4

Suppose $A \in M_n$ has distinct eigenvalues $a_1,\dots,a_n$ and that $A$ commutes with a given matrix $B \in M_n$ so that $AB=BA$.

a. Prove A and B are simultaneously diagonalizable

I was able to show that $B$ is diagonalizable but I cannot figure out how to show that $A$ and $B$ are simultaneously diagonalizable.

Community
  • 1
Fernando Martinez
  • 6,698
  • 19
  • 74
  • 108

2 Answers2

8

Having distinct eigenvalues means that for each eigenvalue $\lambda$, there is a one dimensional subspace $S(\lambda)$ spanned by the respective eigenvector $x$. In this case $Ax=\lambda x$, ans $A(Bx)=B(Ax)=\lambda (Bx)$ so $Bx\in S(\lambda)$ and hence $Bx=\beta x$ for some $\beta$. Therefore $x$ is also an eigenvector of $B$. So $A$ and $B$ have same eigenvectors and therefore can be simultaneously diagonalized.

Arash
  • 11,131
  • So if you have the same eigen vector then they can be simultaneously diagonalzied? – Fernando Martinez Sep 17 '16 at 22:21
  • Yes in this case, try to change the basis of your space to $(x_1,\dots, x_n)$, eigenvectors of $A$ and $B$. $A$ and $B$ in the new basis become diagonal. Writing down the change-of-basis transformation give you the diagonalization. – Arash Sep 17 '16 at 22:27
2

Under these conditions, each eigenspace of $A$ is one-dimensional. Further, $B$ fixes each of these eigenspaces: namely, if $Ax=\lambda x$, then $$A(Bx)=BAx=B\lambda x=\lambda (Bx).$$ It follows that $B$ maps each eigenvector of $A$ to its multiple and hence $B$ is diagonal in a basis consisting of $A$-eigenvectors.

Peter Franek
  • 11,522