# Rotation Matrix in the Plane and its Eigenvalues and Eigenvectors

## Problem 550

Consider the $2\times 2$ matrix
$A=\begin{bmatrix} \cos \theta & -\sin \theta\\ \sin \theta& \cos \theta \end{bmatrix},$ where $\theta$ is a real number $0\leq \theta < 2\pi$.

(a) Find the characteristic polynomial of the matrix $A$.

(b) Find the eigenvalues of the matrix $A$.

(c) Determine the eigenvectors corresponding to each of the eigenvalues of $A$.

## Proof.

### (a) Find the characteristic polynomial of the matrix $A$.

The characteristic polynomial $p(t)$ of $A$ is computed as follows.
Let $I$ be the $2\times 2$ identity matrix.
We have
\begin{align*}
p(t)&=\det(A-tI)\\
&=\begin{vmatrix}
\cos \theta -t & -\sin \theta\\
\sin \theta& \cos \theta -t
\end{vmatrix}\\
&=(\cos \theta -t)^2+\sin^2 \theta \tag{*}\\
&=t^2-(2\cos \theta) t+\cos^2 \theta+\sin^2 \theta\\
&=t^2-(2\cos \theta) t+1
\end{align*}
by the trigonometry identity $\cos^2 \theta+\sin^2 \theta=1$.

Hence the characteristic polynomial of $A$ is
$p(t)=t^2-(2\cos \theta) t+1.$

### (b) Find the eigenvalues of the matrix $A$.

The eigenvalues of $A$ are roots of the characteristic polynomial $p(t)$.
So let us solve
$p(t)=t^2-(2\cos \theta) t+1=0.$ By the quadratic formula, we have
\begin{align*}
t&=\frac{2\cos \theta \pm \sqrt{(2\cos \theta)^2-4}}{2}\6pt] &=\cos \theta \pm \sqrt{\cos^2 \theta -1}\\ &=\cos \theta \pm \sqrt{-\sin^2 \theta}\\ &=\cos \theta \pm i \sin \theta =e^{\pm i\theta}. \end{align*} Hence eigenvalues of A are \[\cos \theta \pm i \sin \theta =e^{\pm i\theta}.

Alternatively, we could have used the equation (*).
Then we have
\begin{align*}
&p(t)=(\cos \theta -t)^2+\sin^2 \theta=0\\
&\Leftrightarrow (\cos \theta-t)^2=-\sin^2 \theta\\
&\Leftrightarrow \cos \theta -t=\pm i \sin \theta
\end{align*}
and thus, we obtain the eigenvalues
$\cos \theta \pm i \sin \theta =e^{\pm i\theta}.$

### (c) Determine the eigenvectors

Let us first find the eigenvectors corresponding to the eigenvalue $\lambda=\cos \theta +i \sin \theta$.
We have
\begin{align*}
A-\lambda I=\begin{bmatrix}
-i \sin \theta & -\sin \theta\\
\sin \theta& -i \sin \theta
\end{bmatrix}.
\end{align*}

If $\theta=0, \pi$, then $\sin \theta=0$ and we have
$A-\lambda I=\begin{bmatrix} 0 & 0\\ 0& 0 \end{bmatrix}$ and thus each nonzero vector of $\R^2$ is an eigenvector.

If $\lambda \neq 0, \pi$, then $\sin \theta \neq 0$.
Thus we have by elementary row operations
\begin{align*}
A-\lambda I=\begin{bmatrix}
-i \sin \theta & -\sin \theta\\
\sin \theta& -i \sin \theta
\end{bmatrix}
\xrightarrow[\frac{1}{\sin \theta} R_2]{\frac{i}{\sin \theta} R_1}
\begin{bmatrix}
1 & -i\\
1& -i
\end{bmatrix}
\xrightarrow{R_2-R_1} \begin{bmatrix}
1 & -i\\
0& 0
\end{bmatrix}.
\end{align*}
It follows that the eigenvectors for $\lambda$ are
$\begin{bmatrix} i \\ 1 \end{bmatrix}t,$ for any nonzero scalar $t$.

The other eigenvalue is the conjugate $\bar{\lambda}$ of $\lambda$.
Since $A$ is a real matrix, the eigenvectors of $\bar{\lambda}$ are the conjugate of those of $\lambda$.
Hence the eigenvectors corresponding to $\bar{\lambda}$ is
$\begin{bmatrix} -i \\ 1 \end{bmatrix}t,$ for any nonzero scalar $t$.

In summary, when $\theta=0, \pi$, the eigenvalues are $1, -1$, respectively, and every nonzero vector of $\R^2$ is an eigenvector.

If $\theta \neq 0, \pi$, then the eigenvectors corresponding to the eigenvalue $\cos \theta +i\sin \theta$ are
$\begin{bmatrix} i \\ 1 \end{bmatrix}t,$ for any nonzero scalar $t$.
The eigenvectors corresponding to the eigenvalue $\cos \theta -i\sin \theta$ are
$\begin{bmatrix} -i \\ 1 \end{bmatrix}t,$ for any nonzero scalar $t$.

## Related Question.

For a similar proble, try the following.

Problem.
For a real number $0\leq \theta \leq \pi$, we define the real $3\times 3$ matrix $A$ by
$A=\begin{bmatrix} \cos\theta & -\sin\theta & 0 \\ \sin\theta &\cos\theta &0 \\ 0 & 0 & 1 \end{bmatrix}.$(a) Find the determinant of the matrix $A$.

(b) Show that $A$ is an orthogonal matrix.

(c) Find the eigenvalues of $A$.

The solution is given in the post ↴
Rotation Matrix in Space and its Determinant and Eigenvalues

### 1 Response

1. 08/28/2017

[…] The solution is given in the post ↴ Rotation Matrix in the Plane and its Eigenvalues and Eigenvectors […]

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