TITLE: If $A^2 = I$, then $A$ is diagonalizable, and is $I$ if $1$ is its only eigenvalue
QUESTION [6 upvotes]: Let $A$ be a square matrix of order $n$ such that $A^2 = I$.

Prove that if $1$ is the only eigenvalue of $A$, then $A = I$.
Prove that $A$ is diagonalizable.


For  (1), I know that there are two eigenvalues which are $1$ and $-1$, how do I go about proving what the question asks me?

REPLY [2 votes]: You are right, eigenvalues are $1$ and $-1$.
For diagonalizable part: 
$$A^2=I$$ Therefore the characteristic polynomial is $( \lambda-1)(\lambda+1)=0$, hence characteristic polynomial splits. So $A$ is diagonalizable.

REPLY [2 votes]: Hint for 2. Assume "optimistically" that every vector $v$ decomposes as a sum of eigenvectors of $A$:
$$
v = v_{1} + v_{-1}.
$$
Apply $A$ to both sides to get a system of two equations in two unknowns, and solve formally for $v_{1}$ and $v_{-1}$.
Armed with the expressions from the preceding paragraph, show that you have actually found eigenvectors of $A$, and therefore have proven that $\mathbf{R}^{n}$ (or whatever field of scalars, so long as the characteristic isn't $2$) decomposes as a sum of eigenspaces of $A$.