Can an Perron eigenvector of a non-symmetric irreducible nonnegative matrix be also a Perron eigenvector of its transpose?

Question

Let $\mathbf{X}$ be nonnegative irreducible matrix such that $\mathbf{X} \ne \mathbf{X}^T$. Let $\mathbf{p}(\mathbf{A})$ denote a right eigenvector corresponding to the Perron root of $\mathbf{A}$, $\rho(\mathbf{A})$. Similarly, $\mathbf{q}(\mathbf{A})$ denotes a left eigenvector corresponding to $\rho(\mathbf{A})$. (In the textbook I'm reading, left eigenvectors are column vectors corresponding to the right eigenvector of $\mathbf{A}^T$.)

A remark in the textbook seems to imply that as long as $\mathbf{X} \ne \mathbf{X} ^ T$, $\mathbf{p}(\mathbf{X}) \ne \alpha \mathbf{p}(\mathbf{X}^T)$ for any constant $\alpha$.

EDIT: As requested by @user1551, here is the exact remark from Fundamentals of Resource Allocation in Wireless Networks:

Finally, as $\mathbf{X}^T\in W_K(\mathbf{X})$ for any $\mathbf{X} \in X_K$, it follows from Theorem 1.14 that $\rho((1-\mu) \mathbf{X} + \mu\mathbf{X}^T)$ is a concave function on $\mu \in [0,1]$. If, in addition, $\mathbf{X}\ne \mathbf{X}^T$, we have $\mathbf{p}(\mathbf{X}) \ne \alpha \mathbf{p}(\mathbf{X^T})$ for any constant $\alpha$.

$X_K$ is the set of nonnegative irreducible $K$-by-$K$ matrices. For $\mathbf{X}\in X_K$, $W_K(\mathbf{X}) = \{\mathbf{Y}\in X_K : \mathbf{q}(Y)\circ \mathbf{p}(\mathbf{Y}) = \mathbf{q}(X)\circ \mathbf{p}(\mathbf{X}) \in \Pi_K^{+}\}$, where $\Pi_K^+$ is the standard simplex restricted to positive values.

In an effort to prove the remark, I started a proof by contradiction. Suppose that $\mathbf{p}(\mathbf{X}^T)\equiv \mathbf{q}(\mathbf{X})$ is also a right Perron eigenvector of $\mathbf{X}$. By definition of left and right Perron eigenvectors, $\mathbf{X}^T \mathbf{q}(\mathbf{X}) = \rho(\mathbf{X})\mathbf{q}(\mathbf{X})$ and $\mathbf{X}\mathbf{q}(\mathbf{X}) = \rho(\mathbf{X})\mathbf{q}(\mathbf{X})$.

Subtracting the two equations, I get $(\mathbf{X}-\mathbf{X}^T)\mathbf{q}(\mathbf{X}) = \mathbf{0}$, and I'm not sure how to proceed. The only conclusion I get from this equation is that $\mathbf{q}(\mathbf{X})$ is in the null space of $\mathbf{X} - \mathbf{X}^T$.

Answer 1

Your contradiction approach will fail because the claim is not false. In terms of simple examples: consider any doubly stochastic matrix, e.g. a permutation matrix and in particular one that is not symmetric (I’d suggest the cyclic shift matrix).

Answer 2

Another argument for why the stated claim is false.

The space of $n \times n$ irreducible Perron matrices $X$ has dimension $n^2$ (if you want, restrict yourself to positive matrices).

Fix $\rho >0$ and $p$ a positive vector. The space of $n \times n$ irreducible Perron matrices $X$ such that $Xp = \rho p$ has dimension $n^2-n$ (we have n linear equations to check).

The space of $n \times n$ irreducible Perron matrices $X$ such that $Xp = \rho p$ and $p^t X = \rho p^t$ has dimension $n^2-2n+1 = (n-1)^2$ (we have $n$ more linear equations to check, the last one being redundant).

The space $n \times n$ irreducible symmetric Perron matrices $X$ such that $Xp = \rho p$ is included in the later, and has dimension $n(n+1)/2-n = n(n-1)/2$. If $n>2$, then $n(n-1)/2 < (n-1)^2$, so that there are non-symmetric Perron matrices with main eigenvector $p$ and main eigen-covector $p^t$.