Left/Right Eigenvectors Let $M$ be a nonsymmetric matrix; suppose the columns of matrix $A$ are the right eigenvectors of $M$ and the rows of matrix $B$ are the left eigenvectors of $M$. 
In one of the answers to a question on left and right eigenvectors it was claimed that $AB=I$. Is that true, and how would you prove it?

No. Let 
$$
M = \begin{bmatrix} 1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 4 & 2 \\
0 & 0 & 0 & 2
\end{bmatrix}
$$
Then two right eigenvectors for $1$ are $e_1$ and $e_2$, and two left eigenvectors for $1$ are $(\sqrt{2}/2) ( e_1 \pm e_2 )$. When you put these into $A$ and $B$, the upper left corner of your product will not be the $2 \times 2$ identity.
I have a bad feeling this question's about to be edited to add the hypothesis that the eigenvalues be distinct...
When that happens, the answer will still be no, for if we permute the rows of $A$ to form $A'$ they'll still all be eigenvectors for $M$, but the product $AB$ will undergo the same row permutation, so that even if $AB = I$, we'll have $A'B \ne I$. 
