non negative matrices and spectrum

Question

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)

Answer 1

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 $SAS^{-1} = -A$.

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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 $SAS^{-1} = -A$.