Diagonal Matrices of Linear Operators Examples 2

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

Recall that if $T \in \mathcal L (V)$ has $\mathrm{dim} (V)$ distinct eigenvalues then there exists a basis $B_V$ of $V$ such that $\mathcal M (T, B_V)$ is a diagonal matrix. More succinctly, that basis $B_V$ is a basis of corresponding nonzero eigenvectors to the distinct eigenvalues. Prove that the converse of this theorem is not true in general.

We want to prove that if there exists a basis $B_V$ of $V$ such that $\mathcal M(T, B_V)$ is a diagonal matrix, then it need not be true that $T$ has $\mathrm{dim} (V)$ distinct eigenvalues.

Consider the vector space $\mathbb{R}^2$. Suppose that $T(x, y) = (x, y)$ and use the standard basis $\{ (1, 0), (0, 1) \}$ in $\mathbb{R}^2$. Note that $T$ is simply the identity matrix. Furthermore, $T$ has only one eigenvalue - $1$ since for any vector $(x, y) \in \mathbb{R}^2$ we have that $T(x, y) = 1(x, y)$. Now:

(1)

\begin{align} \quad T(1, 0) = (1, 0) = 1(1, 0) + 0(0, 1) \\ \quad T(0, 1) = (0, 1) = 0(1, 0) + 1(0, 1) \end{align}

Therefore we have that the matrix of $T$ with respect to $\{ (1, 0), (0, 1) \}$ is:

(2)

\begin{align} \quad \mathcal M (T, \{ (1, 0), (0, 1)\}) = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \end{align}

Clearly $\mathcal M (T, \{ (1, 0), (0, 1)\})$ is diagonal, but $T$ does not have $\mathrm{dim} (\mathbb{R}^2) = 2$ distinct eigenvalues. In fact, the matrix $\mathcal M (T, \{ (1, 0), (0, 1)\})$ shows that $T$ has only one distinct eigenvalue.

Example 2

Let $T$ be noninvertible linear operator on the finite-dimensional vector space $V$. Prove that if there exists a basis $B_V$ of $V$ for which $\mathcal M (T, B_V)$ is a diagonal matrix then $V = \mathrm{null} (T) \oplus \mathrm{range} (T)$.

Suppose that there exists a basis $B_V$ of $V$ for which $\mathcal M (T, B_V)$ is a diagonal matrix. Since $T$ is noninvertible, $0$ is an eigenvalue of $T$. Let $0$, $\lambda_1$, $\lambda_2$, …, $\lambda_m$ be the distinct eigenvalues of $T$ (noting that then $\lambda_1, \lambda_2, ..., \lambda_m \neq 0$). Then we have that:

(3)

\begin{align} \quad V = \mathrm{null} (T - 0I) \oplus \mathrm{null} (T - \lambda_1 I) \oplus ( T - \lambda_2 I) \oplus ... \oplus (T - \lambda_m I) \end{align}

Note that $\mathrm{null} (T - 0I) = \mathrm{null} (T)$. We then only need to show that $\mathrm{range} (T) = \mathrm{null} (T - \lambda_1 I) \oplus ( T - \lambda_2 I) \oplus ... \oplus (T - \lambda_m I)$.

Let $v \in \mathrm{range} (T)$. Since $0, \lambda_1, \lambda_2, ..., \lambda_m$ are distinct eigenvalues of $T$ and there exists a basis $B_V$ such that $\mathcal M (T, B_V)$ is a diagonal matrix, then this implies that there exists a basis of nonzero eigenvectors corresponding to these eigenvalues. Let $\{ v_0, v_1, v_2, ..., v_m \}$ be this basis.

Then we have that for every vector $v \in V$ that then:

(4)

\begin{align} \quad v = a_0v_0 + a_1v_1 + ... + a_mv_m \end{align}

Note that $v_0 \in \mathrm{null} (T)$, $v_1 \in \mathrm{null} (T - \lambda_1I)$, …, $v_m \in \mathrm{null} (T - \lambda_m I)$. since $v_0$ is an eigenvector of $0$, $v_1$ is an eigenvector of $\lambda_1$, …, $v_m$ is an eigenvector of $\lambda_m$. Apply the linear operator $T$ to both sides to get that:

(5)

\begin{align} \quad T(v) = T(a_0v_0 + a_1v_1 + ... + a_mv_m) \\ \quad T(v) = a_0T(v_0) + a_1T(v_1) + ... + a_mT(v_m) \\ \quad T(v) = 0 + a_1 \lambda_1 v_1 + ... + a_m \lambda_m v_m \end{align}

Therefore as we can see, $\mathrm{range} (T) = \mathrm{span} (v_1, v_2, ..., v_m) = \mathrm{null} (T - \lambda_1 I) \oplus ( T - \lambda_2 I) \oplus ... \oplus (T - \lambda_m I)$. Therefore $\mathrm{range} (T) \subseteq \mathrm{null} (T - \lambda_1 I) \oplus ( T - \lambda_2 I) \oplus ... \oplus (T - \lambda_m I)$.

Now let $v \in \mathrm{null} (T - \lambda_1 I) \oplus ( T - \lambda_2 I) \oplus ... \oplus (T - \lambda_m I)$. Then we have that $v$ can be written uniquely as $v = u_1 + u_2 + ... + u_m$ where $u_1 \in \mathrm{null} (T - \lambda_1 I)$, $u_2 \in \mathrm{null} (T - \lambda_2 I)$, …, $u_m \in \mathrm{null} (T - \lambda_m I)$. Therefore $T(u_1) = \lambda_1 u_1$, $T(u_2) = \lambda_2 u_2$, …, $T(u_m) = \lambda_m u_m$.

Thus we have that:

(6)

\begin{align} \quad T \left ( \frac{1}{\lambda_1} u_1 + \frac{1}{\lambda_2} u_2 + ... + \frac{1}{\lambda_m} u_m \right ) = \frac{1}{\lambda_1} T(u_1) + \frac{1}{\lambda_2} T(u_2) + ... + \frac{1}{\lambda_m} T(u_m) \\ \quad T \left ( \frac{1}{\lambda_1} u_1 + \frac{1}{\lambda_2} u_2 + ... + \frac{1}{\lambda_m} u_m \right ) = u_1 + u_2 + ... + u_m \\ \quad T \left ( \frac{1}{\lambda_1} u_1 + \frac{1}{\lambda_2} u_2 + ... + \frac{1}{\lambda_m} u_m \right ) = v \end{align}

Thus $v \in \mathrm{range} (T)$ since $T(v)$ is merely a linear combination of $u_1, u_2, ..., u_m \in V$, and so $\mathrm{null} (T - \lambda_1 I) \oplus ( T - \lambda_2 I) \oplus ... \oplus (T - \lambda_m I) \subseteq \mathrm{range} (T)$.

Thus $\mathrm{range} (T) = \mathrm{null} (T - \lambda_1 I) \oplus ( T - \lambda_2 I) \oplus ... \oplus (T - \lambda_m I)$

Therefore $V = \mathrm{null} (T) \oplus \mathrm{range} (T)$.