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 $T=0$.
Best Answer
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 $T=T^*$...