What is the definition of a singular value over a finite field $\mathcal{F}$ of a matrix ${\bf A}$ in $\mathcal{F}^{m\times n}$? Is there a geometric intuition in the same manner as with the real case where the eigenvalues are the radii of the ellipse $\frac{\|{\bf A}{\bf x}\|^2}{\|{\bf x}\|^2}$?



Singular value decomposition over finite fields?
There is no definition of a singular value of a matrix over a finite field. You could
define it to be a non-zero eigenvalue of $A^TA$, but this does not really work as you might expect.


Over the reals, the eigenvalues of $A^TA$ are non-negative and the smallest singular value
is a measure of how close $A$ is to being a non-invertible matrix. Further, there are
stable algorithms to compute it, whereas we cannot compute the rank of $A$ in a stable fashion.


Over the complex numbers, you would use the eigenvalues of ${\bar A}^TA$ as singular values
instead the eigenvalues of $A^TA$. Over finite fields you might use $\sigma(A^T)A$,
where $\sigma$ is a field automorphism. This is a problem, we have ore choice than we do over the reals and complexes.


There is no difficulty in computing ranks over matrices over finite fields anyway.


Over finite fields, eigenvalues are of limited use. We can get the characteristic polynomial of a matrix, and factor it; the zeros of the factors are the eigenvalues. These eigenvalues
would lie in some extension $E$ of your finite field $F$; if I came along with $E$ and asked
which elements of $E$ were the eigenvalues, you would have a lot of work and your final
answer would depend on exactly how I described $E$. (Even over the reals, eigenvalues
are useful if the matrix is small, or normal, but they are much less use otherwise.
Google 'pseudospectra')


