Linear Algebra/Spectral Theorem

Given a Hermitian matrix , is always diagonalizable. It is also the case that all eigenvalues of are real, and that all eigenvectors are mutually orthogonal. This is given by the "Spectral Theorem": The Spectral Theorem Given any Hermitian matrix , there exists an unitary matrix , and an diagonal matrix of real values such that The columns of are the eigenvectors of , and the diagonal entries of are the corresponding eigenvalues. In essence can be decomposed into a "spectrum" of rank 1 projections: The spectral theorem can in fact be proven without the need for the characteristic polynomial of , or any of the derivative theorems. Proof of the Spectral Theorem The proof will proceed by using induction on . Base Case When , it must be the case that is real, or else is not Hermitian. The spectral decomposition is then simply Inductive Case Let denote the standard basis vector of . Let denote a matrix of 0s. Let be a unit length vector that maximizes (recall that is always real), and let . Let be a unitary matrix where the first column is : . where . It will now be shown that has the form so that the (1,1) entry of is . It will now be shown that the first row and column of is filled with 0s except for the first entry. For an arbitary , consider the parameterized unit vector . where denotes the entry of ( and denote the real and imaginary components respectively). . Unit vector maximizes which implies that is the unit vector that maximizes . Therefore which gives . Now consider the parameterized unit vector . so . Therefore . It has been proven that the first row and column of is filled with 0s except for the first entry. has the form , and by an inductive argument, has the spectral decomposition . Therefore where and