Problem statement:
Diagonalize the following matrix:
$$
\begin{pmatrix}
\cos \theta & \sin \theta \\
\sin \theta & -\cos \theta\\
\end{pmatrix}
$$
My attempt:
Ie found that the eigenvalues of this matrix are $\lambda = -1$ or $\lambda = 1$, so I plugged in $\lambda = -1$:
$$
\begin{pmatrix}
\cos\theta+1 & \sin\theta \\
\sin\theta & -\cos\theta+1 \\
\end{pmatrix}
\begin{pmatrix}
x \\
y \\
\end{pmatrix}
= 0
$$
I tried to solve this, but I got $2x\sin\theta=0$. How do I find the eigenvectors?
Best Answer
For $\;\lambda=-1\;$ :
$$\begin{align*}I&\;\;(\cos\theta+1)x+\sin\theta y=0\iff y=-\frac{\cos\theta+1}{\sin\theta}x\\II&\;\;\sin\theta x-(\cos\theta-1)y=0\end{align*}$$
We need only one equation as these two are linearly dependent (proof?), so
$$II\longrightarrow x=\frac{\cos\theta-1}{\sin\theta}y\;\implies\binom {\frac{\cos\theta-1}{\sin\theta}}{y}\;\;\text{is an eigenvector,}\;\;\theta\neq k\pi\;,\;k\in\Bbb Z$$
Something similar happens if $\;\lambda=1\;$