'Positive-definite' matrices over finite fields

Question

Let $X$ be an $n \times n$ invertible square matrix over some field $\mathbb{F}$, and let $Y = XX^T$ be the product of the matrix with its transpose.

When $\mathbb{F} = \mathbb{R}$, $Y$ is positive-definite, so it is easy to see that for any subset $S \subseteq \{ 1, 2, \dots, n \}$, the matrix $Y_{S,S}$ obtained by considering only the elements with indices in $S$ is also invertible (as it is positive-definite). Moreover, this holds whenever $\mathbb{F}$ is a subfield of a real-closed field.

Now, when $\mathbb{F} = \mathbb{F}_p$, we lack the notion of positive-definiteness. Is is still the case that $Y_{S, S}$ is invertible?

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Context: The reason I ask is because I would like to be able to invert a matrix $X$ using blockwise inversion (it has asymptotic complexity proportional to matrix multiplication, compared with $O(n^3)$ for Gaussian elimination), and that relies on certain minors being invertible. So for an invertible matrix $X$, we can express:

$X^{-1} = X^T(XX^T)^{-1}$

and invert the positive-definite matrix $XX^T$ by blockwise inversion (which is guaranteed to work if we're operating over $\mathbb{R}$).

Answer 1

No, already in the case when $S$ is a set of one element. In this case we ask that the diagonal elements of $X X^T$ are nonzero. These are the sums of squares of row vectors in $X$.

The row vector may be any nonzero vector, so your desired statement implies that, for any $x_1,\dots,x_n$ not all zero, we have $\sum_{i=1}^n x_i^2 \neq 0$.

This is false. It fails already in $\mathbb F_p$ for $n=2$ if $p=3$ mod $4$, as then we can take $x_1=1$, $x_2=\sqrt{-1}$ and it fails for $n=3$ regardles, as the equation $x^2+y^2+z^2=0$ has degree $2$ in $3$ variables and hence necessarily has a nontrivial solution by Chevalley-Warning.

In general, you should find that for $X$ a random $n \times n$ invertible matrix, the $k \times k$ minors of $X X^T$ for $k$ small will be uniformly distributed in $M_k(\mathbb F_p)$ and hence no more likely to be invertible than a random matrix. However, if $k$ is large relative to $n$ then perhaps these matrices are, at the very least, more likely than usual to be invertible.

Answer 2

This is false even for the $2$-dimensional case: $$\begin{pmatrix}a & b \\ c & d\end{pmatrix} \begin{pmatrix}a & c \\ b & d\end{pmatrix} = \begin{pmatrix}a^2+b^2 & ac+bd \\ ac+bd & c^2 + d^2\end{pmatrix},$$ but $a^2 + b^2$ may be $0$ even when $ad-bc \neq 0$.

However, it is true over any commutative ring that the adjugate computes the inverse of an invertible matrix (in the sense that $\det(A) \in R^\times$).

To see this, note that it suffices to check the formula $$A \cdot \operatorname{adj}(A) = \det(A) \cdot \operatorname{id}.$$ But to check a formula like this, we can go to the universal case $A = (X_{ij})$ over the ring $\mathbb Z[X_{ij},\det^{-1}]$: for any ring $R$ and any invertible matrix $B \in \operatorname{GL}_n(R)$, there is a unique ring homomorphism $f \colon \mathbb Z[X_{ij},\det^{-1}] \to R$ such that $f(A) = B$.

In the universal case, we use that $\mathbb Z[X_{ij},\det^{-1}]$ is a characteristic $0$ domain, so contained in a characteristic $0$ field, where we (supposedly) know the result.