# If $A$ is orthogonal and symmetric, there is an orthonormal basis consisting of eigenvectors of $A$

Let $A ∈ M(n, \mathbb{R})$ be an orthogonal symmetric matrix. Show that $\mathbb{R}^n$ has an orthonormal basis consisting of eigenvectors of $A$. What happens if $A$ is assumed to be only symmetric?

In fact, I know this from linear algebra but I could not prove by using some arguments from calculus, analysis.

So I could use some help.

• Really it not seems a calculus problem. Mar 13 '15 at 21:30
• Pretty sure a orthonormal symmetric matrix is diagonal with elements equal to $\pm 1$, by the way
– Ant
Mar 13 '15 at 21:39
• @Ant: If $\Lambda$ is diagonal with $\pm 1$s on the diagonal, then $U \Lambda U^T$ satisfies the hypotheses but is not necessarily diagonal. Mar 13 '15 at 21:45
• Symmetry is enough.
– Vim
Mar 14 '15 at 5:27
• @copper.hat oh you're right! thank you :-)
– Ant
Mar 14 '15 at 10:30

Hint: This is the Spectral Theorem and is valid for any finite vector space with inner product. The proof is given by induction over the dimension of $E$ (vector space) and considering the self-adjoint operator $A : E \to E$.
Suppose $A$ has an eigenvector $v$, then $E=\operatorname{sp} \{v \}$ is $A$ invariant. Suppose $w \in E^\bot$, then since $A$ is symmetric, we see that $E^\bot$ is also $A$ invariant. If we let $v_2,...,v_n$ be a basis for $E^\bot$ then we see that $A$ is block diagonal, with the $1 \times 1$ entry corresponding to the eigenvector $v$ being the corresponding eigenvalue. We now repeat the process on $A \mid_{E^\bot}$. The end result is a basis of eigenvectors.
To see that $A$ has an eigenvector, consider $\lambda = \max_{\|v\|=1} \langle v, A v \rangle$, and let $\hat{v}$ be a maximising unit vector. From Lagrange multipliers, we see that we have $A\hat{v} + \mu {\hat{v}} = 0$ for some $\mu$, and so $\hat{v}$ is an eigenvector (with eigenvalue $\lambda = -\mu$).