18.06.30: Spectral theorem

Lecturer: Barwick

And in the telling of that story I lose my way inside a prepositional phrase. – Wye Oak 18.06.30: Spectral theorem

For 푣, 푤 ∈ C푛, write ⟨푣|푤⟩ ≔ 푣∗푤.

This is a complex number, called the inner product of two complex vectors; it extends the usual dot product, but notices that the linearity in the first coordi- nate is twisted:

⟨훼푣|푤⟩ = 훼⟨푣|푤⟩ but ⟨푣|훼푤⟩ = 훼⟨푣|푤⟩.

The length of a vector 푣 ∈ C푛 is defined by ||푣||2 = ⟨푣|푣⟩; it’s precisely the same as the length of the corresponding vector in R2푛. (Why??) 18.06.30: Spectral theorem

Lemma. An 푛 × 푛 complex matrix 퐵 is Hermitian if and only if, for any 푣, 푤 ∈ C푛, ⟨퐴푣|푤⟩ = ⟨푣|퐴푤⟩.

Proof. If 퐴 is Hermitian, then (퐴푣)∗푤 = 푣∗퐴∗푤 = 푣∗퐴푤.

On the other hand, suppose that for any 푣, 푤 ∈ C푛,

⟨퐴푣|푤⟩ = ⟨푣|퐴푤⟩.

Then when 푣 =푒푖 ̂ and 푤 =푒푗 ̂ , this equation becomes

푖 ∗ ∗ 푗 푎푗푖 = (퐴 ) 푒푗̂ =푒푖 ̂ 퐴 = 푎푖푗. 18.06.30: Spectral theorem

Theorem (Spectral theorem; last big result of the semester). Suppose 퐵 a Her- mitian matrix. Then

(1) The eigenvalues of 퐵 are real.

(2) There is an orthogonal basis of eigenvectors for 퐵; in particular, 퐵 is diago- nalizable over C (and even over R if 퐵 has real entries). 18.06.30: Spectral theorem

Proof. Let’s first see why the eigenvalues of 퐵 must be real. Suppose 푣 ∈ C푛 an eigenvector of 퐵 with eigenvalue 휆, so that 퐵푣 = 휆푣. Then,

휆||푣||2 = 휆⟨푣|푣⟩ = ⟨푣|휆푣⟩ = ⟨푣|퐵푣⟩ = ⟨퐵푣|푣⟩ = ⟨휆푣|푣⟩ = 휆⟨푣|푣⟩ = 휆||푣||2.

Since 푣 ≠ 0, one has ||푣|| ≠ 0, whence 휆 = 휆. 18.06.30: Spectral theorem

Now let’s see about that orthogonal basis of eigenvectors. Using the Funda- mental Theorem of Algebra, write the characteristic polynomial

푝퐵(푡) = (푡 − 휆1) ⋯ (푡 − 휆푛), where 휆1,…, 휆푛 ∈ C are the roots of 푝퐵. We may not assume that the 휆푖’s are distinct!!

Let’s choose an eigenvector 푣1 with eigenvalue 휆1, and consider the hyperplane

푛 푊1 ≔ {푤 ∈ C | ⟨푣1|푤⟩ = 0}. 18.06.30: Spectral theorem

Note that for any 푤 ∈ 푊1, one has

⟨푣1|퐵푤⟩ = ⟨퐵푣1|푤⟩ = ⟨휆1푣1|퐵푤⟩ = 휆1⟨푣1|푤⟩ = 0,

푛 푛 so 퐵푤 ∈ 푊1 as well. Hence the linear map 푇퐵∶ C C restricts to a map 푇1∶ 푊1 푊1. 18.06.30: Spectral theorem

Select, temporarily, a C-basis {푤2,…, 푤푛} of 푊1. Then {푣1, 푤2,…, 푤푛} is a 푛 C-basis of C , and writing 푇퐵 relative to this basis gives us a matrix

휆1 0 퐶1 = ( ), 0 퐵1 where 퐵1 is the (푛 − 1) × (푛 − 1) matrix that represents 푇1 relative to the basis {푤2,…, 푤푛}, and 푝 = (푡 − 휆 ) ⋯ (푡 − 휆 ). 퐵1 2 푛 18.06.30: Spectral theorem

Now we run that same argument again with the (푛 − 1) × (푛 − 1) matrix 퐵1 in place of the 푛 × 푛 matrix 퐵 to get:

▶ an eigenvector 푣2 ∈ 푊1 with eigenvalue 휆2,

▶ the subspace 푊2 ⊂ 푊1 of vectors orthogonal to 푣2,

▶ and an (푛 − 2) × (푛 − 2) matrix 퐵2 that represents 푇퐵 restricted to 푊2. Now we find that 퐵 is similar to

휆1 0 0 퐶2 = ( 0 휆2 0 ). 0 0 퐵2 18.06.30: Spectral theorem

We repeat this argument repeatedly on each new 퐵푖, each time getting:

▶ an eigenvector 푣푖+1 ∈ 푊푖 with eigenvalue 휆푖+1,

▶ the subspace 푊푖+1 ⊂ 푊푖 of vectors orthogonal to 푣푖+1,

▶ and an (푛 − 푖 − 1) × (푛 − 푖 − 1) matrix 퐵푖+1 that represents 푇퐵 restricted to 푊푖+1. 18.06.30: Spectral theorem

At each stage, we find that 퐵 is similar to

휆1 0 ⋯ 0 0 0 휆2 ⋯ 0 0 ( ) 퐶푖+1 = ⋮ ⋮ ⋱ ⋮ ⋮ . 0 0 ⋯ 휆푖 0 ( 0 0 ⋯ 0 퐵푖+1 ) 18.06.30: Spectral theorem

This process eventually stops, when 푖 = 푛. Then we’re left with:

▶ eigenvectors 푣1,…, 푣푛 with eigenvalues 휆1,…, 휆푛, ▶ a string of subspaces

푛 C = 푊0 ⊃ 푊1 ⊃ 푊2 ⊃ ⋯ ⊃ 푊푛 = {0},

with 푣푖+1 ∈ 푊푖, and

푊푖+1 = {푤 ∈ 푊푖 | ⟨푣푖+1|푤⟩},

▶ and a diagonal matrix 퐶푛 = diag(휆1,…, 휆푛) to which 퐵 is similar.