Spectral Theorem

This theorem is made up of two parts

1. All real symmetric matrices are orthonormally diagonalizable.
2. All real symmetric matrices have real eigenvalues.

Proof for Theorem 2

Suppose $\lambda$ and $\vec{v}$ is an eigenvalue and eigenvector for the matrix $A$.

Then, we have that:

$$A\vec{v}=\lambda \vec{v}$$

Multiplying by the transpose of $\vec{v}$, we have that:

$${\vec{v}}^{T}A\vec{v}=\lambda {\vec{v}}^T\vec{v}=\lambda ||\vec{v}|{|}^2$$

Taking the complex conjugate of both sides, we have that:

$$\overline{{\vec{v}}^{T}A\vec{v}}=\overline{\lambda ||\vec{v}|{|}^2}=\overline{\lambda }||\vec{v}|{|}^2$$

But as $A$ is real and symmetric, we have $A=\overline{A}=A^T=\overline{A^T}$, and thus:

$${\vec{v}}^{TA\vec{v}}=\overline{{\vec{v}}^{T}A\vec{v}}$$

And finally

$$\lambda ||\vec{v}|{|}^2=\overline{\lambda }||\vec{v}|{|}^2$$

As $||\vec{v}|{|}^2\ne 0$, we have that $\lambda =\overline{\lambda }$, and thus $\lambda$ is real.