If $ T\in B(H)$ is a finite rank operator, then there are orthonormal vectors $e_1,...,e_n$ and vectors $g_1,...,g_n$ such that $Th=\sum_{i=1}^n (h,e_i )g_i$,

then we can easily see that $T$ is normal iff $g_i=\lambda_i e_i$ for some scalars $\lambda_1,...,\lambda_n$.

Also for every $I=1,...,n$, $e_i\in \ker (T-\lambda_i)$ which shows that $\lambda_i\in \sigma(T)$.

For general case, I want to describe what possible subsets of C could be spectrum of finite rank operators, but I do not know how should do it. Please hint me.

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    $\begingroup$ Thanks. Could you please edit your question to clarify what you are asking? (The answer is the same as for the normal case.) $\endgroup$ – Jonas Meyer Aug 11 '14 at 6:49

Hint: Since $T$ is finite rank, you can pick an orthonormal basis $g_1,\ldots,g_n$ of the range of $T$, $$ Tv=\sum_{j=1}^n \langle Tv,g_j\rangle g_j. $$

Show that $\mbox{ker } T\supset \mbox{span }\{T^*g_1,\ldots,T^*g_n\}^\perp$.

Pick and orthonormal basis of $\mbox{span }\{g_1,\ldots,g_n,T^*g_1,\ldots,T^*g_n\}$, say $e_1,\ldots,e_m$. Show that $m\leq 2n$.

Extend $\{e_j\}$ to an orthonormal basis and show that $Tv=\sum_{i,j=1}^m v_iM_{ji}e_j$, where $v_i=\langle v,e_i\rangle,M_{ij}=\langle Me_j,e_i\rangle$.

Consider the decomposition of your Hilbert space $H$ into $H=A\oplus B$, where $$ A=\mbox{span }\{e_1,\ldots,e_m\},\,\; B=\mbox{span }\{e_{m+1},\ldots\}. $$ Use this decomposition to reduce the analysis of $T$ to the analysis of the matrix $M$.

  • $\begingroup$ Thanks for your hint. I can not understand the last line. Please regard me. $\endgroup$ – niki Aug 11 '14 at 17:47

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