# If $A$ and $B$ are commuting Hermitian matrices, then they have the same eigenvectors?

If $$AB = BA$$ and both $$A$$ and $$B$$ are Hermitian matrices, then I can show that if $$Av = \lambda v$$, then $$ABv = BAv = B\lambda v = \lambda Bv$$. So $$Bv$$ is an eigenvector of $$A$$ as well.

Where I am stuck is how this implies that $$v$$ must also be an eigenvector of $$B$$?

• It doesn't${}$. – Lord Shark the Unknown Apr 25 '19 at 17:58
• The crucial thing to prove is that each eigenspace of $A$ is invariant under $B$. – Lord Shark the Unknown Apr 25 '19 at 18:23
• You should always consider trivial cases to check a statement. Here $A=I$ is commuting with any matrix. But it's quite clear that any non zero vector could not be automatically an eigenvector of $B$. – zwim Apr 25 '19 at 18:43

As others have pointed out, the exact result stated in your question isn't true. What is true is that $$A$$ leaves the eigenspaces of $$B$$ invariant and vice versa. If $$v$$ is an eigenvector of $$A$$ with eigenvalue $$\lambda$$, then $$A(Bv) = (AB)v = (BA)v = B(Av) = B(\lambda v) = \lambda (Bv)$$ Thus, $$Bv$$ is an eigenvector of $$A$$ with eigenvalue $$\lambda$$. Let $$V_\lambda(A)$$ denote the subspace of eigenvectors of $$A$$ with eigenvalue $$\lambda$$. We have shown $$B(V_\lambda(A)) \subset V_\lambda(A)$$. This is the result that holds in general.
If $$V_\lambda(A)$$ is one dimensional, we easily obtain a result that looks like the result you were originally after. Take $$v \in V_\lambda(A)$$. Then $$Bv \in V_\lambda(A)$$, so, since $$V_\lambda(A)$$ is one dimensional, $$Bv = \xi v$$. In words, if $$v$$ is an eigenvector of $$A$$ with nondegenerate eigenvalue $$\lambda$$, then $$v$$ is an eigenvector of $$B$$.
In general, the result you're thinking of is true, you just have to phrase it correctly. It's not the case that every eigenvector of $$A$$ is an eigenvector of $$B$$. The correct statement is that there exists a basis of vectors $$v_1 , \ldots , v_n$$ for the space such that $$v_i$$ is an eigenvector of $$A$$ and $$B$$ for all $$i$$. The truth of this statement requires that $$A$$ and $$B$$ are diagonalizable, which is necessarily true if $$A$$ and $$B$$ are Hermitian.
What you are trying to prove is in fact false. Consider the commuting Hermitian matrices $$A=\begin{pmatrix} 1&0\\ 0&1 \end{pmatrix} \qquad\text{ and }\qquad B=\begin{pmatrix} 1&0\\ 0&2 \end{pmatrix}.$$ Clearly $$(1,1)$$ is an eigenvector of $$A$$ but not of $$B$$.