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 $T=0$.

  • $\begingroup$ Compact operators need not have finite rank. They only need to be the limit of finite rank operators in the operator norm. $\endgroup$ – RKD May 12 '12 at 12:56

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 $T=T^*$...

  • $\begingroup$ 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. $\endgroup$ – Ragib Zaman May 12 '12 at 13:17
  • 2
    $\begingroup$ @RagibZaman: $\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ $\endgroup$ – Nate Eldredge May 12 '12 at 13:24
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    $\begingroup$ Ahh, very naively then =). Thanks Nate! $\endgroup$ – Ragib Zaman May 12 '12 at 13:29

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