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Let $M$ be a symmetric positive definite matrix, and let $a > 0$ be a given constant. Let $\text{vec}(\cdot)$ denote the operator that stacks vertically the columns of a $d \times d$ matrix into a $d^2 \times 1$ vector. Does there exists a matrix $S$ such that

$$ \text{vec}(M) \text{vec}(M)^{\top} + a \, (M \otimes M) = S \otimes S $$ ?

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    $\begingroup$ No such $S$ exists even if $d = 2$ and $M = I$ is the identity matrix. A necessary criterion for the existence of such an $S$ is that each block of the matrix you wrote on the left-hand-side is a multiple of each other, and that criterion fails in this case. $\endgroup$
    – Nathaniel Johnston
    Commented Jul 19, 2021 at 19:14

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@Nathaniel Johnston already answered in comments. On a more general note, detecting whether a matrix is a sum of $r$ Kronecker products is essentially the same problem (up to a permutation of the entries) as determining whether a matrix has rank $r$.

Indeed, given a matrix $M\in\mathbb{F}^{am \times bn}$ (divided into blocks $M_{ij}$ each of size $m\times n$), rearrange its entries to form a matrix $N\in\mathbb{F}^{mn \times ab}$ such that each column of $N$ is the vectorization of a block $M_{ij}$; then a rank-$r$ decomposition of $N = u_1 v_1^T + \dots + u_r v_r^T$ corresponds to a decomposition of $M$ as sum of $r$ Kronecker products $M = \operatorname{vec}^{-1}(v_1) \otimes \operatorname{vec}^{-1}(u_1) + \dots + M = \operatorname{vec}^{-1}(v_r) \otimes \operatorname{vec}^{-1}(u_r)$.

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  • $\begingroup$ What if $a\to \infty$. Can we find a matrix $S_a$ such that $$\text{vec}(M) \text{vec}(M)^{\top} + a (M \otimes M) = a \, (S_a \otimes S_a)$$ ? $\endgroup$
    – Aftermath 12345
    Commented Jul 20, 2021 at 9:16
  • $\begingroup$ @FrédéricOuimet - I'm not sure what would qualify as an answer as $a \rightarrow \infty$ really, but this fails for every single $a > 0$ for the exact same reason that has been given. The term on the right has "rank" (in a slightly unusual sense) 1, while the term on the left has "rank" at least 3. $\endgroup$
    – Nathaniel Johnston
    Commented Jul 20, 2021 at 11:49
  • $\begingroup$ it means, does it hold asymptotically as $a \to \infty$. We could have $S_a = M + a^{-1/2} \text{*** something ***} + O(a^{-1})$. No ? $\endgroup$
    – Aftermath 12345
    Commented Jul 20, 2021 at 13:09
  • $\begingroup$ Bit it fails no matter what $S_a$ is. $\endgroup$
    – Nathaniel Johnston
    Commented Jul 20, 2021 at 13:46
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    $\begingroup$ If I understand you correctly now, you're basically asking if you can approximate a rank-3 matrix by a rank-1 matrix, and the answer is still no---a limit of rank 1 matrices is still rank 1. $\endgroup$
    – Nathaniel Johnston
    Commented Jul 20, 2021 at 14:13

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