Eigenvalues and Eigenvectors Examples 5

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 $U$ and $W$ be nonzero subspaces of $V$ such that $V = U \oplus W$. Let $P$ be a linear operator on $V$ defined by $P(u + w) = u$ for all vectors $u \in U$ and for all vectors $w \in W$. Find all eigenvalues of $P$ and the corresponding eigenvectors, and verify that these eigenvectors are indeed associated with these eigenvalues.

We will linear operators $P$ of this form later on.

Since $V = U \oplus W$ then for every vector $v \in V$ we have that $v = u + w$ where $u \in U$ and $w \in W$. We want to find numbers $\lambda \in \mathbb{F}$ such that:

(1)

\begin{align} \quad P(v) = \lambda v \\ \quad P(u + w) = \lambda (u + w) \\ \quad u = \lambda (u + w) \\ \quad u = \lambda u + \lambda w \end{align}

From the equation above, we have that:

(2)

\begin{align} \quad u = \lambda u \\ \quad 0 = \lambda w \end{align}

From the second equation we have that $\lambda = 0$ or $w = 0$.

If $\lambda = 0$, then we have that $u = 0$. Therefore $\lambda_1 = 0$ is an eigenvalue of $T$ and the corresponding set of eigenvectors is $\{ v = w : w \neq 0 \}$. To verify that this set of vectors are eigenvectors for $\lambda_1 = 0$, we note that if $v = w$ where $w \in W$, then we have that:

(3)

\begin{align} \quad P(v) = P(w) = 0 = 0w = \lambda w = \lambda v \end{align}

Now suppose that instead $w = 0$. Then our system of equations reduces down to:

(4)

\begin{align} \quad u = \lambda u \end{align}

We note that $u \neq 0$, so therefore $\lambda = 1$, and so $\lambda_2 = 1$ is an eigenvalue of $T$ and the corresponding set of eigenvectors is $\{ v = u: \in U : u \neq 0 \}$. To verify that this set of vectors are eigenvectors for $\lambda_2 = 1$, we note that if $v = u$ where $u \in U$, then we have that:

(5)

\begin{align} \quad P(v) = P(u) = u = 1u = \lambda u = \lambda v \end{align}

Example 2

Prove that the right shift operator $T \in \mathcal L (\mathbb{F}^{\infty})$ defined by $T(x_1, x_2, ...) = (0, x_1, x_2, ... )$ has no eigenvalues.

Let $u = (x_1, x_2, ...) \in \mathbb{F}^{\infty}$. We want to find $\lambda \in \mathbb{F}$ such that:

(6)

\begin{align} \quad T(u) = \lambda u \\ \quad T(x_1, x_2, ...) = \lambda(x_1, x_2, ...) \\ \quad (0, x_1, x_2, ...) = \lambda(x_1, x_2, ...) \end{align}

From the equation above, we have that:

(7)

\begin{align} \quad 0 = \lambda x_1 \\ \quad x_1 = \lambda x_2 \\ \quad x_2 = \lambda x_3 \\ \quad \quad \vdots \quad \quad \end{align}

The first equation implies that $\lambda = 0$ or $x_1 = 0$. If $\lambda = 0$, then $x_1 = x_2 = ... = 0$, which is not a nonzero vector in $\mathbb{F}^{\infty}$.

If instead $x_1 = 0$, then this implies that $0 = \lambda x_2$. Therefore $\lambda = 0$ (which we've already seen cannot happen) or $x_2 = 0$. So if $x_2 = 0$, then once again, we have that $0 = \lambda x_3$, etc…, and we get that $x_1 = x_2 = x_3 = ... = 0$ which is not a nonzero vector in $\mathbb{F}^{\infty}$.

Therefore $T$ has no eigenvalues.

Example 3

Find the eigenvalues and corresponding eigenvectors of the left shift operator $T \in \mathcal L (\mathbb{F}^{\infty})$ defined by $T(x_1, x_2, ...) = (x_2, x_3, ... )$.

Let $u = (x_1, x_2, ...) \in \mathbb{F}^{\infty}$. We want to find $\lambda \in \mathbb{F}$ such that:

(8)

\begin{align} \quad T(u) = \lambda u \\ \quad T(x_1, x_2, ...) = \lambda (x_1, x_2, ...)\\ \quad (x_2, x_3, ... ) = \lambda (x_1, x_2, ...) \end{align}

The equation above gives us the following system of equations:

(9)

\begin{align} \quad x_2 = \lambda x_1 \\ \quad x_3 = \lambda x_2 \\ \quad \quad \vdots \quad \quad \end{align}

Note that we can choose $x_1$ however we'd like. We get that:

(10)

\begin{align} \quad x_2 = \lambda x_1 \\ \quad x_3 = \lambda^2 x_1 \\ \quad x_4 = \lambda^3 x_1 \\ \quad \quad \vdots \quad \quad \end{align}

Therefore we have that each $\lambda \in \mathbb{F}$ is an eigenvalue of the left shift operator. The set of corresponding eigenvectors is for each $\lambda$ is $\{ (x_1, \lambda x_1, \lambda^2 x_1, ...) : x_1 \in \mathbb{F} \}$.