# Is there any diagonalisable operator that is not compact?

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

• The identity${}$? – Lord Shark the Unknown Jul 22 '17 at 15:30
• yea, you are right – M.Kardel Jul 24 '17 at 14:17

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.
Let $H$ be an infinite-dimensional Hilert space. Then the identity map is diagonalisable, but not compact.