1
$\begingroup$

Let $T$ be a continuous and bounded self-adjoint compact operator on a Hilbert space H.

I want to prove that if $T^2=0$, then $T=0$.


Is there any thing wrong with the following:

$T^2$ = $TT^*=0$ impiles that all of T's eigenvalues are zero, and so as T is a compact operator and therefore has finite rank, all of T's eigenvalues are zero, and so $T=0$.

  • 0
    Compact operators need not have finite rank. They only need to be the limit of finite rank operators in the operator norm.2012-05-12

1 Answers 1

6

Your proof is not quite right; compact operators need not have finite rank. The idea does work: a compact self-adjoint operator has an orthonormal basis of eigenvectors (this is the spectral theorem), so if all its eigenvalues are zero, it must be the zero operator.

However, there is a much simpler argument that doesn't need the spectral theorem: for any vector $x$, note that $\| Tx \|^2 = \langle Tx, Tx \rangle$. Now use the fact that $T=T^*$...

  • 0
    Somewhat tangentially, naively it seems to me that in any normed space, a continuous operator satisfying $T^2=0$ should have $T=0.$ Can you give me a counterexample? I haven't been able to think of one.2012-05-12
  • 2
    @RagibZaman: $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$2012-05-12
  • 1
    Ahh, *very* naively then =). Thanks Nate!2012-05-12