# Eigenvalues and Eigenvectors Examples 2

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

**Find the eigenvalues of the linear operator $T \in \mathcal (\mathbb{\mathbb{C}}^2)$ defined by $T(x, y) = (-y, x)$ and the corresponding set of eigenvectors for each eigenvalue.**

Let $u = (x, y) \in \mathbb{C}^2$ be a nonzero vector in $\mathbb{C}^2$. We want to find the numbers $\lambda \in \mathbb{F}$ such that:

(1)

The equation above gives us the following system of equations:

(2)

Plugging the second equation into the first equation and dividing both sides by $y$ (which is nonzero since if $y = 0$ then this implies $x = 0$ and so $(x, y) = (0, 0)$) and:

(3)

Therefore the eigenvalues of $T$ are $\lambda_1 = i$ and $\lambda_2 = -i$.

For the eigenvalue $\lambda_1 = i$ we have that $(x, y) = (x, – \lambda_1 x) = (x , -ix)$ where $x \neq 0$ are the corresponding eigenvectors to $\lambda_1 = i$.

For the eigenvalue $\lambda_2 = -i$ we have that $(x, y) = (x, -\lambda_2x) = (x, ix)$ where $x \neq 0$ are the corresponding eigenvectors to $\lambda_2 = -i$.

## Example 2

**Find the eigenvalues of the linear operator $T \in \mathcal (\mathbb{\mathbb{F}}^2)$ defined by $T(x, y) = (y, x)$ and the corresponding set of eigenvectors for each eigenvalue.**

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

(4)

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

(5)

Plugging the second equation into the first equation and we have that $y = \lambda ( \lambda y )$ and so $y = \lambda^2 y$. Once again, we note that $y \neq 0$ (otherwise $x = 0$ and so $u = (0, 0)$). Therefore can divide by $y$ to get that:

(6)

Thus $\lambda_1 = 1$ and $\lambda_2 = -1$ are both eigenvalues of $T$.

For $\lambda_1 = 1$ we have that the set of vectors $(x, y) = (x, \lambda x) = (x, x)$ where $x \neq 0$ are the corresponding eigenvectors of $\lambda_1$.

For $\lambda_2 = -1$ we have that the set of vectors $(x, y) = (x, \lambda x) = (x, -x)$ where $x \neq 0$ are the corresponding eigenvectors of $\lambda_2$.

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