Thread: 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 an eigenvalue of A and the self space associate. Let such that:



Then



Then is invariant by B. Then v is an eigenvector common the A and B, then:





But in this point my demonstration becomes strange, because I assume that exist , but I'm not sure that is correct...

One more time, thanks a lot...

I rewrote my demonstration:

Let two self adjoint operators such that AB = BA, let an eigenvalue of A and a self space associate. Now let not null vector such that:



How AB = BA, then:



Logo is invariant by B. Then v is an eigenvector common on A and B, then exist such that:



How and are real roots of characteristic polynomials of A and B, then and are both not invertible, then:





Then v belong to orthonormal base of eigenvectors of A and B, therefore the base diagonalize A and B simultaneously.

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.