# 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 â‡’ WâŠ¥ is A-invariant.

Want: âˆ€ xâˆˆWâŠ¥, Ax âˆˆ WâŠ¥âŒ©Ax,wâŒª = 0 âˆ€ w âˆˆ W â‡ Orthonormal

AwâˆˆW â‡ W is A-invariant

Since any self-adjoint matrix is ortho-diagonalizable, if A is self-adjoint, then âˆƒ an orthonormal basis Bâˆˆâ„‚n 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, .. vk), A-invariant (this is trivial, but see Appendix A)

then WâŠ¥ is A-invariant, then A is restricted to subspace WâŠ¥

AWâŠ¥: WâŠ¥ â†’ WâŠ¥

Then âˆƒ vâˆˆWâŠ¥, an eigenvector of AWâŠ¥.

But since AWâŠ¥v = Av, 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, .. ,vk, v/ ||v||} is an orthogonal set of eigenvectors.

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

### Conclusion

HTML is really not suited for doing math.

Appendix A