TITLE: Eigenvectors and ''eigenrows''
QUESTION [15 upvotes]: Usually we search the eigenvectors of a matrix $M$ as the vectors that span

a subspace that is invariant by left multiplication by the matrix: $M\vec x= \lambda \vec x$.
If we take the transpose problem $\mathbf x M=\lambda \mathbf x$, where $\mathbf x$ is a row vector, we see that the eigenvalues

are the same, but the ''eigenrows'' are not the transpose of the eigenvector (in general).
E.g., for the matrix 
$$\left[
\matrix{0&1\\2&1}
\right]
$$
we find, for the eigenvalue $\lambda=2$,
$$
\begin{bmatrix}
0&1\\2&1
\end{bmatrix}
\left[
\matrix{1\\2}
\right]=
2\left[
\matrix{1\\2}
\right]
$$
$$\left[
\matrix{1&1}
\right]
\begin{bmatrix}
0&1\\2&1
\end{bmatrix}
=
2\left[
\matrix{1&1}
\right] \;\;.
$$
So my questions are: Is there some relation between these ''right'' and ''left'' eigenspaces of the

same matrix? Is there some reason why the ''eigenrows'' are not so studied as the eigenvectors?

REPLY [8 votes]: There is a relation between left and right eigenspaces, as follows:
If $\lambda$ and $\mu$ are distinct real numbers, then the left $\lambda$ eigenspace is orthogonal to the right $\mu$ eigenspace.
Proof: Suppose that $\vec{v}$ is a row vector in the left $\lambda$ eigenspace and $\vec{w}$ is a column vector in the right $\mu$ eigenspace. Then
$$\vec{v} A \vec{w} = \vec{v} (\mu \vec{w}) = \mu (\vec{v} \cdot \vec{w})$$
and also
$$\vec{v} A \vec{w} = (\lambda \vec{v}) \vec{w} = \lambda (\vec{v} \cdot \vec{w}).$$ 
So
$$\mu (\vec{v} \cdot \vec{w})=\lambda (\vec{v} \cdot \vec{w})$$
and $\vec{v} \cdot \vec{w}=0$.
In particular, if $A$ is symmetric, this shows that the eigenspaces for distinct eigenvalues are orthogonal.