Linear Algebra/Spectral Theorem

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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