Computing eigenvectors using the Schur decomposition

Let’s assume we have computed a Schur decomposition $A=QT{Q}^H$. How can we obtain the eigenvectors?

Let’s focus on the evecs of $T$. For a triangular matrix, finding evecs is relatively straightforward.

$$T=\left(\begin{array}{ccc}{T}_{11}& T_{12}& T_{13}\\ 0& \lambda & T_{23}\\ 0& 0& T_{33}\end{array}\right)$$

Assume we have an eigenvalue/eigenvector pair:

$$Tv=\lambda v$$

This is the linear system we need to solve to find $v$:

$$T-\lambda I=\left(\begin{array}{ccc}{T}_{11}-\lambda I& T_{12}& T_{13}\\ 0& 0& T_{23}\\ 0& 0& T_{33}-\lambda I\end{array}\right)$$

and $(T-\lambda I)v=0.$

Do a back-substitution starting with $v_3$:

$$(T_{33}-\lambda I)v_3=0.$$

So $v_3=0$ assuming that $\lambda$ is not an eval of $T_{33}.$

The next equation is:

$$0v_2+T_{23}0=0.$$

So $v_2$ can be chosen arbitrarily. Let’s choose $v_2=1$.

The last equation is

$$(T_{11}-\lambda I)v_1+T_{12}1+T_{13}0=0.$$

There is a unique solution for $v_1$. The final solution is summarized as:

$$v=\left(\begin{array}{c}-(T_{11}-\lambda I{)}^{-1}{T}_{12}\\ 1\\ 0\end{array}\right)$$

The computational cost is $O(n^2)$ per eigenvector because we only need to solve linear systems with triangular matrices.