Every self-adjoint matrix is unitarily similar to a diagonal matrix I am reading some books and the author stated that:

A self-adjoint $n \times n$ matrix $A$ can be diagonalizabled, more precisely, there is a diagonal matrix $D$ and a unitary matrix $U$ such that $A = UDU^{-1}$.

Actually, I can prove that $A $ is diagonalizable using induction.
However, I do not have any idea of being unitary of $U$.
Could you please give me some material or hint?
Thank you so much.
 A: If you've been able to prove that $A$ is diagonalizable, then the main thing remaining is to prove that eigenvectors corresponding to distinct eigenvalues are orthogonal. This is true because you can inside an eigenspace define an orthogonal basis using orthogonalisation.
Now it is easy to prove that the eigenvalues are real. Let's suppose that $\alpha \neq 0$ is an eigenvalue associated to the eigenvector $u$. Then:
$$\alpha \langle u , u \rangle =  \langle \alpha u , u \rangle = \langle Au , u \rangle = \langle u , A^*u \rangle = \langle u , Au \rangle = \langle u , \alpha u \rangle = \bar{\alpha}\langle u , u \rangle$$ and $\alpha = \bar{\alpha}$ as $\langle u , u \rangle \neq 0$.
Let's prove that that eigenvectors corresponding to distinct eigenvalues are orthogonal. Suppose that $\alpha \neq \beta$ are eigenvalues associated respectively to eigenvectors $u$ and $v$. You have
$$\alpha \langle u , v \rangle =  \langle \alpha u , v \rangle = \langle Au , v \rangle = \langle u , A^*v \rangle = \langle u , Av \rangle = \langle u , \beta u \rangle = \bar{\beta}\langle u , v \rangle = \beta \langle u , v \rangle$$
Based on that we can conclude as desired that $\langle u , v \rangle = 0$ for $\alpha \neq \beta$.
