1. ## [SOLVED]Self adjoint and trace

Hi ALL !!!

First, apology my english !!

Look this problem:

A is a self adjoint operator such that tr(A^2) = 0. Proof that A = 0.

I thought use this property:

<A,B> = tr(B*A)

Because A is a self adjoint operator then I'll have that:

<A,A> = tr(A*A) = tr(AA) = tr(A^2) = 0 => <A,A> = 0 => A = 0.

But, I don't know how show that <A,B> = tr(B*A).

Some one have some idea ???

Thanks a lot !!

2. ## Re: Self adjoint and trace

Originally Posted by Borseti
I thought use this property: <A,B> = tr(B*A)
Why? If $A\in \mathbb{C}^{n\times n}$ is the matrix corresponding to a self adjoint operator then, $A$ is similar to a diagonal matrix $D=\textrm{diag}\;(\lambda_1,\ldots,\lambda_n)\in \mathbb{R}^{n\times n}$ . Now, use that similar matrices have the same trace.

3. ## Re: Self adjoint and trace

following up on previous post:

A is similar to diagonal matrix A' whose diagonal elements are the eigenvalues of A. A'^2 = 0 so the sum of the squares of the eigenvalues of A are zero so the eigenvalues of A are zero so A is zero. (Trace is an invariant).

EDIT: Argument invalid unless you show TrA^2=TrD^2

4. ## Re: Self adjoint and trace

Nice solution !!

Thanks a lot !!!

5. ## Re: Self adjoint and trace

Missing step in previous argument:

D=(P^-1)AP
D^2 = [(P^-1)AP][(P^-1)AP] = (P^-1)(A^2)P

Therefore D^2 similar to A^2 and TrD^2 = TrA^2 = 0.

6. ## Re: Self adjoint and trace

Thanks, I thought the same thing !!!

7. ## Re: Self adjoint and trace

To finish off previous argument, if A and B are similar, Ax=Bx all x. So Ax=Dx=0 all x => A=0.

FrenandoRevilla took first step by pointing out a self-adjoint matrix is diagonalizable.

8. ## Re: Self adjoint and trace

While I do like Fernando's solution, it's clearly the best, I do think your made a good observation. Namely, the map $A\mapsto \tex{tr}(AA^\ast)$ defines a norm (the Frobenius norm or Hilbert-Schmidt norm--the name's to taste) and so $A=0\Leftrightarrow \|A\|=0\Leftrightarrow 0=\text{tr}(AA^\ast)=\text{tr}(A^2)$.

9. ## Re: Self adjoint and trace

Neither you nor Fernando solved the problem, or even came close.

The solution is as I gave it, which I summarize here

Given self-adjoint A and TrA^2=0. Prove A=0.

1) A is self-adjoint therefore diagonalizable, D=PAP^-1 and similar to A.

2) D^2 = P(A^2)P^-1 therefore D^2 is similar to A^2 and their trace is equal.

3) TrD^2 = TrA^2 =0. Trace D^2 is sum of squares of eigenvalues so they are all 0 and D=0.

4) If A and B are similar Ax=Bx all B, so Ax = Dx = 0 all x, so A=0

Edited to change a typo in 4: "all x," not "all D." While I'm here, might as well improve step 4) which is correct but overly sophisticated. A = (P^-1)DP = 0.

10. ## Re: Self adjoint and trace

Originally Posted by Hartlw
Neither you nor Fernando solved the problem, or even came close.
I want to know I had to rewrite this, because my initial post was, put lightly, curt.

I think you need to reexamine the situation.

11. ## Irrelevant Premise Proof

Irrelevant Premise Proof: The Premise is obviously true but irrelevant. Ex:

1) The sky is blue
2) 3 is an integer
Therefore: sqrt2 is rational

Advanced Irrelative Premise Proof: The premises are unnecessarily obtuse. If carefully crafted, they can contain unproven assumptions or need not even be true.

These proofs are rock solid if an attempt is made to question the truth of the premise (the sky is blue).