How do I use the spectral theorem to prove this? How do I prove the following with the spectral theorem?
I know the spectral theorem says that A matrix A is orthogonally diagonalizable (if there exists an orthogonal S such that $S^{-1}AS = S^TAS$ is diagonal) if and only if A is symmetric ($A^T = A$).
 A: Consider an orthonormal basis $\{v_1, \ldots, v_n\}$ of $\mathbb{R}^n$ (a few lines below, we can specify this basis). Now, taking the inner product of any $2$ vectors $Av_i, Av_j$ yields:
$$\langle Av_i, Av_j \rangle = (Av_i)^T \cdot (Av_j) = v_i^T \underbrace{A^T A}_{n\times n} v_j.$$
It is easily proven that the matrix $A^T A $ is positive semi - definite. If we denote $M = A^T A$, we have that $M$ is a symmetric matrix, thus diagonalizable (and non - negative eigenvalues). 
Hence, if $ M = V \Lambda V^T$ for some $n\times n$ orthogonal matrix $V$, it holds that the columns of $V$ form an orthonormal basis of $\mathbb {R}^n$. Actually, we can consider the initial orthonormal basis $\{v_1, \ldots, v_n\}$ the columns of $V$.
Thus, $v_i^T V = [ 0 \quad \cdots \quad 0 \quad 1 \quad 0\quad  \cdots \quad0],$ where the unit lies on the $i-$ position. Also: $$v_i^T V \Lambda = [ 0 \quad \cdots \quad 0 \quad \lambda_i \quad 0\quad  \cdots \quad0].$$
Finally, $$v_i^T V\Lambda V^T v_j = 
\begin{cases} \lambda_i, & i = j\\[2ex]
0, & i\neq j
\end{cases}$$
