4

I am having trouble proving that Hermitian Matrices ($A = A^{*}$) are unitarily diagonalizable ($A = Q^{*}DQ$, where Q is a unitary matrix, $QQ^{*} = I$ and D is a diagonal matrix). I also know that given an hermitian operator T over a finite diminsional vector space V, V has an orthonormal basis which consists of eigenvectors of T.

Any suggestions?

Edit: Following Git Gud's suggestion:

We shall consider the Hermitian operator $L_A$. Let E be the standard basis of V. Also, by hypothesis, we have that there exists an orthonormal basis B for V which consists of eigenvectors of $L_A$. Since the basis consists of only eigenvectors, the basis contains $dim(V)$ linearly independent vectors, so $[L_A]_B$ is diagonal.

We know that $[L_A]_E = A$. Then, $$A = [L_A]_E = Q^{-1}DQ$$ and since Q must be unitary, then $Q^{-1} = Q^*$, by which we have that $$A = Q^*DQ$$

lalaman
  • 297

1 Answers1

13

If you know, and therefore do not need to prove that that there exists a basis of orthonormal vectors of $A$ then this is equivalent to stating that $A$ is unitarily diagonalizable, with the matrix $Q$ having as columns the orthonormal eigenvectors of $A$...

But there is also another way - if you may make use of Schur's theorem... Every $A\in \mathbb{C}_{n \times n}$ is unitarily similar to an upper triangular matrix. So we can find unitary matrix $Q$ so that $Q^*AQ$ is upper triangular. BUT since $A$ is Hermitian, so is $Q^*AQ$ (you can prove this: $(Q^*AQ)^*=Q^*A^*Q=Q^*AQ$), and therefore $Q^*AQ$ must be diagonal.

Christiaan Hattingh
  • 4,745