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 $\mathbb{R}$).



'Positive-definite' matrices over finite fields
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. 


