Real Symmetric Matrices Let $A$ be a real, symmetric, $n$ x $n$ matrix. Suppose $A^m=I$ for some $m$. 
Prove $A^2=I$.
I think I want to use the symmetric implies diagonalizability...and take powers from there...correct?
 A: Since $A$ is real symmetric all its eigenvalues are real, and furthermore there exists an orthogonal matrix $O$ such that
$O^TAO = \Lambda$,
where $\Lambda$ is a diagonal matrix formed by placing the eigenvalues of $A$ on its main diagonal, and zeroes everywhere else.  If we now take the $m^{\text{th}}$ power of both sides of this equation, we obtain
$(O^TAO)^m = \Lambda^m$,
and since
$(O^TAO)^m = O^TA^mO = O^TIO = O^TO = I$,
it follows that
$\Lambda^m = I$,
so for each eigenvalue $\lambda_i$ on the diagonal of $\Lambda$ we have
$\lambda_i^m = 1$.
Now the fact that the $\lambda_i$ are real forces the conclusion that
$\lambda_i = \pm 1$
for each $i$.  Hence 
$\lambda_i^2 = 1$
for each $i$, forcing
$\Lambda^2 = I$.
Then
$O^TA^2O = \Lambda^2 = I$,
whence 
$A^2 = OIO^T = I$.
Quod Erat Demonstrandum, for you classicists out there!
A: Hint
$$\forall\lambda\in\mathbb R,\quad\lambda^m=1\iff\lambda=\pm1\Rightarrow\lambda^2=1$$
A: From $A = PDP^{-1}$ ($D$ is a diagonal matrix) and $A^{m} = I$, we have that
$$P(D^{m} - I)P^{-1} = 0.$$
Thus, $$D^{m} = I.$$
Since each element of $D$ is a real number, diagonal elements of D can be only $1$ or $-1$.
Therefore, $D^{2} = I$ which implies $A^{2} = I.$
