Hi all !!

Someone can show me if that demonstration is going right ??

Let A, B self adjoint operators such that AB = BA. Show that exist unique ortonormal base that diagonalize simultaneously A and B.

Solution:

How AB = BA then A and B are commutative operators. Let $\displaystyle \lambda$ an eigenvalue of A and $\displaystyle E_{\lambda}$ the self space associate. Let $\displaystyle v \in E_{\lambda}$ such that:

$\displaystyle A v = \lambda v$

Then

$\displaystyle AB v = BA v \Rightarrow B \lambda v = \lambda Bv \Rightarrow Bv \in E_{\lambda} $

Then $\displaystyle E_{\lambda}$ is invariant by B. Then v is an eigenvector common the A and B, then:

$\displaystyle Av = \lambda v $

$\displaystyle Bv = \gamma v $

But in this point my demonstration becomes strange, because I assume that exist $\displaystyle v^{\perp}$, but I'm not sure that is correct...

One more time, thanks a lot...