I was trying to prove that, For $n\times n$ matrices $P$ and $Q$, $PQ$ and $QP$ have same eigenvalue(not given $Q$ is invertible or not). My approach is as follows:We have $Q(PQ)Q^{-1}=QP$, from this we can directly see that they have same eigenvalue if $Q$ is invertible. Now, as changing position of entries doesn't change the eigenvalues, hence, I have an idea to prove the rest of the problem.
Can we always find invertible matrix by changing the positions of entries of a non-invertible matrix?
If the above statement is true then we can always get an matrix after rearranging the entries which is non-invertible and has same eigenvalues as the privious non-invertible matrix had and as the reduced matrix(after rearranging entries) $Q'$ satisfy:$$Q'(PQ')Q'^{-1}=Q'P$$hence, eigenvalues of $PQ'$ and $Q'P$ are same which is also equal to eigenvalue of $PQ$.
Now, my question is , is the quoted statement is true?