I am trying to prove that Hermitian Matrices are diagonalizable.
I have already proven that Hermitian Matrices have real roots and any two eigenvectors associated with two distinct eigen values are orthogonal. If $A=A^H;\;\;\lambda_1,\lambda_2$ be two distinct eigenvalues and $v_1,v_2$ be two eigenvectors associated with them.
$$Av_1=\lambda_1v_1\Rightarrow v_1^HAv_1=\lambda_1v_1^Hv_1\Rightarrow \lambda_1=\lambda_1^*.$$
Similarly,
$$Av_1=\lambda_1v_1\Rightarrow v_2^HAv_1=\lambda_1v_2^Hv_1\Rightarrow v_1^H\lambda_2v_2=\lambda_1v_1^Hv_2\Rightarrow v_1^Hv_2=0$$
However, I am not aware of Spectral Theorem. Given this circumstances, how can I prove that Hermitian Matrices are diagonalizable? It should follow from above but the only sufficient condition of a matrix being diagonalizable is to have $dim(A)$ distinct eigenvalues, or existance of $P$ such that $A=P^{-1}DP$. I am not sure how this follows from above conditions.
Best Answer
The proof of this property is not so easy as those of the basic properties of eigenvalues and eigenvectors. It can be shown by induction, or by explicit construction (see eg here)
I like to visualize the property in this way:
We know that an hermitian matrix with $n$ distict eigenvalues has $n$ eigenvectors that are not only LI (as in general matrices) but, more than that, orthogonal. We also know that this matrix is diagonalizable, with unitary $U$ (both properties are easy to prove).
Now, if our hermitian matrix happens to have repeated (degenerate) eigenvalues, we can regard it as a perturbation of some another hermitian matrix with distinct eigenvalues. By a continuity argument, we should see that the matrix perturbation than transforms different (but perhaps close) eigenvalues into coincident ones, cannot make the orthogonal eigenvectors linearly dependent.
Put in other way: an hermitian matrix $A$ with repeated eigenvalues can be expressed as the limit of a sequence of hermitian matrices with distinct eigenvalues. Because all members of the sequence have $n$ orthogonal eigenvectors, by a continuity argument, they cannot end in LD eigenvectors.
This approach leads to a nice intuition, IMO, and it can be formalized. But for a formal proof the other methods are to be preferred.