Consider the Hilbert space $H=l_2$ over the complex field and $A:H\rightarrow H.$
How to show whether this operator is normal? self-adjoint? unitary?
$A(x)=(x_1,0,0,\frac{1}{2}x_4,0,0,0,0,\frac{1}{3}x_9,0,...,0,\frac{1}{n}x_{n^2},0,...)$
Could you please help.
First you should convince yourself that $A$ is bounded (it is in fact compact as it is the limit of finite rank operators). Then you probably should compute this adjoint. So consider for any $x,y \in l^2$, $ (A(x),y) = x_1 \bar{y_1} + \frac{1}{2}x_4 \bar{y_4} + \cdots \frac{1}{n}x_{n^2} \bar{y_{n^2}} + \cdots = (x,A^*(y))$.
From this, you can see that $A^*(x) = A(x)$. It follows that $A$ is also normal as all self-adjoint operators are normal. $A$ cannot be unitary as it is not invertible, namely, the vector $(0,1,0,0,...)$ is not in its range.
Another way to see that $A$ is self-adjoint is to note that since it is compact, its spectrum consists of only eigenvalues (and possibly $0$). But inspection shows that its spectrum is just $\{1,1/2,1/3,....,0\}$, which is real. So $A$ is self-adjoint.