[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 !!(Bow)

Re: Self adjoint and trace

Quote:

Originally Posted by

**Borseti** I thought use this property: <A,B> = tr(B*A)

Why? If $\displaystyle A\in \mathbb{C}^{n\times n}$ is the matrix corresponding to a self adjoint operator then, $\displaystyle A$ is similar to a diagonal matrix $\displaystyle D=\textrm{diag}\;(\lambda_1,\ldots,\lambda_n)\in \mathbb{R}^{n\times n}$ . Now, use that similar matrices have the same trace.

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

Re: Self adjoint and trace

Nice solution !! (Clapping)

Thanks a lot !!!

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.

Re: Self adjoint and trace

Thanks, I thought the same thing !!!

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.

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 $\displaystyle A\mapsto \tex{tr}(AA^\ast)$ defines a norm (the Frobenius norm or Hilbert-Schmidt norm--the name's to taste) and so $\displaystyle A=0\Leftrightarrow \|A\|=0\Leftrightarrow 0=\text{tr}(AA^\ast)=\text{tr}(A^2)$.

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.

Re: Self adjoint and trace

Quote:

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.