Let $S$ be invertible matrix whose columns are the left eigenvectors of $A$. Then, prove that the columns of $(S^*)^{-1}$ are right eigenvectors.

Question

Let $S$ be invertible matrix whose columns are the left eigenvectors of $A\in M_n(C)$. Then, prove that the columns of $(S^*)^{-1}$ are right eigenvectors.

Here $^* $ is complex conjugate transpose and $^t$ is transpose

Since the columns of $S$ are left eigenvectors of $A$, so let

$$(S[:,I])^*A= k_i (S[:,I])^*$$

where $S= [(S[:,1]) , (S[:,2])\ldots (S[:,n])]$

Note that $$S^* = [(S[:,1])^* , (S[:,2])^*\ldots (S[:,n])^*]^t = B (say)$$

Now, we want to show that,

$AB^{-1}[:,i] = m_iB^{-1}[:,I]$ for some $m_i$

here $B^{-1}[:,i] \neq 0$ because $S$ is invertible.

How can I proceed from here?