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We know that every diagonalisable operator $T$ on an infinite dimensional Hilbert space $H$ is normal. The converse is not necessarily true. In fact, every compact normal operator is diagonalisable. We find out that a normal operator on an infinite dimensional Hilbert space can not be of finite rank. I am searching for some examples to find the diagonalisable operators which are not compact. please help

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    $\begingroup$ The identity${}$? $\endgroup$ – Lord Shark the Unknown Jul 22 '17 at 15:30
  • $\begingroup$ yea, you are right $\endgroup$ – M.Kardel Jul 24 '17 at 14:17
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The identity is the easiest example (as suggested by both Lord Shark the Unknown and Jose Carlos Santos).

Here is a bit more. Let $H$ be an infinite dimensional Hilbert space and let $\{e_i\}_{i=1}^{\infty}$ be an orthonormal basis of $H$. Let $\{\lambda_i\}_{i=1}^{\infty}\subset\mathbb{R}$ be a sequence of bounded real numbers. Consider the bounded linear operator $A:H\to H$ given by $$Ax=\sum_{i=1}^{\infty}\lambda_i(x,e_i)e_i.$$ Clearly $A$ above is diagonalizable. Furthermore, you can show that $A$ is diagonalizable if and only if it has a representation of the above form. Finally, you can show that $A$ is compact if and only if $\lambda_i\to 0$ as $i\to\infty$. In particular, you have a representation of all of the non-compact diagonalizable operators.

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  • $\begingroup$ Yes, you are right $\endgroup$ – M.Kardel Jul 24 '17 at 14:18
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Let $H$ be an infinite-dimensional Hilert space. Then the identity map is diagonalisable, but not compact.

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