The rigidity of a matrix $M\in \mathbb{F}^{n\times n}$ is the minimum number of entries that need to be changed in $M$ to reduce its rank to $r$. Formally, it is defined as:

$$R_{M}^{\mathbb{F}}(r) = \min_{A} \{sparsity(A) | A\in \mathbb{F}^{n\times n}, rank(M+A)\leq r\}.$$

I was wondering is someone could point me to a proof of the fact that for a random matrix $M\in \mathbb{F}^{n\times n},$, with high probability, $ R_{M}^{\mathbb{F}}(r) \geq \frac{(n-r)^2}{\log n}.$

  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Jun 12, 2023 at 14:08
  • $\begingroup$ If I had to prove this I would proceed in this very boring way: (1) Determine the expectation of R_M^F(r); (2) Determine the variance of R_M^F(r); (3) Pick up a book titled "Concentration inequalities" and look for an inequality in the book that seems applicable. $\endgroup$
    – Stef
    Commented Jun 14, 2023 at 16:05

1 Answer 1


For simplicity, let us first consider the case $\mathbb{F}=\mathbb{F}_2$. Every non-rigid matrix can be specified by a rank-$r$ matrix $L$ and a matrix $S$ of total sparsity $s$. Since every low-rank matrix $R$ can be written as a product of $n\times r$ and $r\times n$ matrices, the number of matrices of rank $\leq r$ is at most $2^{2nr}$. Every sparse matrix $S$ is specified by a list of at most $s$ ones in the matrix, and, thus, assuming $s\leq n^2/2$, the number of matrices $S$ is at most $\binom{n^2}{\leq s}\leq n^2 \binom{n^2}{s} \leq 2^{O(s\log{n})}$. This implies that the total number of non-rigid matrices is at most $2^{O(nr+s\log{n})}$. Therefore, for $s<O(n(n-2r)/\log{n})$, we have that a random matrix is rigid.

This argument extends to all finite fields. The bound can be improved to $(n-r)^2/\log{n}$. Moreover, for many settings of the parameters, the factor of $\log{n}$ can be shaved off, too. Finally, a slight modification of this argument gives a tight bound of $(n-r)^2$ for the case when $\mathbb{F}$ is infinite. All of these extensions can be found, for example, in Theorem 1.10 here.

  • $\begingroup$ why is the number of sparse matrices not $\binom{n^2}{\leq s}$, also where do you get the upper bound on this combinatorial quantity $\endgroup$
    – atul ganju
    Commented Jul 19, 2023 at 17:20
  • $\begingroup$ You're absolutely right that this should be $\binom{n^2}{\leq s}$, I've fixed this, and explained the combinatorial bound. $\endgroup$ Commented Jul 26, 2023 at 16:08

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