# Diagonal Matrices of Linear Operators Examples 3

Recall from the Diagonal Matrices of Linear Operators page that if $V$ is a finite-dimensional vector space and $T \in \mathcal L (V)$, then $T$ is said to be diagonalizable if there exists a basis $B_V$ such that $\mathcal M (T, B_V)$ is a diagonal matrix.

We saw that if $T$ has $\mathrm{dim} V$ distinct eigenvalues then there exists a basis $B_V$ of $V$ of eigenvectors corresponding to these $\mathrm{dim} V$ eigenvalues.

We also saw a chain of equivalent statements regarding $T$ being diagonalizable.

We will now look at some more problems regarding diagonal matrices of linear operators.

## Example 1

**Let $T$ be a linear operator over the finite-dimensional vector space $V$. Prove that if $T$ has $\mathrm{dim} (V)$ distinct eigenvalues and $S$ is a linear operator on $V$ with the same eigenvectors as $T$ then $ST = TS$.**

Let $V$ be a finite-dimensional vector space and let $\mathrm{dim} (V) = n$. Let $\lambda_1, \lambda_2, …, \lambda_n$ be the distinct eigenvalues of $T$. Then there exists nonzero vectors $u_1, u_2, …, u_n$ corresponding to these eigenvalues. Furthermore, $\{ u_1, u_2, …, u_n \}$ is a linearly independent set with $\mathrm{dim} (V) = n$ vectors and is hence a basis of $V$. So for every vector $v \in V$ we have that $v = a_1u_1 + a_2u_2 + … + a_nu_n$. Furthermore, since $u_1, u_2, …, u_j$ are eigenvectors of $T$ then for $j = 1, 2, …, n$ we have that:

(1)

Suppose that $S$ has the same eigenvectors as $T$. Then for some eigenvalues $\mu_1, \mu_2, …, \mu_n$ we have that for $j = 1, 2, …, n$:

(2)

Therefore we have that:

(3)

We also have that:

(4)

Therefore $ST = TS$.

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