Left and right eigenvectors perpendicular to each other

Question

I just read in a textbook on numerical methods that you can always have that the right eigenvectors of a matrix can be taken as orthonormal to the left eigenvectors for a diagonalisable matrix. This seems suprising to me. I would have assumed that you need to have that the matrix is normal or hermitian, but is this also possible for all matrices?

Answer 1

I've never seen this before, but it's quite true.

What is a standard exercise is that when $\lambda\ne\mu$, with $x$ a $\lambda$-eigenvector for $A$ and $y$ a $\mu$-eigenvector for $y$, then $x\cdot y = 0$ (compute $Ax\cdot y$ two ways). But it wasn't obvious to me what happened with the $\lambda$ eigenvectors for both.

Consider the diagonalization $P^{-1}AP=\Lambda$. Then, transposing this equation, we obtain $Q^{-1}A^\top Q=\Lambda$, where $Q=(P^{-1})^\top$. The columns of $P$ and $Q$ are eigenvectors of $A$ and $A^\top$, respectively. But let's look at $Q^\top P$: the $ij$-entry will compute the dot product of the $i^{\text{th}}$ eigenvector of $A^\top$ and the $j^{\text{th}}$ eigenvector of $A$. But $Q^\top P = P^{-1}P = I$, which is the desired result.