# 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 $A$.

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## 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

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