• Dec 1st 2011, 08:48 AM
Borseti
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...
• Dec 1st 2011, 12:06 PM
Borseti
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.
• Dec 1st 2011, 10:16 PM
Hartlw