Eigenvalues and Eigenvectors Examples 3

# Eigenvalues and Eigenvectors Examples 3

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 $T \in \mathcal L(\mathbb{F}^3, \mathbb{F}^3)$ be defined by $T(x, y, z) = (3y, 2x, 0)$. Find all eigenvalues of $T$.

Let $u = (x, y, z)$. Then we want to find $\lambda \in \mathbb{F}$ such that:

(1)
\begin{align} \quad T(u) = \lambda u \\ \quad T(x, y, z) = \lambda (x, y, z) \\ \quad (3y, 2x, 0) = \lambda (x, y, z) \end{align}

From the equation above we get the following system of equations:

(2)
\begin{align} \quad 3y = \lambda x \\ \quad 2x = \lambda y \\ \quad 0 = \lambda z \end{align}

The third equation implies that $\lambda = 0$ or $z = 0$.

If $\lambda = 0$ then all three equations are satisfied, so $\lambda_1 = 0$ is an eigenvalue of $T$. We will also then have that $x = y = 0$.

Thus the set of vectors $(0, 0, z)$ such that $z \neq 0$ are the corresponding eigenvectors to $\lambda_1 = 0$.

If $z = 0$, then we must still satisfy the first two equations. From the second equation we have that $x = \frac{\lambda y}{2}$. Substituting this into the first equation and we have that:

(3)
\begin{align} \quad 3y = \lambda \left ( \frac{\lambda y}{2} \right ) \\ \quad 6y = \lambda^2 y \\ \end{align}

We note that $y \neq 0$ otherwise $x = y = 0$ and so $u = (0, 0, 0)$. Therefore we can divide both sides by $y$ to get that $\lambda_2 = \sqrt{6}$ and $\lambda_3 = -\sqrt{6}$ are both eigenvalues of $T$.

The set of vectors $\left (x, \frac{\lambda x}{3} ,0 \right ) = \left ( x, \frac{\sqrt{6}}{3} x, 0 \right )$ where $x \neq 0$ are the corresponding eigenvectors to $\lambda_1 = \sqrt{6}$.

The set of vectors $\left ( x, \frac{\lambda x}{3}, 0 \right ) = \left ( x, - \frac{\sqrt{6}}{3}x, 0 \right )$ where $x \neq 0$ are the corresponding eigenvectors to $\lambda_2 = -\sqrt{6}$.

## Example 2

Let $T$ be a linear operator on $V$ be such that every vector $v \in V$ is an eigenvector to some eigenvalue of $T$ Prove that then $T = aI$ where $a \in \mathbb{F}$.

Since every vector $v \in V$ is an eigenvector to some eigenvalue of $T$, then for every $v \in V$ we have that $T(v) = \lambda v$ where $\lambda$ is an eigenvalue of $T$. Note that if $v = 0$ then any choice of $\lambda$ will suffice this equation.

Therefore, suppose that $u, v \in V \setminus \{ 0 \}$. We have that $T(u) = \lambda u$ and $T(v) = \mu v$. We want to show that $\lambda = \mu$.

Suppose that $\{ u, v \}$ is a linearly dependent set of vectors. Then $v$ is a scalar multiple of $u$, so $v = bu$ for some $b \in \mathbb{F}$ and thus:

(4)
\begin{align} \quad \mu v = T(v) = T(bu) = bT(u) = b \lambda u = \lambda (bu) = \lambda v \end{align}

The equation above implies that $\mu = \lambda$.

Now suppose that $\{ u, v \}$ is a linearly independent set of vectors. For the vector $u + v$ let $\gamma$ be the eigenvalue such that $T(u + v) = \gamma (u + v)$. Then we have that:

(5)
\begin{align} \quad \gamma u + \gamma v = \gamma (u + v) = T(u + v) = T(u) + T(v) = \lambda u + \mu v \end{align}

Since $\{ u, v \}$ is a linearly independent set of vectors, the equation above implies that $\gamma = \lambda$ and $\gamma = \mu$, so $\lambda = \mu$.

Therefore $T$ must be a scalar multiple of the identity operator, that is, $T = aI$ for $a \in \mathbb{F}$.

## Example 3

Let $T$ be an operator on the finite-dimensional vector space $V$. Prove that every vector $v \in V$ is an eigenvector of $T$ if and only if $ST = TS$ for all linear operators $S \in \mathcal L (V)$.

$\Rightarrow$ Suppose that every vector $v \in V$ is an eigenvector of $T$. Then from example 2, we must have that $T = aI$ for some $a \in \mathbb{F}$. Therefore we have that:

(6)
\begin{align} \quad (ST)(v) = S(T(v)) = S(aI(v)) = S(av) = a S(v) \end{align}
(7)
\begin{align} \quad (TS)(v) = T(S(v)) = aI(S(v)) = aS(v) \end{align}

Therefore $ST = TS$.

$\Leftarrow$ Now suppose that $ST = TS$. If this is true for every linear operator $S \in \mathcal L (V)$ then this implies that $\mathcal M (S) \mathcal M (T) = \mathcal M (T) \mathcal M (S)$, i.e, matrix commutativity holds for all $S$. Therefore we must have that $T = a I$ for some $a \in \mathbb{F}$. Therefore every vector $v \in V$ is an eigenvector of $T$ (that belongs to the eigenvalue $a$).