Left eigenvectors of a complex nonsymmetric matrix and the diagonalization transformation Let $A$ be a diagonalizable complex nonsymmetric matrix. Let $D=S^{-1}AS$ be a diagonalization of $A$, so $D$ is a diagonal matrix.
The columns of $S$ are (right) eigenvectors of $A$. What is the relationship between left eigenvectors of $A$ and $S^{-1}$?
 A: The rows of $S^{-1}$ are the left eigenvectors of $A$:
Assuming $A,D,S\in\mathbb C^n$ with $D=S^{-1}AS$ and $D$ is diagonal. By multiplying with $S^{-1}$ from the right, we get
\begin{align}
DS^{-1} = S^{-1}A.\tag{1}
\end{align}
Let $v_1,\dots,v_n\in\mathbb C^n$ be the row vectors such that of $S^{-1}$
$$
S^{-1} = \begin{bmatrix}v_1\\\vdots\\v_n\end{bmatrix}
$$
which makes $v_i$ the $i$th row of $S^{-1}$.
Let $\lambda_1,\dots,\lambda_n\in\mathbb C$ be the diagonal elements of $D$. (This makes them the Eigenvalues including muliplicities of $A$.) This means that we have
$$
D = \begin{bmatrix}
\lambda_1 & \\
          & \lambda_2 \\
          &            & \ddots \\
          &            &        & \lambda_n
\end{bmatrix}.
$$
Now we can write (1) in the following form:
$$
\begin{bmatrix}\lambda_1v_1\\\vdots\\\lambda_nv_n\end{bmatrix} 
= D S^{-1} = S^{-1} A = \begin{bmatrix}v_1A\\\vdots\\v_nA\end{bmatrix} 
$$
Note that $v_iA\in\mathbb C^n$ is well defined row vector for all $i$.
By taking the $i$th row of the above equation we get
$$
\lambda_i v_i = v_i A,
$$
where $v_i$ is the $i$th row of $S^{-1}$ and $\lambda_i$ is the $i$th Eigenvalue. This makes the rows of $S^{-1}$ the left Eigenvectors of $A$.
A: Assume first that $A$ is normal, i.e. that it can be diagonalized by unitary matrices (as a consequence of the spectral theorem).
Since $S$ is unitary, its inverse $S^{-1}$ is its conjugate transpose.
We know that the left eigenvectors of $A$ are the conjugate transpose of the right eigenvectors of $A$.
Since the right eigenvectors of $A$ are the columns of $S$, it follows that the left eigenvectors of $A$ are the rows of the conjugate transpose of $S$.
But as was mentioned at the outset, the conjugate transpose of $S$ is just $S^{-1}$ (since $S$ is unitary).
Therefore, the left eigenvectors of $A$ are the rows of $S^{-1}$.
Another way to think of this is that $$(S^{-1} A S)^* = S^* A^* (S^{-1})^* = S^{-1} A^* S$$ is the diagonalization of the conjugate transpose of $A$, and we know that a vector is a left eigenvector for $A$ if and only if its conjugate is a right eigenvector for the conjugate transpose of $A$, namely $A^*$.
See for example here for more detail.

Now consider the case where $A$ is diagonalizable but not normal (examples exist, see enter link description here). Then we have: $$D = S^{-1} A S$$ for $S$ NOT unitary.
It still holds that $y$ is a left eigenvector of $A$ if and only if it is the conjugate of a right eigenvector for $A^*$: $$y^T A = \lambda y^T \iff A^* \bar{y} = \bar{\lambda} \bar{y}$$
So let us examine the diagonalization of $A^*$: $$(S^{-1} A S)^* = S^* A^* (S^{-1})^*$$
Note that all of the steps are the same here except that we cannot simplify at the end, since $S$ is not unitary.
If $w$ is a right eigenvector of $A^*$, then it is a column of $(S^{-1})^*$. Therefore $\bar{w}$ is a left eigenvector of $A$.
Hence $y$ is a left eigenvector of $A$ if and only if $\bar{y}$ is a column of $(S^{-1})^*$.
