Eigenvalues and Eigenvectors Examples 4

Recall from the Eigenvalues and Eigenvectors page that the number $\lambda \in \mathbb{F}$ is said to be an eigenvalue of the linear operator $T \in \mathcal L (V)$ if $T(u) = \lambda u$ for some nonzero vector $u \in V$. The nonzero vectors $u$ such that $T(u) = \lambda u$ are called eigenvectors corresponding to the eigenvalue $\lambda$.

We will now look at some examples regarding eigenvalues of linear operators and eigenvectors corresponding to eigenvalues.

Example 1

Let $T \in \mathcal ( \wp (\mathbb{R} )$ be defined by $T(p(x)) = p'(x)$ for all $p(x) \in \wp (\mathbb{R})$. Find all eigenvalues of $T$ and their corresponding eigenvectors.

We want to find numbers $\lambda \in \mathbb{R}$ such that:

(1)

\begin{align} \quad T(p(x)) = \lambda p(x) \\ \quad p'(x) = \lambda p(x) \end{align}

For $p(x) = a_0 + a_1x + a_2x^2 + ... + a_nx^n$ we have that:

(2)

\begin{align} \quad a_1 + 2a_2x + ... + na_nx^{n-1} = \lambda a_0 + \lambda a_1x + \lambda a_2x^2 + ... + \lambda a_nx^n \end{align}

We thus have that:

(3)

\begin{align} \quad a_1 = \lambda a_0 \\ \quad 2a_2 = \lambda a_1 \\ \quad \quad \vdots \quad \quad \\ \quad na_n = \lambda a_{n-1} \\ \quad 0 = \lambda a_n \end{align}

From the last equation we see that $\lambda = 0$ or $a_n = 0$.

If $\lambda = 0$, then all of the equations above are satisfied, so $\lambda_1 = 0$ is an eigenvalue of $T$. Furthermore, we will have that $a_1 = a_2 = ... = a_n = 0$, and $a_0 \in \mathbb{R}$. Therefore the set of polynomials $\{ p(x) = a_0 : a_0 \in \mathbb{R} \: , a_0 \neq 0 \}$ are the corresponding eigenvectors to $\lambda_1 = 0$.

Now if $a_n = 0$, then this implies that $a_{n-1} = ... = a_2 = a_1 = a_0 = 0$ which means that $p(x)$ is a zero vector.

Therefore the only eigenvalue is $\lambda = 0$.

Example 2

Let $T \in \mathcal L ( \wp_3 (\mathbb{R}))$ be defined by $T(p(x)) = xp'(x)$. Find all eigenvalues of $T$ and their corresponding eigenvectors.

Let $p(x) = a_0 + a_1x + a_2x^2 + a_3x^3$. We want to then find $\lambda \in \mathbb{F}$ such that:

(4)

\begin{align} \quad T(p(x)) = \lambda p(x) \\ \quad xp'(x) = \lambda p(x) \\ \quad x(a_1 + 2a_2x + 3a_3x^2) = \lambda (a_0 + a_1x + a_2x^2 + a_3x^3) \\ \quad a_1x + 2a_2x^2 + 3a_3x^3 = \lambda a_0 + \lambda a_1x + \lambda a_2x^2 + \lambda a_3x^3 \end{align}

From above, we obtain the following system of equations:

(5)

\begin{align} \quad 0 = \lambda a_0 \\ \quad a_1 = \lambda a_1 \\ \quad 2a_2 = \lambda a_2 \\ \quad 3a_3 = \lambda a_3 \end{align}

The first equation above implies that either $\lambda = 0$ or $a_0 = 0$.

If $\lambda = 0$ then all of the equations are satisfied and so $\lambda_1 = 0$ is an eigenvalue. We also have that $a_1 = a_2 = a_3 = 0$, and so the polynomials in $\{ p(x) = a_0 : a_0 \in \mathbb{R} \: a_0 \neq 0 \}$ are the corresponding eigenvalues.

If $a_0 = 0$, then the system of equations from above reduces down to:

(6)

\begin{align} \quad a_1 = \lambda a_1 \\ \quad 2a_2 = \lambda a_2 \\ \quad 3a_3 = \lambda a_3 \end{align}

Now suppose that $a_1 \neq 0$. Then $\lambda = 1$. This then implies that $a_2 = a_3 = 0$, and so $\lambda_2 = 1$ is an eigenvalue of $T$ and $\{ p(x) = a_1x : a_1 \in \mathbb{R} \: a_1 \neq 0 \}$ is the set of corresponding eigenvectors.

Now suppose that $a_1 = 0$. Then the system of equations from above reduces down to:

(7)

\begin{align} \quad 2a_2 = \lambda a_2 \\ \quad 3a_3 = \lambda a_3 \end{align}

Now suppose that $a_2 \neq 0$. Then $\lambda = 2$. Then this implies that $a_3 = 0$, and so $\lambda_3 = 2$ is an eigenvalue of $T$, and $\{ p(x) = a_2x^2 : a_2 \in \mathbb{R} \: a_2 \neq 0 \}$ is the set of corresponding eigenvectors to the eigenvalue $\lambda_3 = 2$.

Lastly suppose that $a_2 = 0$. Then the system of equations from above reduces down to:

(8)

\begin{align} \quad 3a_3 = \lambda a_3 \end{align}

Suppose that $a_3 \neq 0$. Then $\lambda = 3$. Thus we have that $\lambda_4 = 3$ is an eigenvalue of $T$, and $\{ p(x) = a_3x^3 : a_3 \in \mathbb{R} \: a_3 = 0 \}$ is the set of corresponding eigenvectors to the eigenvalue $\lambda_4 = 3$.

Note that if $a_3 = 0$ then since $a_0 = a_1 = a_2 = a_3 = 0$, we have that $p(x) = 0$ which not a nonzero vector.