# 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 $S^{-1}$?

• This is confusing. If $A$ is not supposed symmetric (or Hermitian, or normal) there is no reason that $S$ could be taken to be unitary. (In fact it cannot if $A$ is no normal.) Jun 2, 2016 at 13:58
• Also you don't need to (indeed are not supposed to) Captalise All Words In The Title. Jun 2, 2016 at 14:02
• @MarcvanLeeuwen Unitarity is an edit by William, not in the original question. Jun 2, 2016 at 14:15
• OK, so I rolled back to (almost) the original question. Jun 2, 2016 at 14:19
• A note for anyone computing $S^{-1}$ with Matlab or Lapack: the definition of the left eigenvector adopted there uses a conjugate of the vector defined in the two answers below. Therefore, the left eigenvectors computed by these libraries are the conjugates of the rows of $S^{-1}$. Jun 6, 2016 at 11:43

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 $A$.

• This is assuming that $A$ is normal, whereas not all diagonalizable matrices are normal. Jun 2, 2016 at 14:10
• @William Why? S/he just multiplied $D=S^{-1}AS$ on the right by $S^{-1}$ and took the $i^\text{th}$ row of both sides. ( I think there's a typo column->row in the second sentence ). Jun 2, 2016 at 14:50
• Yes, I fixed the typo. Jun 2, 2016 at 15:13
• There appear to be a lot of steps missing, so I don't quite see the argument, but we can't assume that the rows of $S^{-1}$ are orthonormal unless $S$ is unitary, i.e. unless $A$ is normal. Jun 2, 2016 at 15:48
• I increased the verbosity of my answer. The matter should be crystal clear now :) Jun 3, 2016 at 7:32

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 $(S^{-1})^*$.

• I think $D^* = S^*A^*(S^{-1})^*$ is the diagonalization of the conjugate transpose of $A$. Jun 2, 2016 at 13:40
• Isn't $S$ unitary only when $A$ is Hermitian? Jun 2, 2016 at 13:51
• No, $S$ is unitary if and only if $A$ is normal (spectral theorem -- see chapter 7 of Axler, Linear Algebra done right, for example), and there are plenty of normal matrices which are not unitary. Since my post assumed that $A$ is normal, I added a section which considers the case when $A$ is diagonalizable but not normal. Jun 2, 2016 at 14:12
• I took a simple matrix to Matlab A=[1+i,2-i,3;1,i,0.5i;5i,7,-2], diagonalized it, but haven't seen any relation between $S^{-1}$ and the left eigenvectors computed by Matlab. Jun 2, 2016 at 14:37
• To tie your conclusion to the other answer, if $\bar{y}$ is a column of $(S^{-1})^{*}$, then $y$ is a row of $S^{-1}$, so it's true even if $A$ is not normal. Jun 2, 2016 at 23:01