Proof that any self-adjoint matrix is diagonalizable.

(Sorry. If you don’t know what this is, please ignore it. It’s not important. Really.)

Setup: If A is self-adjoint, and W is an A-invariant subspace ⇒ W^⊥ is A-invariant.

Want: ∀ x∈W^⊥, Ax ∈ W^⊥〈Ax,w〉 = 0 ∀ w ∈ W ⇐ Orthonormal

Given:〈Ax,w) =〈x,Aw〉 ⇐ Self-Adjoint

Aw∈W ⇐ W is A-invariant

then〈x∈W^⊥, Aw∈W〉= 0

Since any self-adjoint matrix is ortho-diagonalizable, if A is self-adjoint, then ∃ an orthonormal basis B∈ℂⁿ made out of eigenvectors such that [A]_B

Want: k=n (that is, an orthonormal basis made out of eigenvectors).

Proof by contradiction: Suppose k less then n

Given:

W=span(v₁, v₂, .. v_k), A-invariant (this is trivial, but see Appendix A)

then W⊥ is A-invariant, then A is restricted to subspace W^⊥

A_W^⊥: W^⊥ → W^⊥

Then ∃ v∈W^⊥, an eigenvector of A_W^⊥.

But since A_W^⊥v = A_v, v is an eignevector to of A perpendicular to W.

We assumed that S is maximal, but we ended up with a contradiction, since the set {v₁, v₂, .. ,v_k, v/ ||v||} is an orthogonal set of eigenvectors.

So k must be equal to n. As a result, A is orthodiagonalizable.

Conclusion

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Appendix A

If λ₁≠λ₂,〈v₁,v₂〉must be 0

Here is how: Av₁=λ₁v₁, Av₂=λ₂v₂

Given that: λ₁〈v₁,v₂〉 = 〈λ₁v₁,v₂〉= 〈Av₁,v₂〉= 〈v₁,Av₂〉= 〈v₁,λ₂v₂〉=λ₂〈v₁,v₂〉⇒ λ₁〈v₁,v₂〉= λ₂〈v₁,v₂〉

Since λ₁≠λ₂, 〈v₁,v₂〉=0

.

QED