• December 1st 2011, 09:48 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 $\lambda$ an eigenvalue of A and $E_{\lambda}$ the self space associate. Let $v \in E_{\lambda}$ such that:

$A v = \lambda v$

Then

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

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

$Av = \lambda v$

$Bv = \gamma v$

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

One more time, thanks a lot...
• December 1st 2011, 01:06 PM
Borseti
I rewrote my demonstration:

Let $A,B: E \rightarrow E$ two self adjoint operators such that AB = BA, let $\lambda$ an eigenvalue of A and $E_{\lambda}$ a self space associate. Now let $v \in E$ not null vector such that:

$Av = \lambda v$

How AB = BA, then:

$AB = BA \Rightarrow ABv = BAv \Rightarrow AB v = B \lambda v \Rightarrow B \lambda v = \lambda B v \therefore B v \in E_{\lambda}$

Logo $E_{\lambda}$ is invariant by B. Then v is an eigenvector common on A and B, then exist $\gamma$ such that:

$B v = \gamma v$

How $\lambda$ and $\gamma$ are real roots of characteristic polynomials of A and B, then $A - \lambda \textrm{I}$ and $B - \gamma \textrm{I}$ are both not invertible, then:

$Av = \lambda v \Rightarrow (A - \lambda \textrm{I}) v = 0$

$Bv = \gamma v \Rightarrow (B - \gamma \textrm{I})v = 0$

Then v belong to orthonormal base $\mathbb{B} \subset E$ of eigenvectors of A and B, therefore the base $\mathbb{B}$ diagonalize A and B simultaneously.
• December 1st 2011, 11:16 PM
Hartlw