TITLE: Left/Right Eigenvectors
QUESTION [6 upvotes]: 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?

REPLY [5 votes]: Most counter-examples found here have to do with the inherent ambiguities in defining the eigenvectors. There are essentially two kinds:

For repeating eigenvalues, the eigenvectors spanning their space are not uniquely defined. Only their subspace is. Any linearly indepedent basis for their space will do (extreme example $M=I$: any full rank matrix $A$ qualifies as matrix of right singular vectors. Likewise, any full rank matrix $B$ can be chosen independently as a valid basis of left singular vectors).
If we exclude this by forcing distinct eigenvalues we still have each eigenvector only being unique up to a scalar multiple, since any $q$ satisfying $M q = \lambda q$ will also lead to $\alpha q$ satisfying the same equation for any $\alpha \neq 0$. Forcing $\left\|q\right\|=1$ eliminiates this only partially, in the real-valued case there is still a sign ambiguity (e.g., Roberts reply), in the complex-valued case a phase ambiguity.

These ambiguities aside, let us not forget that left singular vectors of $M$ are right singular vectors of $M^T$. So if $M$ has an EVD of the form $M=A \Lambda A^{-1}$, then $M^T = A^{-T} \Lambda A^T$.  This shows that $B = A^{-1}$ does qualify as a pair of left/right singular vectors. It's only that if we compute $A$ and $B$ independently, we are not guaranteed to pick the pair of inverses among the set of ambiguities. If eigenvalues are distinct we might get a diagonal matrix (due to the scaling ambiguities), if they are not we might get something pretty arbitrary.