The spectral radius of $A$ and its transpose

Question

Let $A$ be a non-negative irreducible $n\times n$ matrix. Then the function $$f(t)=\rho(tA+(1-t)A^T)$$ is increasing on $[0,1/2]$, and is decreasing on $[1/2,1]$.

Here are the notations.

1. $A$ is non-negative if any entry of $A$ is greater than or equal to $0$.

2. $A$ is irreducible if $A$ is not reducible; and $A$ is reducible if there exists a permutation matrix $P$ such that $$P^T AP=\begin{pmatrix} B&0\\ C&D\end{pmatrix},$$ or equivalently, there exists a permutation $\sigma$ of $\{1,2,\cdots,n\}$ and a $1\leq k\leq n-1$ such that the sub-matrix of $A$ in rows $\sigma(1),\cdots,\sigma(k)$ and columns $\sigma(k+1),\cdots,\sigma(n)$ being $0$.

3. $A^T$ is the transpose of $A$.

4. $\rho(A)$ is the spectral radius of $A$, that is, the largest modulus of the eigenvalues of $A$.

And now I have no idea on it. However, it is intuitively right. As there are more symmetry in the matrix, the spectral radius becomes larger.

Answer 1

$f(t)$ does attain maximum at $t=\frac12$. However, I am not so sure about the monotonicity of $f$ on $[0,\frac12]$ or $[\frac12,1]$.

Proof. As $A$ is nonnegative and irreducible, so is $B_t = tA+(1-t)A^T$. Let $u$ be a unit Perron eigenvector $B_t$. From $u^TAu = (u^TAu)^T = u^TA^Tu$, we get $$ \rho(B_t) = u^TB_tu = u^TB_{1/2}u. $$ Since $B_{1/2}$ is nonnegative, irreducible and real symmetric, we have $\rho(B_{1/2}) = \max_{\|v\|=1}v^TB_{1/2}v$. Hence $\rho(B_t)\le\rho(B_{1/2})$.