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 $1$?