Linear Algebra - Eigenvectors and Eigenvalues We are given the following problem:
Consider the matrix $$ A = \left[\begin{array}{rrr}
    \cos\theta & \sin\theta\\
    \sin\theta & -\cos\theta
    \end{array}\right]
$$, where $\theta \in \mathbb R$.
a) Show that $A$ has an eigenvector in $\mathbb{R^2}$ with eigenvalue $1$.
We start the problem by taking the difference of the original matrix with the product of lambda and the identity:
$$A-\lambda I = \left[\begin{array}{rrr}
    \cos\theta -\lambda & \sin\theta\\
    \sin\theta & -\cos\theta -\lambda
    \end{array}\right]
$$
We then find its determinant:
$$det(A-\lambda I) = (\cos\theta -\lambda)(-\cos\theta -\lambda) - \sin\theta^2$$
which gives us: $\lambda^2 - 1$ and eigenvalues of $\lambda_1 = 1, \lambda_2 = -1$
Using $\lambda_1=1$ we have:
$$B(1) = \left[\begin{array}{rrr}
    \cos\theta -1 & \sin\theta\\
    \sin\theta & -\cos\theta -1
    \end{array}\right]
$$
How do I reduce it and answer the problem: Show that $A$ has an eigenvector in $\mathbb{R^2}$ with eigenvalue $1$?
 A: You want to solve the equation
$$\left(\begin{array}{cc} \cos\theta - 1 & \sin\theta \\ \sin\theta & -\cos\theta -1\end{array}\right)\left(\begin{array}{cc} x \\ y\end{array}\right) = \left(\begin{array}{cc} 0 \\ 0\end{array}\right).$$
Since this is an eigenvalue equation, the two rows are linearly dependent. Let's work with the first row without loss of generality. Then
$$x(\cos\theta-1) + y\sin\theta = 0.$$
That is to say that
$$ x = \frac{\sin\theta}{1-\cos\theta}y.$$
Thus your eigenvector is nothing more than
$$\left(\begin{array}{cc} \dfrac{\sin\theta}{1-\cos\theta} \\ 1\end{array}\right).$$
If you wanted to work with the second equation, you could relate the two eigenvectors you get by using the Pythagorean identity.
A: there is no need to compute the eigenvalues. the reason is that the matrix $A$ represents the reflection on the line $ y = \tan(\theta/2) x.$  therefore the eigenvalues are $1, -1.$ the corresponding eigenvectors are $\pmatrix{\cos(\theta/2)\\\sin(\theta/2)}, \pmatrix{-\sin(\theta/2)\\\cos(\theta/2)}.$
