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This answer is basically the same as Paul Garrett's. --- First I'll state the question as follows.

Let $V$ be a finite dimensional vector space over a field $K$, and let $S$ and $T$ be diagonalizable endomorphisms of $V$. We say that $S$ and $T$ are simultaneously diagonalizable if (and only if) there is a basis of $V$ which diagonalizes both. The theorem is

$S$ and $T$ are simultaneously diagonalizable if and only if they commute.

If $S$ and $T$ are simultaneously diagonalizable, they clearly commute. For the converse, I'll just refer to Theorem 5.1 of The minimal polynomial and some applications by Keith Conrad. [Harvey Peng pointed out in a comment that the link to Keith Conrad's text was broken. I hope the link will be restored, but in the meantime here is a link to the Wayback Machine version.]

EDIT. The key statement to prove the above theorem is Theorem 4.11 of Keith Conrad's text, which says:

Let $A: V \to V$ be a linear operator. Then $A$ is diagonalizable if and only if its minimal polynomial in $F[T]$ splits in $F[T]$ and has distinct roots.

[$F$ is the ground field, $T$ is an indeterminate, and $V$ is finite dimensional.]

The key point to prove Theorem 4.11 is to check the equality $$V=E_{\lambda_1}+···+E_{\lambda_r},$$ where the $\lambda_i$ are the distinct eigenvalues and the $E_{\lambda_i}$ are the corresponding eigenspaces. One can prove this by using Lagrange's interpolation formula: put $$f:=\sum_{i=1}^r\ \prod_{j\not=i}\ \frac{T-\lambda_j}{\lambda_i-\lambda_j}\ \in F[T]$$ and observe that $f(A)$ is the identity of $V$.

This answer is basically the same as Paul Garrett's. --- First I'll state the question as follows.

Let $V$ be a finite dimensional vector space over a field $K$, and let $S$ and $T$ be diagonalizable endomorphisms of $V$. We say that $S$ and $T$ are simultaneously diagonalizable if (and only if) there is a basis of $V$ which diagonalizes both. The theorem is

$S$ and $T$ are simultaneously diagonalizable if and only if they commute.

If $S$ and $T$ are simultaneously diagonalizable, they clearly commute. For the converse, I'll just refer to Theorem 5.1 of The minimal polynomial and some applications by Keith Conrad. [Harvey Peng pointed out in a comment that the link to Keith Conrad's text was broken. I hope the link will be restored, but in the meantime here is a link to the Wayback Machine version. Edit: original link just updated.]

EDIT. The key statement to prove the above theorem is Theorem 4.11 of Keith Conrad's text, which says:

Let $A: V \to V$ be a linear operator. Then $A$ is diagonalizable if and only if its minimal polynomial in $F[T]$ splits in $F[T]$ and has distinct roots.

[$F$ is the ground field, $T$ is an indeterminate, and $V$ is finite dimensional.]

The key point to prove Theorem 4.11 is to check the equality $$V=E_{\lambda_1}+···+E_{\lambda_r},$$ where the $\lambda_i$ are the distinct eigenvalues and the $E_{\lambda_i}$ are the corresponding eigenspaces. One can prove this by using Lagrange's interpolation formula: put $$f:=\sum_{i=1}^r\ \prod_{j\not=i}\ \frac{T-\lambda_j}{\lambda_i-\lambda_j}\ \in F[T]$$ and observe that $f(A)$ is the identity of $V$.

This answer is basically the same as Paul Garrett's. --- First I'll state the question as follows.

Let $V$ be a finite dimensional vector space over a field $K$, and let $S$ and $T$ be diagonalizable endomorphisms of $V$. We say that $S$ and $T$ are simultaneously diagonalizable if (and only if) there is a basis of $V$ which diagonalizes both. The theorem is

$S$ and $T$ are simultaneously diagonalizable if and only if they commute.

If $S$ and $T$ are simultaneously diagonalizable, they clearly commute. For the converse, I'll just refer to Theorem 5.1 of The minimal polynomial and some applications by Keith Conrad.

EDIT. The key statement to prove the above theorem is Theorem 4.11 of Keith Conrad's text, which says:

Let $A: V \to V$ be a linear operator. Then $A$ is diagonalizable if and only if its minimal polynomial in $F[T]$ splits in $F[T]$ and has distinct roots.

[$F$ is the ground field, $T$ is an indeterminate, and $V$ is finite dimensional.]

The key point to prove Theorem 4.11 is to check the equality $$V=E_{\lambda_1}+···+E_{\lambda_r},$$ where the $\lambda_i$ are the distinct eigenvalues and the $E_{\lambda_i}$ are the corresponding eigenspaces. One can prove this by using Lagrange's interpolation formula: put $$f:=\sum_{i=1}^r\ \prod_{j\not=i}\ \frac{T-\lambda_j}{\lambda_i-\lambda_j}\ \in F[T]$$ and observe that $f(A)$ is the identity of $V$.

This answer is basically the same as Paul Garrett's. --- First I'll state the question as follows.

Let $V$ be a finite dimensional vector space over a field $K$, and let $S$ and $T$ be diagonalizable endomorphisms of $V$. We say that $S$ and $T$ are simultaneously diagonalizable if (and only if) there is a basis of $V$ which diagonalizes both. The theorem is

$S$ and $T$ are simultaneously diagonalizable if and only if they commute.

If $S$ and $T$ are simultaneously diagonalizable, they clearly commute. For the converse, I'll just refer to Theorem 5.1 of The minimal polynomial and some applications by Keith Conrad. [Harvey Peng pointed out in a comment that the link to Keith Conrad's text was broken. I hope the link will be restored, but in the meantime here is a link to the Wayback Machine version.]

EDIT. The key statement to prove the above theorem is Theorem 4.11 of Keith Conrad's text, which says:

Let $A: V \to V$ be a linear operator. Then $A$ is diagonalizable if and only if its minimal polynomial in $F[T]$ splits in $F[T]$ and has distinct roots.

[$F$ is the ground field, $T$ is an indeterminate, and $V$ is finite dimensional.]

The key point to prove Theorem 4.11 is to check the equality $$V=E_{\lambda_1}+···+E_{\lambda_r},$$ where the $\lambda_i$ are the distinct eigenvalues and the $E_{\lambda_i}$ are the corresponding eigenspaces. One can prove this by using Lagrange's interpolation formula: put $$f:=\sum_{i=1}^r\ \prod_{j\not=i}\ \frac{T-\lambda_j}{\lambda_i-\lambda_j}\ \in F[T]$$ and observe that $f(A)$ is the identity of $V$.

This answer is basically the same as Paul Garrett's. --- First I'll state the question as follows.

Let $V$ be a finite dimensional vector space over a field $K$, and let $S$ and $T$ be diagonalizable endomorphisms of $V$. We say that $S$ and $T$ are simultaneously diagonalizable if (and only if) there is a basis of $V$ which diagonalizes both. The theorem is

$S$ and $T$ are simultaneously diagonalizable if and only if they commute.

If $S$ and $T$ are simultaneously diagonalizable, they clearly commute. For the converse, I'll just refer to Theorem 5.1 of The minimal polynomial and some applications by Keith Conrad.

EDIT. The key statement to prove the above theorem is Theorem 4.11 of Keith Conrad's text, which says:

Let $A: V \to V$ be a linear operator. Then $A$ is diagonalizable if and only if its minimal polynomial in $F[T]$ splits in $F[T]$ and has distinct roots.

[$F$ is the ground field, $T$ is an indeterminate, and $V$ is finite dimensional.]

The key point to prove Theorem 4.11 is to check the equality $$V=E_{\lambda_1}+···+E_{\lambda_r},$$ where the $\lambda_i$ are the distinct eigenvalues and the $E_{\lambda_i}$ are the corresponding eigenspaces. One can prove this by using Lagrange's interpolation formula: put $$f:=\sum_{i=1}^r\ \prod_{j\not=i}\ \frac{T-\lambda_j}{\lambda_i-\lambda_j}\ \in F[T]$$ and observe that $f(A)$ is the identity of $V$.

This answer is basically the same as Paul Garrett's. --- First I'll state the question as follows.

Let $V$ be a finite dimensional vector space over a field $K$, and let $S$ and $T$ be diagonalizable endomorphisms of $V$. We say that $S$ and $T$ are simultaneously diagonalizable if (and only if) there is a basis of $V$ which diagonalizes both. The theorem is

$S$ and $T$ are simultaneously diagonalizable if and only if they commute.

If $S$ and $T$ are simultaneously diagonalizable, they clearly commute. For the converse, I'll just refer to Theorem 5.1 of The minimal polynomial and some applications by Keith Conrad.

EDIT. The key statement to prove the above theorem is Theorem 4.11 of Keith Conrad's text, which says:

Let $A: V \to V$ be a linear operator. Then $A$ is diagonalizable if and only if its minimal polynomial in $F[T]$ splits in $F[T]$ and has distinct roots.

[$F$ is the ground field, $T$ is an indeterminate, and $V$ is finite dimensional.]

The key point to prove Theorem 4.11 is to check the equality $$V=E_{\lambda_1}+···+E_{\lambda_r},$$ where the $\lambda_i$ are the distinct eigenvalues and the $E_{\lambda_i}$ are the corresponding eigenspaces. One can prove this by using Lagrange's interpolation formula: put $$f:=\sum_{i=1}^r\ \prod_{j\not=i}\ \frac{T-\lambda_j}{\lambda_i-\lambda_j}\ \in F[T]$$ and observe that $f(A)$ is the identity of $V$.