left and right eigenvectors

Turning to the consequences of symmetry, in the technical sense of a symmetric matrix, the usual analysis observes that for left and right eigenvectors (writing vectors as columns means that rows are transposes of columns)

$$\begin{eqnarray*} M X & = & \lambda X \\ Y^T M & = & \mu Y^T, \end{eqnarray*}$$

so $(Y^T M X)$ takes alternative values according to how the associative law is applied:

$$\begin{eqnarray*} \lambda Y^T X = \mu Y^T X. \end{eqnarray*}$$

Accordingly, either $\lambda = \mu$ or else $Y^T X = 0$, and the vectors are orthogonal. Complications arise when there are several linearly independent eigenvectors with a common eigenvalue, which happens when $M$'s fixed subspaces have higher dimension than 1, but it is all just a matter of going ahead and constructing a basis. Since we wouldn't want a zero vector for an eigenvector (trivial and uninteresting alternative) it could be supposed that left and right eigenvectors belonging to the same eigenvalue could be scaled to make $Y^T X = 1$.

This result holds for any matrix $M$, symmetric or not, and can be summarized by saying that if we make up two matrices, one a column of left eigenrows and the other a row of right eigencolumns, the two matrices are inverses. Or at least partial inverses, because we still don't know how many eigenvectors there actually are, and maybe there are not enough to complete a basis.

If $M$ is itself a symmetric matrix, simply transposing the defining equation $M X = \lambda X$ to get $X^T M^T = X^T M = \lambda X^T$ shows that eigencolumns are eigenrows, keeping the same eigenvalue. Supposing there were enough eigenvectors to make a basis, $B$, a matrix of eigenvectors, would satisfy $B^T = B^{-1}$, a condition expressed by saying that $B$ is orthogonalorthogonal matrix.