Tensor rank decomposition for some vectors

I have a problem with solving of following problem.
We have a vector space $$V$$ over field $$k$$ with basis $$v_1, v_2$$
Rank of a vector $$v$$ of $$V \otimes V \otimes V$$ is minimum length of decomposition of $$v$$ into linear combination of rank-1 vectors
I need to solve the following:

• Prove that vector $$v = v_1 \otimes v_1 \otimes v_1 + v_1 \otimes v_2 \otimes v_2 + v_2 \otimes v_1 \otimes v_2$$ has rank $$3$$

• Vector $$t = v_1 \otimes v_1 \otimes v_1 - v_2 \otimes v_2 \otimes v_1 + v_1 \otimes v_2 \otimes v_2 + v_2 \otimes v_1 \otimes v_2$$. Show decomposition of $$t$$ into sum of two rank-1 vectors when $$k = \Bbb C$$. Prove that tensor rank of $$t$$ is $$3$$ when $$k = \Bbb R$$

Tried to look for an information but found nothing usefull. I thought about supposing that we can decompose $$v$$ into $$e_{11} \otimes e_{12} \otimes e_{13} + e_{21} \otimes e_{22} \otimes e_{23}$$ and supposing that $$e_{ij}$$ form basis for each $$j$$ that allows us to rewrite $$v$$ in terms of $$e_{ij}$$. After doing this i got a large system with a large amount of variables that i'm unable to work with. I have been thinking about this problem for a several weeks already and do not know how to even think about it.

• Is this an exercise from a class? From a textbook? Dec 13 '19 at 21:23
• @Omnomnomnom It's from a class. I met this problem before and never was able to solve it. Usually i try to solve problems by myself but this one really frustrated me. Dec 13 '19 at 21:28
• Yeah it's a doozy... have you discussed any tricks for tensors in class? Any way to provide a lower bound to a tensor rank? I'm sure that you must have discussed the case of rank-$1$ tensors in $V \otimes V$. Dec 13 '19 at 21:42
• Based on some googling, this article seems to have some interesting ideas. I found the article from the references of this textbook, which I've only just been skimming through Dec 13 '19 at 21:44
• As an array, your first matrix has (frontal) slices $$X_1 = \pmatrix{1&0\\0&0}, \quad X_2 = \pmatrix{0&1\\1&0}$$ Dec 13 '19 at 21:46

For the purposes of the following, I identify $$\sum_{ijk}x_{ijk} v_{i} \otimes v_j \otimes v_k$$ with a 3-dimensional array $$X$$. For the purposes of your question over a 2-dimensional $$V$$, we have $$X = [X_1 |X_2] = \left[ \begin{array}{cc|cc} x_{111} & x_{121} & x_{112} & x_{122}\\ x_{211} & x_{221} & x_{212} & x_{222} \end{array}\right].$$

Question 1:

For your problem, we have $$X_1 = \pmatrix{1&0\\0&0}, \quad X_2 = \pmatrix{0&1\\1&0}.$$ We will simply apply the result explained below. Since exchanging slices of a tensor will not change its rank, we'll exchange the roles of $$X_1$$ and $$X_2$$ since $$X_2$$ is the invertible slice. We find that $$X_1X_2^{-1} = \pmatrix{0&1\\0&0}.$$ Since this matrix fails to be diagonalizable, $$X$$ must be a tensor of rank $$3$$.

Question 2:

Your second problem can be approached similarly. We now have $$X_1 = \pmatrix{1&0\\0&-1},\quad X_2 = \pmatrix{0&1\\1&0}.$$ Applying the result below, we find that $$X_2 X_1^{-1} = \pmatrix{0&-1\\1&0}.$$ Because this matrix is diagonalizable with strictly complex eigenvalues, we can conclude that $$X$$ has a rank of at least $$3$$ if we restrict ourselves to real coefficients, and a rank of $$2$$ if we allow complex coefficients.

It remains to be shown, however, that the rank of $$X$$ over $$\Bbb R$$ is not more than $$3$$. To do this, simply observe that we can get this tensor by adding a rank-1 tensor to the tensor given in the first question.

Regarding the presentation of $$t$$ as a rank-2 tensor: if we follow the construction from the proof I present below, then we note that $$X_2 X_1^{-1} = K\Lambda K^{-1}$$, where $$\Lambda = \pmatrix{i\\&-i}, \quad K = \pmatrix{i&1\\1&i}.$$ So, we find that $$X = a_1 \otimes b_1 \otimes c_1 + a_2 \otimes b_2 \otimes c_2$$ where $$a_1,a_2$$ are the columns of $$K$$, $$b_1,b_2$$ are the rows of $$K^{-1}X_1$$, and we have $$c_1 = (1,i), c_2 = (1,-i)$$.

Now, here is an adaptation of the statement and proof of Lemma 1 of this paper.

Claim: Let $$X$$ be a real-valued $$p \times p \times 2$$ array with $$p \times p$$ slices $$X_1$$ and $$X_2$$. Suppose that $$X_{1}^{-1}$$ exists. The following statements hold:

• If $$X_2X_1^{-1}$$ has $$p$$ real eigenvalues and is diagonalizable, then $$X$$ has rank $$p$$ over $$\Bbb R$$
• If $$X_2X_1^{-1}$$ has at least one pair of complex eigenvalues, then $$X$$ has rank $$p$$ over $$\Bbb C$$ and rank at least $$p+1$$ over $$\Bbb R$$
• If $$X_2X_1^{-1}$$ is not diagonalizable, then $$X$$ has rank at least $$p+1$$ over $$\Bbb C$$.

Proof: First, note that since $$X_1$$ is invertible, the rank of $$X$$ must be at least $$p$$.

Proof of i: Now, suppose that we have $$X_2 X_1^{-1} = K \Lambda K^{-1}$$, where $$\Lambda = \operatorname{diag}(\lambda_1,\dots,\lambda_p)$$. If we take $$A = K, \quad B^T = K^{-1}X_1, \quad C_1 = I_p, \quad C_2 = \Lambda,$$ then we find that $$X_1 = AC_1B^T, \quad X_2 = AC_2B^T.$$ This corresponds to a rank-$$p$$ decomposition of the matrix $$X$$. In particular: if we take $$a_j$$ to denote the $$j$$th column of $$A$$ and $$b_j$$ to denote the $$j$$th column of $$B$$, then we have $$X_1 = AC_1B^T = \sum_{j=1}^p c_{1,i} \, a_ib_i^T, \quad X_2 = AC_2B^T = \sum_{j=1}^p c_{2,i} \, a_i b_i^T.$$ Correspondingly, we have $$X = \left(\sum_{j=1}^p c_{1,j} \, a_j \otimes b_j\right) \otimes e_1 + \left(\sum_{j=1}^p c_{2,j} \, a_j \otimes b_j\right) \otimes e_2\\ = \sum_{j=1}^p a_j \otimes b_j \otimes (c_{1,j} e_1) + \sum_{j=1}^p a_j \otimes b_j \otimes (c_{2,j}e_2)\\ = \sum_{j=1}^p a_j \otimes b_j \otimes (c_{1,j}e_1 + c_{2,j}e_2).$$ In the above, $$e_1 = (1,0)$$ and $$e_2 = (0,1)$$.

Proof of ii and iii: It suffices to prove that if $$X$$ is a rank-$$p$$ tensor and $$X_1$$ is invertible, then $$X_2X_1^{-1}$$ must be diagonalizable. Indeed, if $$X$$ is a rank-$$p$$ tensor, then we can take $$X_1 = AC_1B^T, \quad X_2 = AC_2B^T$$ by reversing the above sequence of equations. It follows that $$X_2X_1^{-1} = (AC_2B^T)(AC_1B^T)^{-1} = AC_2 B^T B^{-T} C_1^{-1} A^{-1} = A (C_2 C_1^{-1})A^{-1}.$$ So, $$X_2X_1^{-1}$$ is indeed diagonalizable (and diagonalizable over $$\Bbb R$$ when $$A,B,C$$ are real). The conclusion follows.

• That's a really good answer, I didn't really expect anything like this. I guess this proof works over any field, doesn't it? Also while it answers my question I want to ask the following: There was proved that if rank-p decomposition exists then matrix is diagonalizable. But how does "if matrix have two complex eigenvalues then it have rank p over C" follows from it? I see how it's proved for real coefficients when p = 2 (in this case it's equivalent to the fact that it's diagonalizable) Dec 14 '19 at 10:44
• Yes, I agree that the proof seems to work over any field. The point I'm making with the proof of ii and iii is that we can go for a proof by contradiction. Given that $X_2X_1^{-1}$ has at least two (strictly) complex eigenvalues, suppose that $X$ has a tensor decomposition with real coefficients. Then by reversing the process in the proof of i, we obtain a decomposition $X_2X_1^{-1} = A \Lambda A^{-1}$ where $\Lambda$ is a real diagonal matrix, which would imply that $X_2X_1^{-1}$ has no strictly complex eigenvalues. Dec 14 '19 at 12:24
• And if we allow for complex coefficients, then we can simply apply the proof from i to see that $X$ has rank $p$. Dec 14 '19 at 12:29
• If you prefer, the real result is that given that the slice $X_1$ is invertible, $X$ has rank $p$ over $\Bbb F$ if and only if $X_2X_1^{-1}$ is diagonalizable over $\Bbb F$. Dec 14 '19 at 12:31