h3 = i4 = h4 = 0. Then 𝓖j is of the form

[10000e2h2j2000j3000j40100000i2000000000010000g2000000000001000000000000].

If one of j3, j4 is equal to zero, then the mode-3 rank of 𝓖j is equal to 3. If j3 and j4 are all equal to zeros, then the mode-3 rank of 𝓖j is equal to 2. If j3 and j4 are nonzero, by subtracting j4/j3 times the third slice from the fourth slice, we can turn j4 into zero. Then the mode-3 rank of 𝓖j is equal to 3. Consequently, if i3 = h3 = i4 = h4 = 0, we can’t turn 𝓖j into canonical form.

The fourth case is i3h3i4h4≠ 0. It follows from e2i3 = h3g2, e2i4 = h4g2 that i3h4 = i4h3. This means the vectors (h3, i3) and (h4, i4) are proportional. If the vectors (h3, j3) and (h4, j4) are also proportional, then the mode-3 rank of 𝓖j is not equal to 4. Therefore, we conclude that h3j4h4j3. Similarly to the discussion of the case that i3 = h3 = 0,i4h4≠ 0. If we add a restriction condition h2i4 = i2h4 or h2i3 = i2h3, we can turn h2 and i2 into zero. Hence, if i3h3i4h4≠ 0, under conditions h3j4h4j3, h2i4 = i2h4 or h3j4h4j3, h2i3 = i2h3, we can turn 𝓖j into canonical form.

Based on the above argument, we draw the conclusion that if exgx ≠ 0, the remaining seven of ep, fp, gp (p = 2, 3, 4) are equal to zeros, under conditions iy = hy = 0, izhz ≠ 0, jy ≠ 0, hxiz = ixhz or iyhyizhz ≠ 0, hxiy = ixhy, hyjzjyhz, where x = 2, y = 3, z = 4 or x = 2, y = 4, z = 3, we can turn 𝓖j into canonical form. □

Theorem 4.4

If three of ep, fp, gp (p = 2, 3, 4) are not equal to zeros, the remaining six of them are equal to zeros, there have 5 combinations in T1, i.e. e2e3e4, g2g3g4, e2f2g2, e3f3g3, e4f4g4. For each combinations, 𝓖j can be turned into canonical form under the following conditions:

exeyez0,ix=iy=iz=0,h~yj~zj~yh~z;gxgygz0,hx=hy=hz=0,i~yj~zj~yi~z;exfxgx0,iy=hy=0,izhz0,jy0;(exfxgx0,iyhyizhz0,jyhzhyjz);wherex,y,z{2,3,4}andxyz.

Proof

Firstly, we discuss the combination that e2e3e4 ≠ 0, the remaining six of ep, fp, gp (p = 2, 3, 4) are equal to zeros. The combination g2g3g4 can be similar as in the discussion with e2e3e4.

If e2e3e4≠ 0,f2 = f3 = f4 = g2 = g3 = g4 = 0, according to (9), we have e3i2 = i3e2, e4i2 = e2i4, e4i3 = e3i4. Because e2e3e4 ≠ 0, so we must have i2, i3, i4 being zero or nonzero at the same time. This yields two possibilities. The first case is i2 = i3 = i4 = 0. The second case is i2i3i4 ≠ 0. Similarly to the discussion of the situation e2e3 ≠ 0 in Theorem 4.3, only for the case that i2 = i3 = i4 = 0 we can turn 𝓖j into canonical form. Now, we show if i2 = i3 = i4 = 0, under what conditions we can turn 𝓖j into canonical form. In fact, if i2 = i3 = i4 = 0, then 𝓖j is of the form

[10000e2h2j20e3h3j30e4h4j4010000000000000000100000000000000001000000000000].

Through subtracting e3/e2 times the second slice from the third slice and e4/e2 times the second slice from the fourth slice, we can turn e3 and e4 into zeros. Then 𝓖j is of the form

[10000e2h2j200h~3j~300h~4j~4010000000000000000100000000000000001000000000000],

where 3 = h3e3h2/e2, 3 = j3e3j2/e2, 4 = h4e4h2/e2, 4 = j4e4j2/e2. According to the mode-3 rank of 𝓖j is 4, then 3434. Thus we draw the conclusion that if exeyez ≠ 0, the remaining six of ep, fp, gp (p = 2, 3, 4) are equal to zeros, under the conditions ix = iy = iz = 0, yzy z, where {x, y, z} ∈ {2, 3, 4} and xyz, we can turn 𝓖j into canonical form.

Next, we discuss the combination that e2f2g2 ≠ 0, the remaining six of ep, fp, gp (p = 2, 3, 4) are equal to zeros. The combinations e3f3g3, e4f4g4 can be similar as in the discussion with e2f2g2.

If e2f2g2≠ 0, e3 = e4 = f3 = f4 = g3 = g4 = 0, according to (9), we have e2i3 = h3g2, e2i4 = h4g2. Because e2g2 ≠ 0, then i3 and h3 are zero or nonzero at the same time, i4 and h4 are zero or nonzero at the same time. This yields four possibilities. The proof of these four cases is almost identical as the case that e2g2 ≠ 0 in Theorem 4.3, the major change is that h2 can be turned into zero by e2, i2 can be turned into zero by f2. Consequently, if exfxgx≠ 0, the remaining six of ep, fp, gp(p = 2, 3, 4) are zeros, under conditions iy = hy = 0, izhz ≠ 0, jy ≠ 0 or iyhyizhz ≠ 0, hyjzjyhz, where x = 2, y = 3, z = 4 or x = 2, y = 4, z = 3, we can turn 𝓖j into canonical form. □

Theorem 4.5

If four of ep, fp, gp (p = 2, 3, 4) are not equal to zero, the remaining five of them are equal to zeros, there have 6 combinations in T1, i.e. e2e3f2f3, e2e4f2f4, e3e4f3f4, f2f3g2g3, f2f4g2g4, f3f4g3g4. For each combination, 𝓖j can be turned into canonical form under the following conditions:

exeyfxfy0,iz=0,ix,iybeingzeroornonzeroatsametime,h~yjzj~yhz;fxfygxgy0,hz=0,hx,hybeingzeroornonzeroatsametime,j~yizi~yjz;wherex,y,z{2,3,4}andxyz.

Proof

Here, we only discuss the combination that e2e3f2f3 ≠ 0, the remaining five of ep, fp, gp (p = 2, 3, 4)are equal to zeros. The combinations e2e4f2f4, e3e4f3f4, f2f3g2g3, f2f4g2g4, f3f4g3g4 can be similar as in the discussion with e2e3f2f3.

If e2e3f2f3≠ 0, e4 = f4 = g2 = g3 = g4 = 0, according to (9), we have i4 = 0 and e2i3 = e3i2. The proof of this case is almost identical as the combination that e2e3 ≠ 0 in Theorem 4.3, the major change is that i2 (or i3) can be turned into zero by f2 (or f3) if i2 (or i3) is nonzero.

Thus we conclude that if e2e3f2f3≠ 0, the remaining five of ep, fp, gp(p = 2, 3, 4) are equal to zeros, under conditions iz = 0, ix, iy being zero or not at the same time, yjzyhz, where x = 2, y = 3, z = 4 or x = 3, y = 2, z = 4, we can turn 𝓖j into canonical form. □

Theorem 4.6

If six of ep, fp, gp (p = 2, 3, 4) are not equal to zero, the remaining three of them are equal to zeros, there we have 5 combinations in T1, i.e. e2e3e4f2f3f4, f2f3f4g2g3g4, e2f2g2e3f3g3, e2f2g2e4f4g4, e3f3g3e4f4g4. For each combination, 𝓖j can be turned into canonical form under the following conditions:

exeyezfxfyfz0,ix,iy,izbeingzeroornonzeroatsametime,h~yj~zj~yh~z;fxfyfzgxgygz0,hx,hy,hzbeingzeroornonzeroatsametime,j~yi~zi~yj~z;exfxgxeyfygy0,h~y=i~y=0,hziz0,j~y0;(exfxgxeyfygy0,h~yi~y0,hz=iz=0,jz0);(exfxgxeyfygy0,h~yi~yhziz0,h~yjzj~yhz);wherex,y,z{2,3,4}andxyz.

Proof

Firstly, we discuss the combination that e2e3e4f2f3f4 ≠ 0, the remaining three of ep, fp, gp (p = 2, 3, 4) are equal to zeros. The combination f2f3f4g2g3g4 can be similar as in the discussion with e2e3e4f2f3f4. If e2e3e4f2f3f4≠ 0, g2 = g3 = g4 = 0, according to (9), we have e2i3 = e3i2, e4i2 = e2i4, e4i3 = e3i4. The proof of this combination is almost identical as the combination that e2e3e4 ≠ 0 in Theorem 4.4, the major change is that i2 (or i3 or i4) can be turned into zero by f2 (or f3 or f4) if i2 (or i3 or i4) is nonzero. Thus we conclude that if e2e3e4f2f3f4≠ 0, the remaining three of ep, fp, gp (p = 2, 3, 4) are equal to zeros, under conditions ix, iy, iz being zero or not at the same time, yzyz, where x, y, y ∈ {2, 3, 4} and xyz, we can turn 𝓖j into canonical form.

Next, we discuss the combination that e2f2g2e3f3g3 ≠ 0, the remaining three of ep, fp, gp (p = 2, 3, 4)are equal to zeros. The combinations e2f2g2e3f3g3, e2f2g2e4f4g4, e3f3g3e4f4g4 can be similar as in the discussion with e2f2g2e3f3g3. If e2f2g2e3f3g3≠ 0, e4 = f4 = g4 = 0, according to (9), we have e3i2e2i3+h3g2g3h2 = 0, e2i4 = h4g2 and e3i4 = h4g3. Because e2e3g2g3 ≠ 0 and e3/e2 = g3/g2 = α, then we obtain e2ĩ3 = g23, where ĩ3 = i3α i2, 3 = h3α h2. On the other hand, because e2e3g2g3 ≠ 0, so 3 and ĩ3 are zero or nonzero at the same time, h4 and i4 are zero or nonzero at the same time. This yields four possibilities. These four situations can be discussed as the situation that e2f2g2 ≠ 0, the remaining six of ep, fp, gp (p = 2, 3, 4) are equal to zeros.

Consequently, if exfxgxeyfygy≠ 0, the remaining three of ep, fp, gp (p = 2, 3, 4) are equal to zeros, under conditions y = ĩy = 0, hziz ≠ 0, y ≠ 0, or yĩy ≠ 0, hz = iz = 0, jz ≠ 0, or yĩyhziz ≠ 0, yjzyhz, where x = 2, y = 3, z = 4, or x = 2, y = 4, z = 3, we can turn 𝓖j into canonical form. □

Theorem 4.7

If ep, fp, gp (p = 2, 3, 4) are all nonzero, 𝓖j can be turned into canonical form under the following conditions:

exeyezfxfyfzgxgygz0,h~y=i~y=0,h~z=i~z0,j~y0;(exeyezfxfyfzgxgygz0,h~yi~yh~zi~z0,j~yh~zh~yj~z);wherex,y,z{2,3,4}andxyz.

Proof

According to the second conclusion of Property 2.1, if ep, fp, gp ( p = 2, 3, 4 ) are all nonzero, the vectors (ep, fp, gp) (p = 2, 3, 4) being proportional. Then we can turn e3, e4, f3, f4, g3, g4 into zeros due to the vectors (ep, fp, gp) (p = 2, 3, 4) are proportional. Then we obtain the following for the last three slices of 𝓖j:

[0e2h2j200h~3j3~00h~4j~400f2i2000i~3000i~4000g200000000000000000000]

where 3 = h3α h2, 3 = j3α j2, 4 = h4β h2, 4 = j4β j2 and α = e3/e2, β = e4/e2.

Now we show that equality e2ĩ3 = g23 and e2ĩ4 = g24 holds. Firstly, we write e3 = α e2, g3 = α g2, e4 = β e2, g4 = β g2. Next, by substituting them into the first two equations of (9), we obtain e2ĩ3 = g23 and e2ĩ4 = g24. Because e2g2 ≠ 0, then ĩ3 and 3 are zero or nonzero at the same time, ĩ4 and 4 are zero or nonzero at the same time. This yields four possibilities. These four situation can be discussed just as the situation that e2e3e4f2f3f4 ≠ 0, the remaining three of ep, fp, gp(p = 2, 3, 4) are equal to zeros.

Consequently, if exfxgxeyfy, gyezfzgz are nonzero, under conditions y = ĩy = 0,z = ĩz ≠ 0,y ≠ 0 or yĩyzĩz ≠ 0, yzyz, we can turn 𝓖j into canonical form. □

5 Concluding remarks

In this paper, we have studied under what conditions we can turn 𝓖j into canonical form (6) if some of the upper triangular entries of the last three slices of 𝓖j are zeros. In addition, we have shown how to turn 𝓖j into canonical form under these conditions. In the future, it will be interesting to explore the connection between our three order generalization of the Jordan canonical form and eigenvectors for three order arrays.

Acknowledgement

Research supported in part by National Natural Science Foundations of China (No. 11526083, 11571905, 11601134), Foundation of Henan Educational Committee (No. 15A110030), Youth Science Foundation of Henan Normal University (No.2014QK06), Research Fund for the Doctoral Program of Henan Normal University (No.qd14147) and Guangdong Provincial Engineering Technology Research Center for Data Science (No. 2016KF01).

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About the article

Received: 2018-03-04

Accepted: 2018-06-13

Published Online: 2018-08-08


Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 897–912, ISSN (Online) 2391-5455, DOI: https://doi.org/10.1515/math-2018-0078.

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© 2018 Cui and Li, published by De Gruyter. This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License. BY-NC-ND 4.0

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