# Given 2 matrices generate a reducible algebra, show they have a common eigenvector

Two matrices A, B generate an algebra.... the span of all words made with A and B... example of an element of the algebra: $A^kB^nA^m + I + B^sA^q$ etc... (exponents are all nonnegative). This algebra is given to be reducible. Let W be the nontrivial subspace which this algebra maps to itself.

Please look at Chris' definition of reducible algebra here: What is a reducible algebra?

which applies to this question.

How do I show that A and B must have a common eigenvector?

clues I've gathered: 1) I know that W contains an eigenvector of A, as well as an eigenvector of B. But I don't know how to use this to show that there is a common eigenvector.

2) I know there's a matrix S such that $S^{-1}AS$ and $S^{-1}BS$ are both block diagonal of the form: $\left( \begin{array}{ccc} Q & R \\ 0 & T \\ \end{array} \right)$

Thanks.

-

With your definitions of algebra and reducibility, $A$ and $B$ do not neccessarily share a common eigenvector. For instance, consider $A=\operatorname{diag}(1,2,3,4)$ and $B=R\oplus R$, where $R$ is the rotation matrix $\pmatrix{0&-1\\ 1&0}$.