Eigenvalues of a 2x2 matrix A such that $A^2$=I I have no idea where to begin.
Let A be a $2\times 2$ matrix such that $A^2= I$, where $I$ is the identity matrix.  Find all the eigenvalues of $A$.
I know there are a few matrices that support this claim, will they all have the same eigenvalues?
 A: If $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $$v=Iv=A^2v=A(Av)=A(\lambda v)=\lambda(Av)=\lambda^2 v.$$
Thus, if $\lambda$ is an eigenvalue of $A$ and $A^2=I$ then $\lambda^2=1$.  This gives only two possibilities for $\lambda$, $\pm 1$.
Notice, that we never assumed that $A$ is $2\times 2$.  Indeed, if $A$ is any square matrix and $A^2=I$ then the only possible eigenvalues of $A$ are $\pm 1$.
A: Let $A$ be an arbitrary $2\times 2$ matrix, and $a,b,c,d\in\mathbb{R}$ such that:
$$A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$$
Then we have:
$$A^2=I\Longrightarrow\begin{pmatrix}a^2+bc&ab+bd\\ac+cd&bc+d^2\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}$$
Therefore, any matrix of the form:
$$A=\begin{pmatrix}a&b\\c&-a\end{pmatrix}\text{ or }A=\begin{pmatrix}\pm1&0\\0&\pm1\end{pmatrix}$$
will satisfy the condition $A^2=I$. In the first case its eigenvalues are then given by:
$$\begin{vmatrix}a-\lambda&b\\c&-a-\lambda\end{vmatrix}=0\Longrightarrow \lambda^2=a^2+bc=1$$
Therefore $\lambda=\pm 1$. The second case also gives us $\lambda=\pm1$.
