Assume that $A, B\in\mathbb{R}^{n\times n}$ are both diagonalizable then $AB=BA$ implies that $A$ and $B$ are Simultaneously Diagonalizable.
I have seen a proof of this statement that goes as follows:
Let $u$ be an eigenvector of A such that $Au=\lambda u$ then consider $ABu=BAu=\lambda Bu$ which implies that $Bu$ is also an eigenvector of A with eigenvalue $\lambda$. The author proceeds to claim that this implies that $Bu=\gamma u$ that is that $Bu$ has to be a multiple of $u$ concluding that $u$ is also an eigenvector of B. The issue I am having is how does one conclude that $Bu=\gamma u$? Why can't $Bu$ be any other vector of A. The only way I see this happening is if the multiplicity of every eigenvalue of $A$ is 1 and therefore the statement follows. Can anyone offer any clarification as to how $Bu=\gamma u$ is guaranteed without the multiplicity restriction?
