non negative matrices and spectrum Let $r$ be the Perron Frobenius eigenvalue of a non negative matrix $A$. It seems that it's true that if $r, -r \in Spec(A)$ then $-A = S\cdot A \cdot S^{-1}$ with $S$ a diagonal matrix with $\pm 1$ , but I have no clue how to prove it (the reciprocal is trivial)
 A: This will hold if $A$ is irreducible as a consequence of the Perron Frobenius theorem for non-negative matrices.
In particular: with statement 6,7, and 8 here, we can deduce that the period $h$ of $A$ is even, and we can further suppose without loss of generality that $A$ has the form
$$
A = \begin{pmatrix}
O & A_1 & O & O & \ldots & O \\
O & O & A_2 & O & \ldots & O \\
\vdots & \vdots &\vdots & \vdots & & \vdots \\
O & O & O & O & \ldots & A_{h-1} \\
A_h & O & O & O & \ldots & O
\end{pmatrix},
$$
where $O$ denotes a zero matrix and the blocks along the main diagonal are square matrices. Suppose that the $j$th diagonal $0$-block is has size $k_j \times k_j$. Then, let $S$ denote the diagonal matrix
$$
S = \pmatrix{I_{k_1}\\ & -I_{k_2}\\ && \ddots \\ &&& I_{k_{h-1}}\\ &&&& -I_{k_{h}} },
$$
where $I_k$ denotes the identity matrix of size $k$. Using block-matrix multiplication, verify that $SAS^{-1} = -A$.

The statement does not necessarily hold if $A$ is reducible. As an example, consider
$$
A = \pmatrix{0&2&0\\2&0&0\\0&0&1}.
$$
In this case, $A$ and $-A$ do not have the same eigenvalues, so there is no $S$ such that $SAS^{-1} = -A$.
