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

For example, matrix MAT is full-rank in $GF(2^8)$, why MAT is also full-rank in $GF(2^{16})$ and $GF(2^{32})?$

Thanks

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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 ...

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.

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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.

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Hints:

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