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Two matrices $A, B$ can be simultaneously diagonalized if there is an invertible matrix that diagonalizes both of them. In other words, if there is a (possibly complex) basis in which both matrices are diagonalized.
a) If $A$ and $B$ can be simultaneously diagonalized, show that $A B=B A$.

b) Conversely, if $A B=B A$, and if one of these matrices, say $A$, has distinct eigenvalues (so the eigenspaces all have dimension one), show they can be simultaneously diagonalized.

SUGGESTION: Say $\lambda$ is an eigenvalue of $A$ and $\vec{v}$ a corresponding eigenvector: $A \vec{v}=\lambda \vec{v}$. Show that $B \vec{v}$ satisfies $A(B \vec{v})=\lambda B \vec{v}$ and deduce that $B \vec{v}=c \vec{v}$ for some constant $c$ (possibly zero). Thus, the eigenvectors of $A$ are also eigenvectors of $B$. Why does this imply that (in this case where $A$ has distinct eigenvalues) in a basis where $A$ is diagonal, so is $B$ ?

REMARK: This result extends to any two commuting $n \times n$ matrices $A$ and $B$, assuming that $A$ and $B$ can each be diagonalized.
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