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

Definition 1.1

([1]). Let 𝓐 be a m^th-order n-dimensional tensor. The mode-k matrix (or k-th matrix unfolding)A_(k) ∈ ℝ^n×nm−1is a matrix containing the elementa_i1i2…im.

Definition 1.2

([2]). The multilinear rank of anI × J × Karray 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.

Lemma 1.3

([13]). For a group ofd_j = 2 diverging rank-1 terms, the limit 𝓧_jcan be written as 𝓧_j = (S_j, T_j, U_j) ⋅ 𝓖_jwithS_j, T_j, U_jof rank 2, and 2 × 2 × 2 array 𝓖_jgiven by

we have rank(𝓖_j) = 3. Here, we denote the 2 × 2 × 2 array 𝓖_jwith 2 × 2 slicesG₁andG₂as [G₁ |G₂]. (3) is referred to as the canonical form of a boundary array ofS₂(2, 2, 2).

Lemma 1.4

([6]). For a group ofd_j = 3 diverging rank-1 terms, and min(I, J, K) ≥ 3, almost all limits 𝓧_jwith multilinear rank(3, 3, 3) can be written as 𝓧_j = (S_j, T_j, U_j) ⋅ 𝓖_jwithS_j, T_j, U_jof rank 3, and 3 × 3 × 3 array 𝓖_jgiven by

where * denotes a nonzero entry. We have rank(𝓧_j) = rank(𝓖_j) = 5.

Lemma 1.5

([8]). For a group ofd_j = 4 diverging rank-1 terms, and min(I, J, K) ≥ 4, almost all limits 𝓧_jwith multilinear rank(4, 4, 4) can be written as 𝓧_j = (S_j, T_j, U_j) ⋅ 𝓖_jwithS_j, T_j, U_jof rank 4, and 4 × 4 × 4 array 𝓖_jgiven by

where * denotes a nonzero entry. We have rank(𝓧_j) = rank(𝓖_j) ≥ 7.

Remark 1.6

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

Example 1.7

Let 𝓖_jbe 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.,

By subtracting 3 times the first slice of 𝓖_jfrom 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, 𝓖_jis of the form

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

In each slice of the above 𝓖_j, 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

Example 1.8

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

Using the same method as Example 1.7, we can obtain

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

Remark 1.9

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

whereē, f̄, ḡ, h̄, īmay 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 𝓖_j. Because the first slice of 𝓖_jis an identity matrix, it follows that the mode-1 rank of 𝓖_jis always 4, no matter what valuesē, f̄, ḡ, h̄, ītake.

The mode-2 matrix of 𝓖_jis

Since each row of this matrix has a nonzero entry 1, it follows that mode-2 rank of 𝓖_jis always 4, no matter what valuesē, f̄, ḡ, h̄, ītake.

The mode-3 matrix of 𝓖_jis

For this matrix, if there exists a row that its elements are all zeros, then the mode-3 rank of 𝓖_jdoes not equal to 4. This contradicts the fact that the mode-3 rank of 𝓖_jis 4. Consequently, ē, f̄, ḡcan’t be zero at the same time, andh̄, īcan’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 matrixA=(a11a12a13a14a21a22a23a24a31a32a33a34a41a42a43a44),we call the main diagonal elements of matrix A as the 1st diagonal elements , anda₁₂, a₂₃, a₃₄as the 2nd diagonal elements, anda₁₃, a₂₄as the 3rd diagonal elements, anda₁₄as the 4th diagonal element. The kth diagonal element ofAis called zero if the elements of the kth diagonal are all zeros. The kth diagonal elements ofAis called nonzero if the kth diagonal ofAhas a nonzero element.

Property 2.1

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

and we can obtain the following three conclusions: (1)a_p = b_p = c_p = d_p(p = 2, 3, 4) hold for almost all 𝓖_j;

(2) The equations

hold for almost all 𝓖_j; (3) The equations

holds for almost all 𝓖_j.

Proof

The proof of Lemma 3.3 in [8] has shown that 𝓖_j 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, (e_p, f_p, g_p) 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 (h₃ − αh₂,i₃ − αi₂) and (h₄ − βh₂, i₄ − βi₂) 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 e₃i₂ − e₂i₃+h₃g₂ − g₃h₂ = 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 e₄i₂ − e₂i₄+h₄g₂ − g₄h₂ = 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 e₄i₃ − e₃i₄+h₄g₃ − g₄h₃ = 0.

Remark 2.2

According to the first conclusion of Property 2.1, if 𝓖_jsatisfies the conditions of Lemma 1.5, by subtractinga_ptimes the first slice of 𝓖_jfrom slicep(p = 2, 3, 4), then it can be further turned into

Remark 2.3

According to the second conclusion of the Property 2.1, iff₂f₃f₄ ≠ 0, the vectors (e_p, f_p, g_p) (p = 2, 3, 4) are proportionate. Iff₂ = f₃ = f₄ = 0, the vectors (e_p, f_p, g_p) (p = 2, 3, 4)are disproportionate.

Property 2.4

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

Proof

We traverse e_p, f_p, g_p (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 e_p, f_p, g_p (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 e_p, f_p, g_p (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 e_p, f_p, g_p (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 e_p, f_p, g_p (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 e_p, f_p, g_p (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 e_p, f_p, g_p (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 e_p, f_p, g_p (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 e_p, f_p, g_p (p = 2, 3, 4) are all nonzero, i.e., e₂e₃e₄f₂f₃f₄g₂g₃g₄. □

Remark 2.5

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

Theorem 4.1

If one ofe_p, f_p, g_p(p = 2, 3, 4)is not equal to zero, the remaining eight of them are equal to zeros, there are 6 combinations inT₁, i.e., e_p, g_p(p = 2, 3, 4). Ife_x ≠ 0, the other ofe_p, f_p, g_p(p = 2, 3, 4)are equal to zeros, under conditionsi_x = i_y = i_z = 0,h_yj_z ≠ j_yh_z, or if g_x ≠ 0, the other ofe_p, f_p, g_p(p = 2, 3, 4)are equal to zeros, under conditionsh_x = h_y = h_z = 0, j_yi_z ≠ i_yj_z, wherex, y, z ∈ {2, 3, 4} andx ≠ y ≠ z, we can turn 𝓖_jinto canonical form (6).

Proof

Here, we only consider the combination that e₂ ≠ 0, the remaining eight of e_p, f_p, g_p (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 e_p, f_p, g_p (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 e_p, f_p, g_p (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 𝓖_j is of the form

Because the mode-3 rank of 𝓖_j is 4, then h₃j₄ ≠ j₃h₄. Next, we show why the mode-3 rank of 𝓖_j 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 𝓖_j is 3. Thus we draw the conclusion that h₃j₄ ≠ j₃h₄.

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 𝓖_j into canonical form, i₂ can’t be turned into zero. Therefore, we must have i₂ equal to zero.

Now we prove how we turn 𝓖_j into canonical form if e₂ ≠ 0,e₃ = e₄ = f₂ = f₃ = f₄ = g₂ = g₃ = g₄ = 0, under conditions that i₂ = i₃ = i₄ = 0 and h₃j₄ ≠ j₃h₄. In fact, under these conditions, 𝓖_j have the following form

Firstly, we discuss how to standardize the last two slices. Because h₃j₄ ≠ j₃h₄, 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

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 𝓖_j 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 e_x ≠ 0, the remaining eight of e_p, f_p, g_p (p = 2, 3, 4) are equal to zeros, under conditions i_x = i_y = i_z = 0 and h_yj_z ≠ j_yh_z, where x = 2, y = 3, z = 4 or x = 2, y = 4, z = 3, we can turn 𝓖_j into canonical form. □

Remark 4.2

In fact, theorem 4.1 contains the following six cases.

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

1. Ife₂ ≠ 0, e₃ = e₄ = f₂ = f₃ = f₄ = g₂ = g₃ = g₄ = 0, under conditionsi₂ = i₃ = i₄ = 0 andh₃j₄ ≠ j₃h₄, we can turn 𝓖_jinto canonical form.

2. Ifg₂ ≠ 0, e₂ = e₃ = e₄ = f₂ = f₃ = f₄ = g₃ = g₄ = 0, under conditionsh₂ = h₃ = h₄ = 0 andj₃i₄ ≠ i₃j₄, we can turn 𝓖_jinto canonical form.

   Letx = 3, y = 2, z = 4 orx = 3, y = 4, z = 2.

3. Ife₃ ≠ 0, e₂ = e₄ = f₂ = f₃ = f₄ = g₂ = g₃ = g₄ = 0, under conditionsi₂ = i₃ = i₄ = 0 andh₂j₄ ≠ j₂h₄, we can turn 𝓖_jinto canonical form.

4. Ifg₃ ≠ 0, e₂ = e₃ = e₄ = f₂ = f₃ = f₄ = g₂ = g₄ = 0, under conditionsh₂ = h₃ = h₄ = 0 and j₂i₄ ≠ i₂j₄, we can turn 𝓖_jinto canonical form.

   Letx = 4, y = 2, z = 3 orx = 4, y = 3, z = 2.

5. Ife₄ ≠ 0, e₂ = e₃ = f₂ = f₃ = f₄ = g₂ = g₃ = g₄ = 0, under conditionsi₂ = i₃ = i₄ = 0 andh₂j₃ ≠ j₂h₃, we can turn 𝓖_jinto canonical form.

6. Ifg₄ ≠ 0, e₂ = e₃ = e₄ = f₂ = f₃ = f₄ = g₂ = g₃ = 0, under conditionsh₂ = h₃ = h₄ = 0 andj₂i₃ ≠ i₂j₃, we can turn 𝓖_jinto canonical form.

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

   ex≠0,theotherarezero,ix=iy=iz=0,hyjz≠jyhz;gx≠0,theotherarezero,hx=hy=hz=0,iyjz≠jyiz;

   wherex, y, z ∈ {2, 3, 4} andx ≠ y ≠ z.

   The following theorem is also written in a similar way, and we don’t repeat it. For example, e_xe_y ≠ 0, i_x = i_y = i_z = 0, h̃_yj_z ≠ j̃_yh_z, wherex = 2, y = 3, z = 3, in Theorem 4.3 represents ife₂ ≠ 0, e₃ ≠ 0, e₄ = f₂ = f₃ = f₄ = g₂ = g₃ = g₄ = 0, under conditionsi₂ = i₃ = i₄ = 0 andh̃₃j₄ ≠ j̃₃h₄, can we turn 𝓖_jinto canonical form.