left and right eigenvalues On the Stochastic Matrices article in Wikipedia there's a claim that left and right eigenvalues of a square matrix are the same. I tried looking this up, but can't find an explanation, only for hermitian matrices with real eigenvalues. Is it correct?
 A: The left eigenvalues of a matrix are the zeroes of its minimal polynomial.
The right eigenvalues of a matrix are the zeroes of its minimal polynomial.
A: In essence, it's just saying that 

If A is a square matrix, then its eigenvalues are equal to the eigenvalues of its transpose $A^T$, since they share the same characteristic polynomial, i.e. $det(\lambda I-A) = det(\lambda I-A^T)$.

To see this, let $v^TA=\gamma v^T$ where $(\gamma,v)$ are a pair of left eigenvalue and eigenvector. Since $v^TA=\gamma v^T \iff A^Tv=\gamma v$, $\gamma$ is also one right eigenvalue of $A^T$. As stated above, eigenvalues of a square matrix and its transpose are the same. Therefore, the left and right eigenvalues of a square matrix are the same.
A: It can also be proved as following:
$AV=V \Lambda$, where $V$ has its column vectors as the right eigenvectors of $A$ and $\Lambda$ contains the (right) eigenvalues of $A$ on its diagonal. Multiply both sides by $V^{-1}$ on the right end, we have $A=V\Lambda V^{-1}$. Then multiply both sides by $V^{-1}$ again on the left end, we have $V^{-1}A=\Lambda V^{-1}$.   The rows of $V^{-1}$ actually contains the left eigenvectors of $A$. We thus show that the eigenvalues for left and right eigenvectors are the same.
A: Consider a square matrix $\mathbf{A}$ and its right eigenpair ($\lambda$, $\mathbf{u}$). Then, it holds
\begin{equation*}
\mathbf{Au}=\lambda\mathbf{u} \iff (\mathbf{A}-\lambda\mathbf{I})\mathbf{u}=\mathbf{0}
\end{equation*}
Thus, $\mathbf{A}-\lambda\mathbf{I}$ has at least one zero singular value, and, therefore, it has at least one left singular vector, $\mathbf{v}$, such that  $\mathbf{v}^T(\mathbf{A}-\lambda\mathbf{I})=\mathbf{0}^T \iff \mathbf{v}^T\mathbf{A}=\lambda \mathbf{v}^T$.
