GLn group action rank

Question

Having the group action $GL(n,K) \times K^{n\times m} \to K^{n\times m} \; (g,A)↦gA $. The rank is an invariant but not an separating one (why?), how then does the orbits look like? If we acted also from the other side it would be JNF, but so its applying Gauss and for me it seems like only the rank is invariant?

Answer 1

It is not true that the rank is the only invariant. Note that the kernel of $A$ (or equivalently, the row-space) is also invariant.

Each orbit under this action corresponds to a single reduced row echelon form matrix.

To see that the rank is not an orbit-separating invariant, verify that the matrices $$ \pmatrix{1&0&0\\0&1&0}, \quad \pmatrix{1&0&0\\0&0&1} $$ do not lie in the same orbit.