Why is a full-rank matrix in a finite field also full-rank in an expanded finite field?

Question

E.g., let matrix $M$ be full-rank in $GF(2^8)$. Why is $M$ also full-rank in $GF(2^{16})$ and $GF(2^{32})?$

Answer 1

A matrix is of full rank, if and only if it has a maximum size square submatrix with a non-zero determinant. The zero element of an extension field is the same as that of the subfield, so ...

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Note that the same argument works for all extension fields. For example, if a matrix has full rank over $\mathbb{Q}$ it has full rank over $\mathbb{R}$ and $\mathbb{C}$ as well.

Answer 2

Hints:

1. the rank can be caracterized with (non-vanishing) determinants,
2. determinants are polynomial functions of the coefficients.

Answer 3

It's worth remarking that there is no need to use determinants here. A matrix with entries in a field $F$ has rank $r$ if and only if it can be transformed using $F$-elementary row and column operations to a matrix which has $I_{r \times r}$ in the "top left corner" and zeros elsewhere. By $F$-elementary, I mean that the allowed operations are: interchanging rows (or columns); multiplying rows or columns by a non-zero element of $F$, and replacing a row (or similarly column) by itself + an $F$-multiple of another one. This only depends on the smallest field containing all entries of the matrix, so moving to a larger field does not affect the rank.