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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.

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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: http://www.math.wisc.edu/~ellenber/Math204Lects/Week10.pdf)

I like to visualize the property thus: we know that, if we have $n$ distict eigenvalues we have $n$ eigenvectors that are non only LI (as in a general matrix) but, more than that, orthogonal; that the matrix is diagonalizable here (with unitary $U$) is trivial.

Now, if the hermitian matrix has repeated (degenerate) eigenvalues, it can be regarded as a *perturbation* of another hermitian matrix with distinct eigenvalues. And, by a continuity argument, we should see that the matrix pertubation than transforms different (but perhaps close) eigenvalues into coincident ones, cannot make the orthogonal eigenvectors linearly dependent.

This can give you a nice intutition, but for a formal proof, the other methods are to be preferred.

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