No idea how to prove this property about symmetric matrices This is from homework, so please hints only.

Suppose $A$ is symmetric such that all of its eigenvalues are 1 or -1. Prove that $A$ is orthogonal.

The converse is really easy, but I really have no idea how to do this. Any hints?
 A: Hint: A symmetric matrix can be orthogonally diagonalized, i.e. $A = QDQ^T$ for some orthogonal matrix $Q$. What can you say about $AA^T$ and $A^TA$?
A: Here is a proof without using the fact that real symmetric matrices are orthogonally diagonalisable.
By the given conditions, the minimal polynomial of $A$ must have the form $(x-1)^r (x+1)^s$. Since $r\le2\lceil r/2\rceil$ and $s\le2\lceil s/2\rceil$, we get $(A-I)^{2\lceil r/2\rceil}(A+I)^{2\lceil s/2\rceil}=0$. As $A$ is symmetric, it follows that
$$
\left((A-I)^{\lceil r/2\rceil}(A+I)^{\lceil s/2\rceil}\right)^T
\left((A-I)^{\lceil r/2\rceil}(A+I)^{\lceil s/2\rceil}\right)=0.\tag{1}
$$
However, for any real matrix $B$, if $B^TB=0$, then $\|Bx\|^2=x^TB^TBx=0$ for all $x$ and hence $B=0$. Therefore $(1)$ implies that $(A-I)^{\lceil r/2\rceil}(A+I)^{\lceil s/2\rceil}=0$. Since the minimal polynomial must divide any annihilating polynomial, it follows that $r\le \lceil r/2\rceil$ and similarly for $s$. Therefore $r,s\in\{0,1\}$. Consequently, $(x-1)(x+1)$ annihilates $A$, i.e. $A^2=I$. Yet, $A^2=A^TA$. Hence $A$ is orthogonal.
A: Edited Saturday 16 November 2013 10:03 PM PST
Well, it seems the "hints" have had their desired effect, so I'm editing this post to be an answer, pure and simple.
That being said, try this:
since $A$ is symmetric, there exists orthogonal $O$ such that $O^TAO = \Lambda$, with $\Lambda$ diagonal and $\Lambda_{ii} = \pm 1$ for all $i$.  Then $\Lambda^T\Lambda = I$ and since $O^TA^TO = \Lambda^T$, we have $O^TA^TOO^TAO = O^TA^TAO = \Lambda^T\Lambda$ = I.
Thus $A^TA = OIO^T = OO^T = I$ and $A$ is orthogonal. QED
Whew!  That feels better!
Well, now I've certainly said too much! ;- )!!!
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
