Proving Eigenvalue I'm stuck at this part of a question: 
Let $A$ be a square matrix of order $n$ such that $A^2=I$
Prove that if lambda is an eigenvalue of $A$, then $\lambda=1$ or $\lambda=-1$.
I know im suppose to get $A^2-I=0 \implies (A+I)(A-I)=0$ but how do I continue from there? Any help is greatly appreciated!
 A: Recall that $\lambda$ is an eigen value iff there exists a non-zero eigen-vector $x$ such that
$$Ax = \lambda x$$
In our case, we have that
$$Ax = \lambda x \implies A^2x = A(Ax) = A(\lambda x) = \lambda(A x) = \lambda (\lambda x) = \lambda^2 x$$
Since $A^2 = I$, we have that $A^2x = x$.
Further, $x$ is a non-zero vector, and hence we have that $\lambda^2 = 1 \implies \lambda = \pm1$
A: From $(A-I)(A+I)=0$, you just have to take the determinant of both sides of this equation to get:
$$
0=\det((A-I)(A+I)) = \det(A-I)\det(A+I),
$$
therefore, either $\det(A-I)=0$ or $\det(A+I)=0$, which implies that either $\lambda=1$ or $\lambda=-1$.
Maybe the last implication was not so clear: if $\lambda$ is an eigenvalue of $A$, then for all $v \neq 0$: 
$$
Av = \lambda v,
$$
therefore, we can write
\begin{align*}
Av = \lambda v = \lambda I v &\Longleftrightarrow Av - \lambda I v = 0 \\
&\Longleftrightarrow (A-\lambda I)v = 0.
\end{align*}
But the last equation has a solution only if $\det(A-\lambda I)$ = 0. So if $\det(A-\lambda I) = 0$ then $\lambda$ is an eigenvalue of $A$. 
