Necessary and sufficient conditions for left and right eigenvectors to be equal Suppose I have a matrix $A$ such that for some eigenvalue $\lambda$, the left eigenvector corresponding to $\lambda$ is equal to the conjugate transpose of the right eigenvector corresponding to $\lambda$. That is, I have $\lambda$, $\mathbf u$ and $\mathbf v$ such that
$$
A\mathbf u = \lambda \mathbf u, \qquad \mathbf v A = \mathbf v \lambda\quad\text{and}\quad \mathbf u = \mathbf v^*.
$$
Obviously this will be the case for all eigenvalues if $A$ is Hermitian (or symmetric in the real case). However, I'm interested in the case where $A$ is not symmetric, and the relation holds only for some particular eigenvalue, not for all of them.
I am interested in what properties the matrix $A$ must have in order for this to be the case. Are there any simple necessary and sufficient conditions in terms of the elements of $A$?
Although I have stated the question more generally, I am actually interested in the case where $A$ has real, non-negative elements and is irreducible, and where $\lambda$ is the Perron-Frobenius eigenvalue. So if it helps to assume that $\lambda$ is real or that the elements of $\mathbf{u}$ and $\mathbf{v}$ are positive then please do so.
 A: I am answering the question about the general case.
Let $A=B+C$ denote the decomposition of $A$ into hermitian and antihermitian part.
The assumption that $u$ and $u^*$ are (right and left)eigenvectors with respect to $\lambda$ is then equivalent to the equations
$$Bu+Cu=\lambda u,~Bu-Cu=\lambda^*u.$$
These equations are equivalent to
$$Bu=\Re(\lambda)u,~Cu=i\Im(\lambda)u,$$
where $\Re(\lambda)$ and $\Im(\lambda)$ are the real and imaginary part.
In other words, $u$ is an eigenvector of both $B$ and $C$.
I somewhat doubt that there is a more explicit expression in the entrys of $A$ (in the general case) because the whole situation is unchanged when undergoing a unitary similarity transformation.
A: First, notice that if $A$ is irreducible with non-negatives entries then $A^t$ is also irreducible. The eigenspaces associated to the Perron-Frobenius eigenvalue, $\lambda$, of $A$ and $A^t$ are one dimensional, by Perron-Frobenius theorem. Let $B=A-\lambda Id$ and $k$ its order.
Let $\mathrm{adj}(C)$ be the adjugate of $C$, i.e., $C\ \mathrm{adj}(C)=\det(C)Id$.
Since $B$ and $B^t$ have rank $k-1$ then  $\mathrm{adj}(B)$ and $\mathrm{adj}(B^t)$ have rank 1.
Notice that  $$B\ \mathrm{adj}(B)=\det(B)Id=0_{k\times k}.$$
Therefore, the non-null columns of  $\mathrm{adj}(B)$ are multiples of the Perron-Frobenius eigenvector of $A$ associated to $\lambda$. Analogously, the non-null columns of  $\mathrm{adj}(B^t)$ are multiples of the Perron-Frobenius eigenvector of $A^t$ associated to $\lambda$
Thus, $A,A^t$ have the same Perron-Frobenius eigenvector if and only if $$B^t\mathrm{adj}(B)=0.$$
