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

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 $$\mathbf{X} - \mathbf{X}^T$$.

• "A remark in the textbook seems to imply that...". What exactly does the remark say? – user1551 May 11 '20 at 9:20
• I included an edit to include the remark. @user8675309 provided a good counterexample. For doubly stochastic matrix, the $\rho$ function is linear on $\mu \in [0,1]$. – owovrokfop May 11 '20 at 9:33

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 $$p^t$$.