(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‡’ W^{âŠ¥} is A-invariant.

**Want:** âˆ€ xâˆˆW^{âŠ¥}, Ax âˆˆ W^{âŠ¥}âŒ©Ax,wâŒª = 0 âˆ€ w âˆˆ W â‡ Orthonormal

**Given:**âŒ©Ax,w) =âŒ©x,AwâŒª â‡ Self-Adjoint

AwâˆˆW â‡ W is A-invariant

thenâŒ©xâˆˆW^{âŠ¥}, AwâˆˆWâŒª= 0

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(v_{1}, v_{2}, .. v_{k}), A-invariant (this is trivial, but see Appendix A)

then WâŠ¥ is A-invariant, then A is restricted to subspace W^{âŠ¥}

A_{WâŠ¥}: W^{âŠ¥} â†’ W^{âŠ¥}

Then âˆƒ vâˆˆW^{âŠ¥}, an eigenvector of A_{WâŠ¥}.

But since A_{WâŠ¥}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 {v_{1}, v_{2}, .. ,v_{k}, 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**

If Î»_{1}â‰ Î»_{2},âŒ©v_{1},v_{2}âŒªmust be 0

Here is how: Av_{1}=Î»_{1}v_{1}, Av_{2}=Î»_{2}v_{2}

Given that: Î»_{1}âŒ©v_{1},v_{2}âŒª = âŒ©Î»_{1}v_{1},v_{2}âŒª= âŒ©Av_{1},v_{2}âŒª= âŒ©v_{1},Av_{2}âŒª= âŒ©v_{1},Î»_{2}v_{2}âŒª=Î»_{2}âŒ©v_{1},v_{2}âŒªâ‡’ Î»_{1}âŒ©v_{1},v_{2}âŒª= Î»_{2}âŒ©v_{1},v_{2}âŒª

Since Î»_{1}â‰ Î»_{2}, âŒ©v_{1},v_{2}âŒª=0

.

QED