I'm sorry if this question is a duplicate.
Suppose $(e_{n})_{n\in\mathbb{N}}$ is the usual orthonormal basis of $\ell^{2}(\mathbb{N})$. We can define an operator $v\colon H\to H$ by $ve_{n}:=e_{n+1}$. I'm trying to prove that its adjoint is given by the left shift (that annihilates $e_{1}$). So I computed $$\langle e_{m},v^{*}e_{n}\rangle=\langle ve_{m},e_{n}\rangle=\langle e_{m+1},e_{n}\rangle=\delta_{m+1,n}.$$ From this I deduced that $v^{*}e_{1}$ is orthogonal to each basis vector $e_{m}$, which implies that $v^{*}e_{1}=0$. But I am missing the formal argument to conclude that $v^{*}e_{n}=e_{n-1}$ for $n\geq2$. Any help is greatly appreciated!
For $n \geq 2$ we have $ \langle e_m, v^{*}e_n \rangle=\delta _{m+1,n}=\langle e_m, e_{n-1} \rangle$ for all $m$. If $x$ and $y$ are such that $\langle e_m, x \rangle= \langle e_m, y \rangle$ for all $m$ the $x=y$. Hence $v^{*}e_n=e_{n-1}$.