Diagonal Matrices of Linear Operators Examples 3

# 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)
\begin{align} \quad T(u_j) = \lambda_1 u_j \end{align}

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)
\begin{align} \quad S(u_j) = \mu_j u_j \end{align}

Therefore we have that:

(3)
\begin{align} \quad (ST)(v) = S(T(a_1u_1 + a_2u_2 + ... + a_nu_n)) = S(a_1T(u_1) + a_2T(u_2) + ... + a_nT(u_n)) \\ \quad = S(a_1 \lambda_1 u_1 + a_2 \lambda_2 u_2 + ... + a_n \lambda_n u_n) \\ \quad = a_1 \lambda_1 S(u_1) + a_2 \lambda_2 S(u_2) + ... + a_n \lambda_n S(u_n) \\ \quad (a_1 \lambda_1 \mu_1) u_1 + (a_2 \lambda_2 \mu_2) u_2 + ... + (a_n \lambda_n \mu_n) u_n \end{align}

We also have that:

(4)
\begin{align} \quad (TS)(v) = T(S(a_1u_1 + a_2u_2 + ... + a_nu_n)) T(a_1S(u_1) + a_2S(u_2) + ... + a_nS(u_n)) \\ \quad = T(a_1 \mu_1 u_1 + a_2 \mu_2 u_2 + ... + a_n \mu_n u_n ) \\ \quad = a_1\mu_1 T(u_1) + a_2 \mu_2 T(u_2) + ... + a_n \mu_n T(u_n) \\ \quad = (a_1 \mu_1 \lambda_1) u_1 + (a_2 \mu_2 \lambda_2) u_2 + ... + (a_n \mu_n \lambda_n) u_n \end{align}

Therefore $ST = TS$.