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[Cross-posted from MS after 8 days without reply.] I have a real matrix $R_{ij} \in \mathbb{R}^{n \times m}$ whose entries are sampled iid from an absolutely continuous distribution $D$; a fixed real 3-dimensional tensor $T_{ijk} \in \mathbb{R}^{n \times n \times m}$; and any matrix $A_{ij}$ which satisfies $$ A_{ij} = R_{ij} + \sum_k T_{ijk} A_{ik} \,. $$ I would like to show that $A_{ij}$ is almost surely full-rank. I know that the set of matrices with rank-deficient matrices has zero measure (so $R_{ij}$ is almost surely full-rank), and tried to use the fact that $rank(AA^T) = rank(A^TA) = rank(A)$, or to re-express $$ \sum_k A_{ik}(\delta_{jk} - T_{ijk}) = R_{ij} \,, $$ to no avail. In the case of $n = m$, I was hoping to take the determinant on both sides and conclude, but don't know how to take the determinant because of the 3D tensor on the LHS. Any ideas?

Edit: if I moreover assume that the matrix $S_i = T_{ijk}$, for each fixed $i$, has a spectrum bounded by 1, we can expand the equation to get $$ A_{ij} = \sum_k R_{ik}\sum_{n = 0}^{\infty} (S_i^n)_{ijk} = \sum_k R_{ik}(I-S_i)^{-1}_{ijk} \,, $$ from which we conclude that the entries of $A_{ij}$ are also absolutely continuous and independent, and thus has full-rank with probability 1. Is this correct? Can we somehow generalise this without assuming a bounded spectrum?

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  • $\begingroup$ In which sense this is a recurrence? What is a time variable? Looks like just a system of equations. $\endgroup$ Commented Jul 10 at 22:27
  • $\begingroup$ In addition to Fedor Petrov's comment: Why are you saying that $A$ is defined by that matrix equation that you refer to as "the recurrence relation"? The matrix equation is a system of $nm$ scalar linear equations with $nm$ unknowns. The determinant of such a system may be $0$, and then the system may have no or infinitely many solutions (actually, no solutions with probability $1$). $\endgroup$ Commented Jul 11 at 3:01
  • $\begingroup$ Apologies, the word recurrence was used incorrectly; I removed it everywhere! $\endgroup$ Commented Jul 11 at 7:02
  • $\begingroup$ Couldn't you just scale the equation such that $S_i = R_{ijk}$ has spectrum bounded by $1$. Since the distribution of the entries of $R$ are absolutely continuous, the distribution might not be scaling invariant, but the property that $R$ is of full rank with probability $1$ might be preserved. $\endgroup$ Commented Jul 13 at 17:40

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So you have a linear map $\tau$ from ${\rm Mat}(m,n)$ to itself and $A$ is defined by $\tau A=R$, i. e., $A=\tau^{-1} R$ (if $\tau$ is not invertible, then with probability 1 your $A$ does not exist). This is of course full rank with probability 1, since the map $\tau$ just multiplies Lebesgue measure by a non-zero constant.

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  • $\begingroup$ Thank you Fedor! That's exactly what I was looking for. $\endgroup$ Commented Jul 17 at 19:38

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