Orthogonality and Identity Matrix

Question

If $A$ is a $m\times n$ matrix of full rank $m$, and $k\neq k'$ are invertible $m\times m$ matrices, do we have : $A=k^{-1} k' A$ implies $k^{-1} k' = I$?

It looks obvious true at the first sight, but then I realize that it may be false. If you rearrange the previous equation, then $(I-k^{-1} k' )A=0$. What if $(I-k^{-1} k' )$ is just orthogonal to $A$?

Also, if A is singular (not full rank), then may be $k\neq k'$ but still $kA=k'A$. I found an example. But could anyone explain the reason behind?

Answer 1

Let's fix some bases on our vector spaces so that we can identify matrices to linear morphisms.

If $A$ is surjective then for any $x$, $x=Ay$ for some vector $y$, but we have $A(y)=g^{-1}g'A(y)$ i.e. $x=g^{-1}g'(x)$ whence $g=g'$.

For a counter example when $A$ is singular, think of $A$ 'very' singular (like $A=0$ for instance).

Answer 2

if $A$ is a $m\times n$ matrix.

If $rank(A)=m,$ then $m=n$, that is, the matrix is a square matrix is a special case, and the answer is obvious.

If $m < n,$ taking the first $m$ columns from the equation, the new equation still holds, reducing to the case of $m=n$.

If $A$ is rank deficient, consider the special case where $A = 0.$