
Self adjoint
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...

Re: Self adjoint
I rewrote my demonstration:
Let $\displaystyle A,B: E \rightarrow E$ two self adjoint operators such that AB = BA, let $\displaystyle \lambda$ an eigenvalue of A and $\displaystyle E_{\lambda}$ a self space associate. Now let $\displaystyle v \in E$ not null vector such that:
$\displaystyle Av = \lambda v$
How AB = BA, then:
$\displaystyle AB = BA \Rightarrow ABv = BAv \Rightarrow AB v = B \lambda v \Rightarrow B \lambda v = \lambda B v \therefore B v \in E_{\lambda}$
Logo $\displaystyle E_{\lambda}$ is invariant by B. Then v is an eigenvector common on A and B, then exist $\displaystyle \gamma$ such that:
$\displaystyle B v = \gamma v$
How $\displaystyle \lambda$ and $\displaystyle \gamma$ are real roots of characteristic polynomials of A and B, then $\displaystyle A  \lambda \textrm{I}$ and $\displaystyle B  \gamma \textrm{I}$ are both not invertible, then:
$\displaystyle Av = \lambda v \Rightarrow (A  \lambda \textrm{I}) v = 0$
$\displaystyle Bv = \gamma v \Rightarrow (B  \gamma \textrm{I})v = 0$
Then v belong to orthonormal base $\displaystyle \mathbb{B} \subset E$ of eigenvectors of A and B, therefore the base $\displaystyle \mathbb{B}$ diagonalize A and B simultaneously.

Re: Self adjoint
It looks like you are saying this:
Let u be the eigenvectors of A
Au = su
A(Bu) = s(Bu).......A and B commute
Bu = pu......... Bu is an eigenvector of A
u is an eigenvector of B
The eigenvectors of A and B are the same.
The matrix whose column vectors are u diagonalizes A and B.