Diagonal Matrices of Linear Operators Examples 1

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 problems regarding diagonal matrices of linear operators.

Example 1

Reprove that if $T \in \mathcal L(V)$ is such that $T$ has $\mathrm{dim} (V)$ distinct eigenvalues then there exists a basis $B_V$ of $V$ such that $\mathcal M (T, B_V)$ is diagonal.

Suppose that $T$ has $\mathrm{dim} (V) = n$ distinct eigenvalues. Let $\lambda_1, \lambda_2, ..., \lambda_n \in \mathbb{F}$ be these eigenvalues. Let $v_1, v_2, ..., v_n$ be corresponding nonzero eigenvectors to these eigenvalues. Then then set $\{ v_1, v_2, ..., v_n \}$ is linearly independent.

Furthermore, this set has $\mathrm{dim} (V) = n$ vectors in it, and so this set is the "right size", which implies that $\{ v_1, v_2, ..., v_n \}$ is a basis of $V$.

The columns of $\mathcal M (T, B_V)$ are determined by $T$ applied to each basis vector in $\{ v_1, v_2, ..., v_n \}$. We have that for each $j = 1, 2, ..., n$ that:

(1)

\begin{align} \quad T(v_j) = \lambda_j v_j \end{align}

Therefore, all entries off of the main diagonal are zero. Furthermore, the entry in row $j$ column $j$ is the eigenvalue $\lambda_j$ for each $j = 1, 2, ..., n$.

Example 2

Consider the linear map $T \in \mathcal (\mathbb{R}^2)$ defined by $T(x, y) = (3x + 4y, 2y)$ for every $(x, y) \in \mathbb{R}^2$. Find a basis of $\mathbb{R}^2$ for which $T$ has a diagonal matrix.

If we use the standard basis $\{ (1, 0), (0, 1) \}$ of $\mathbb{R}^2$, then we have that:

(2)

\begin{align} \quad T(1, 0) = (3, 0) = 3(1, 0) + 0(0, 1) \\ \quad T(0, 1) = (4, 2) = 4(1, 0) + 2(0, 1) \end{align}

Therefore we have that the matrix of $T$ with respect to the standard basis on $\mathbb{R}^2$ is:

(3)

\begin{bmatrix} 3 & 4\\ 0 & 2 \end{bmatrix}

We note that this matrix is already upper triangular, and so the eigenvalues of $T$ are $\lambda_1 = 3$ and $\lambda_2 = 2$.

Note that $T$ therefore has $\mathrm{dim} (\mathbb{R}^2)$ distinct eigenvalues, $\lambda_1$ and $\lambda_2$ so there exists a basis $B$ of $\mathbb{R}^2$ for which $\mathcal M (T, B)$ is upper triangular. We can find such a matrix by finding corresponding nonzero eigenvectors to the eigenvalues that we have already found.

We note that if $u = (x, y) \in \mathbb{R}^2$ then:

(4)

\begin{align} \quad T(u) = \lambda u \\ \quad T(x, y) = \lambda (x, y) \\ \quad (3x + 4y, 2y) = \lambda (x, y) \\ \quad (3x + 4y, 2y) = (\lambda x, \lambda y) \end{align}

From above we have that:

(5)

\begin{align} \quad 3x + 4y = \lambda x \\ \quad 2y = \lambda y \end{align}

Therefore for $\lambda_1 = 3$ we have that the system above reduces to:

(6)

\begin{align} \quad 3x + 4y = 3x \\ \quad 2y = 3y \end{align}

Therefore the corresponding set of eigenvectors to the eigenvalue $\lambda_1 = 3$ are $\{ (x, 0) \in \mathbb{R}^2 : x \in \mathbb{R} \: x \neq 0 \}$. Choose the vector $(1, 0)$.

For the eigenvalue $\lambda_2 = 2$ we have that the earlier system from above reduces to:

(7)

\begin{align} \quad 3x + 4y = 2x \\ \quad 2y = 2y \end{align}

The first equation tells us that $x = -4y$. Therefore the corresponding set of eigenvectors to the eigenvalue $\lambda_2 = 2$ are $\{ (-4y, y) \in \mathbb{R}^2 : y \in \mathbb{R} \: y \neq 0 \}$. Choose the vector $(-4, 1)$.

Therefore $\{ (1, 0), (-4, 1) \}$ should be a basis $B$ for which $\mathcal M (T, B_V)$ is upper triangular. Let's check this:

(8)

\begin{align} \quad T(1, 0) = (3, 0) = 3(1, 0) + 0(-4, 1) \end{align}

(9)

\begin{align} \quad T(-4, 1) = (-8, 2) = 0(1, 0) + 2(-4, 1) \end{align}

Therefore we have that the matrix $T$ with respect to the basis $B = \{ (1, 0), (-4, 1) \}$ is:

(10)

\begin{align} \quad \mathcal M (T, B) = \begin{bmatrix} 3 & 0 \\ 0 & 2 \end{bmatrix} \end{align}

Thus, $\mathcal M (T, B)$ is indeed an upper triangular matrix.