Let $A$ and $B$ be an $n \times n$ matrices.
Suppose that all the eigenvalues of $A$ are distinct and the matrices $A$ and $B$ commute, that is $AB=BA$.

Then prove that each eigenvector of $A$ is an eigenvector of $B$.

(It could be that each eigenvector is an eigenvector for distinct eigenvalues.)

Hint.

Each eigenspace for $A$ is one dimensional.

Proof.

Since $A$ has $n$ distinct eigenvalues, the characteristic polynomial for $A$ factors into the product of degree $1$ polynomials.

Thus, the algebraic multiplicity of each eigenvalue is, $1$ and hence the geometric multiplicity is also $1$.
(The geometric multiplicity is always less than or equal to the algebraic multiplicity and greater than 0 by definition.)

Thus the dimension of each eigenspace, which is the geometric multiplicity, is $1$.

Let $\lambda$ be an eigenvalue of the matrix $A$ and let $\mathbf{x}$ be the eigenvector corresponding to $\lambda$.
Since the eigenspace $E_{\lambda}$ for $\lambda$ is one dimensional and $\mathbf{x}\in E_{\lambda}$ is a nonzero vector in it, the vector $\mathbf{x}$ is a basis.
That is, we have $E_{\lambda}=\{t\mathbf{x} \mid t\in \C \}$.

Now we multiply $A\mathbf{x}=\lambda \mathbf{x}$ by the matrix $B$ on the left and obtain
\begin{align*}
BA\mathbf{x}&=\lambda B\mathbf{x}\\
\iff \,\,\,\, AB\mathbf{x}&=\lambda B\mathbf{x} \text{ since } AB=BA.
\end{align*}

This implies that $B\mathbf{x} \in E_{\lambda}=\{t\mathbf{x} \mid t\in \C \}$. Therefore there exists $t\in \C$ such that $B \mathbf{x}= t\mathbf{x}$.

Hence the vector $\mathbf{x}$ is also an eigenvector corresponding to the eigenvalue $t$ of the matrix $B$.
This completes the proof.

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