I'm doing a work about Perron-Frobenius theorem, and I'm trying to give a proof of it. I'm stuck, because I found that an irreducible matrix can't have a row or a column of zeros. I understand that this is a property that primitive matrix have, but why an irreducible matrix verify this too?
Thanks!
A matrix $A$ is reducible if there is a permutation matrix $P$ such that $$\tag{1} P^TAP=\begin{bmatrix}A_{11} & A_{12} \\ 0 & A_{22}\end{bmatrix}. $$ If $A$ would have a zero row, then it could be permuted to $$ P^TAP=\begin{bmatrix}A_{11} & A_{12} \\ 0 & 0_{1\times 1}\end{bmatrix}, $$ by some $P$, which is (1) with $A_{22}=0$. Similarly, if $A$ has a zero column, it can be permuted to $$ P^TAP=\begin{bmatrix}0_{1\times 1} & A_{12} \\ 0 & A_{22}\end{bmatrix}, $$ which is (1) with $A_{11}=0$.
Hence, if $A$ has a zero row/column, it is reducible. Reversing this implication shows that an irreducible matrix cannot have a zero row and/or column.
