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Source: AMM, proposed by Daniel Goffinet

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#1 h Jul 1, 2025, 10:38 AM

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$(a)$ Prove that a (square) matrix over a field $F$ is singular if and only if it is a product of nilpotent matrices.

$(b)$ If $F=\mathbb{C}$ , prove that the number of nilpotent factors can be bounded independently of the size of the matrix.

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#2 h Jul 5, 2025, 9:40 PM

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$\textbf{Part 1.}$ There is an easy case; the one where $F$ is algebraically closed and where, in the normal Jordan form of the considered matrix $A$ , there is at least one nilpotent Jordan block of dimension $1$ .
Then $A$ is similar to $T$ , an upper-triangular matrix with last column $0_n$ .
Let $B=[0_n,SubMatrix(T,[1..n],[1..n-1])]$ (the first column is $0_n$ ).
Then $T=BJ^T$ , a product of $2$ nilpotent matrices and we are done. $\square$

$\textbf{Part 2.}$ In fact, for any field $F$ , any singular matrix $A\in M_n(F)$ is nilpotent or otherwise is the product of $2$ nilpotent matrices.
cf. Hattingh, "a note on products of nilpotent matrices".
https://arxiv.org/pdf/1608.04666v3

This post has been edited 1 time. Last edited by loup blanc, Jul 5, 2025, 9:48 PM

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