Permutation Matrix P and Diagonal Entries of PA I am trying to the problem below and I have no idea how to begin. I want to show that:
If $A\in\mathbb{R}^{n\times n}$ is nonsingular, then there exist a permutation matrix $P$ such that $PA$ has nonzero diagonal entries.
I know what permutation matrices are but how to connect with this, I have no clue. Can someone please help me out here.

Here's an outline . . .

Suppose $A$ is non-singular. Then $\det(A)$ is nonzero.

But $\det(A)$ is the sum of signed products of generalized diagonals, hence at least one generalized diagonal, $D$ say, must have all nonzero entries.

Let $Q$ be the permutation matrix corresponding to $D$, and let $P = Q^{-1}$.

Since $PQ = I_n$, it follows that entries of the diagonal of $PA$ are some permutation of the entries of $D$, hence are all nonzero.
