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