How can I use the spectral theorem to prove this corollary ("spectral theorem for matrices")?
Let $M$ be a symmetric matrix or order $n$. Then $M$ is diagonalizable. Also, there exist an orthogonal matrix $P$ of order $n$ such that $M=P^{-1}DP$, where $D$ is a diagonal matrix.
The latest version of Proofs from THE BOOK (5th edition) has a nice proof of this theorem (in chapter 7) orginally published by Herb Wilf.
In short, given symmetric matrix $A$, he defines the functions $$f_A: Q \mapsto Q^\top A Q$$ and $$\text{Od}: A \mapsto \sum_{i\neq j}a_{ij}.$$
He then proves that there exists an orthogonal matrix $Q_0$ such that $f_A\circ \text{Od}(Q_0) = 0$.
Suddenly our problem can be solved with analytical techniques! We can show that there exists a minimum for $f_A \circ \text{Od}$, and we can get a contradiction if we assume that this minimum would be greater than zero.
I have made this prezi from this proof: https://prezi.com/aexypzd25tmd/spectral-theorem-from-proofs-of-the-book/
This prezi was meant for a presentation, but I think it may be understandable on its own as well.
You can prove it using Schur's unitary triangularization theorem which states that for every square matrix there exists some unitary matrix $U$ such that $$U^*MU=T$$ wih $T$ an upper triangular matrix with diagonal elements the eigenvalues of $M$. The unitary matrix $U$ is created in an iterative manner as follows.
Assume $\lambda_1\cdots,\lambda_n$ eigenvalues of $M$. Let $v_1$ normalized eigenvector of $M$ that corresponds to eigenvalue $\lambda_1$. Then we can create an orthonormal basis comprising of $v_1,z_2,\ldots,z_n$ (Gram-Schmidt orthonormalization process). Define now $$V_1=\left[\matrix{v_1& z_2 & \cdots & z_n}\right]$$ Then $$V_1^*MV_1=\left[\matrix{\lambda_1 & *\\ 0 & M_1}\right]$$ with $M_1$ square with dimensions $(n-1)\times (n-1)$ having eigenvalues $\lambda_2\cdots,\lambda_n$.
Let $v_2\in\mathbb{C}^{n-1}$ normalized eigenvector of $M_1$ that corresponds to eigenvalue $\lambda_2$. Then we can similarly create an orthonormal basis comprising of $v_2,\cdots$. Those vector can be used as columns of a unitary matrix $U_2$ such that $$U^*_{2}M_1U_2=\left[\matrix{\lambda_2 & *\\ 0 & M_2}\right]$$ with $M_2$ square with dimensions $(n-2)\times (n-2)$ having eigenvalues $\lambda_3\cdots,\lambda_n$. If we now define $$V_2=\left[\matrix{1 & 0\\ 0 & U_2}\right]$$ then $$V_2^*V_1^*MV_1V_2= \left[\matrix{\lambda_1 & * & *\\ 0 & \lambda_2 & * \\ 0 & 0 & M_2}\right]$$ Repeating in this way we can prove $$V_{n-1}^*\cdots V_2^*V_1^*MV_1V_2\cdots V_{n-1}= \left[\matrix{\lambda_1 & * & * & *\\ 0 & \lambda_2 & * & *\\ 0 & 0 & \ddots & *\\ 0& 0 & \cdots & \lambda_n}\right]$$ i.e. $U^*MU=T$ with unitary $U=V_1V_2\cdots V_{n-1}$.
Then for a Hermitian $M$ (symmetric if $M$ has real elements) it is easy to prove that an upper triangular $T$ satisfying $$TT^*=U^*MUU^*M^*U=U^*MM^*U=U^*M^*MU=U^*M^*UU^*MU=T^*T$$ must be diagonal.
