# If $P$ be permutation matrix then $({P^T}AP)x\mathop = \limits^? \left( {\begin{array}{*{20}{c}} * \\ 0 \\ \end{array}} \right)$

• Let $A\in M_n$ is nonnegative(all $a_{ij}\ge0$).
• $x\in C^n$ be eigenvector of $A$ with $r ≥ 1$ positive entries and $n − r$ zero entries .
• There is $P\in M_n$(permutation matrix) such that $Px = \left( {\begin{array}{*{20}{c}} {x'} \\ 0 \\ \end{array}} \right)$ where $x' > 0$.

Can we say that $({P^T}AP)x\mathop = \limits^? \left( {\begin{array}{*{20}{c}} * \\ 0 \\ \end{array}} \right)$ .

## 1 Answer

No, this won't hold for general matrices $A$.

For instance, you can take $n=2$, and $x = \left( {\begin{array}{*{20}{c}} 0 \\ 1 \\ \end{array}} \right)$. Then the permutation matrix you are looking for is $P = \left( {\begin{array}{*{c}{c}} 0 & 1 \\ 1 & 0\\ \end{array}} \right)$, and $Px = \left( {\begin{array}{*{20}{c}} 1 \\ 0 \\ \end{array}} \right).$

Now take $A = \left( {\begin{array}{*{c}{c}} 1 & 1 \\ 1 & 1\\ \end{array}} \right)$. Then $APx = A(Px) = \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ \end{array}} \right)$, and so $(P^TAP)x = P^T(APx) = \left( {\begin{array}{*{20}{c}} 1 \\ 1 \\ \end{array}} \right)$.

• I edited this post. Please see. – user304079 Jan 10 '16 at 15:56