Describing $T^*$ Let $H$ be a Hilbert space, let $(e_n)$ be an orthonormal series in $H$ and $(\lambda_n)$ be a scalar bounded series.
Let $T$ be the (only) operator in $B(H)$ such that:
$Te_n=\lambda_n e_n$ for all $n\in N$, and $Tx=0$ for all $x$ orthogonal to $(e_n)$.
Describe $T^*$.
It seems related to Let $Te_n=e_{n+1}$ show that $T:H\to H$ is isometric.
But not much similar.
 A: Let me prove that $T^*$ vanishes on the complement of the span of the $e_n$'s (call this orthogonal complement $X$, i.e. $X:=(\text{span}\{ e_n \ : \ n\in \mathbb{N} \})^\perp$). Fix some $x\in X$, then we have for all $y\in X$
$$ \langle T^*(x), y\rangle = \langle x, T(y) \rangle = \langle x, 0 \rangle = 0. $$
Thus, $T^*(x)$ is in the orthogonal complement of $X$. Now we also test on the $e_n$'s and find
$$ \langle T^*(x), e_n \rangle = \langle x, T(e_n) \rangle = \lambda_n \langle x, e_n \rangle = 0 $$
as $X$ is the orthogonal complement of the span of the $e_n$'s. Hence, $T^*(x)$ orthogonal to all elements in $H$ and thus equal to zero.
Now you can try to play a similar game to determine the values of $T^*(e_n)$. You will get $T^*(e_n)= \overline{\lambda}_n e_n$ (bar means complex conjugation).
Added: Let me also do the second case. We have
$$ \langle T^*(e_n) - \overline{\lambda}_n e_n, e_m \rangle
= \langle e_n, T(e_m) \rangle - \lambda_n \langle e_n, e_m \rangle
= \lambda_m \langle e_n, e_m \rangle - \lambda_n \langle e_n, e_m \rangle
= \lambda_m \delta_{n,m} - \lambda_n \delta_{n,m} =0. $$
On the other hand, we have for $x$ in the orthogonal complement of $X$
$$  \langle T^*(e_n) - \overline{\lambda}_n e_n, x \rangle
= \langle e_n, T(x) \rangle - \lambda_n \langle e_n, x \rangle
= \langle e_n, 0 \rangle - \lambda_n 0 =0. $$
So again, $T^*(e_n) - \overline{\lambda}_n e_n$ is orthogonal to all elements in $H$ and hence we have $T^*(e_n) - \overline{\lambda}_n e_n=0$, which implies $T^*(e_n) = \overline{\lambda}_n e_n$.
