Nullity of the same matrix over a finite and infinite field

Question

Is there a relation between the nullity of the same matrix over a finite field and a infinite field?

Or is there a prime power q so there the differents of nullity of a matrix over, let's say the rational numbers and the field $\Bbb{F_q}$ is small?

Answer 1

Let $p$ be a prime; then the nullity of \begin{bmatrix} p & 2p \\ p & 3p \end{bmatrix} is zero over $\mathbb{Q}$ (and every infinite field of characteristic $0$); on the other hand, the matrix has nullity two over the $p$-element field (and every extension thereof).

If $K$ is an extension of the field $F$, then the rank and nullity of a matrix over $F$ don't change if we consider it as a matrix over $K$, because the rank can be computed with Gaussian elimination and is the same over $F$ and over $K$.

Answer 2

Your question makes most sense if you ask it about a matrix of integers, say $A$. It is slightly easier to talk about the rank, so I will do that: the Rank-Nullity Theorem lets one translate to nullities.

Given such a matrix matrix $A$ we can find (by the Smith Normal Form algorithm) invertible matrices of integers $P,Q$ of determinant $\pm 1$ such that $PAQ=D$ where $$ D=\begin{pmatrix} d_1 & 0 & \dots & 0 & \dots & 0\\ 0 & d_2 & \dots & 0 & \dots& 0\\ \vdots&\vdots&\ddots & 0 &\dots& 0\\ 0 & 0 &\dots& d_k & \dots &0\\ \vdots &\vdots& & \vdots& & \vdots\\ 0 & 0 &&0&&0 \end{pmatrix} \text{ (all other entries $0$)} $$ with non-zero integers $d_1|d_2|\dots|d_k$.

If now $K$ is a field of characteristic $0$, the rank of $A$ is $k$.

If $K$ is a field of characteristic $p$ then the rank of $A$ is $\ell$ where $\ell$ is the least $s$ such that $p\not|d_s$.