Question on the self-adjointness of an operator Given a Hilbert space (separable) $\mathcal{H}$ with an orthonormal basis $\{e_i\}_{i=1}^{\infty}$, define an operator $T$ with domain $\mathcal{D}(T)$ equal to the span of $\{e_i\}$ by $Te_i:=\lambda_ie_i$, for some $λ_i∈R$.
I am trying to show that $T$ is an essentially self-adjoint operator. That is, $T^*=\overline{T}$. In order to do so, I only need to prove that $(T^{*} \pm i)$ is injective. Or, equivalently, ker$(T^{*} \pm i)=\{0\}.$
But I'm not entirely sure how to show that ker$(T^{*} \pm i)=\{0\}.$
Any help or hint will be very useful. Thanks.
 A: The domain of $T^*$ is
$$
D(T^*)=\{y\in H:\ x\longmapsto\langle Tx,y\rangle\ \text{ is bounded on }D(T) \}.
$$
So we want to analyze the boundedness of the map
$$
\phi(x)=\langle Tx,y\rangle=\sum_k\lambda_kx_ky_k,\qquad x\in D(T).
$$
The sum is finite because $x$ has finitely many nonzero terms. Using Cauchy Schwarz,
$$
|\phi(x)|^2\leq\|x\|^2\,\sum_k|\lambda_k|^2\,|y_k|^2.
$$
So
$$
D(T^*)\supset\{y:\ \sum_k|\lambda_k|^2\,|y_k|^2<\infty\}.
$$
Conversely, if $y\in D(T^*)$ then there exists $c>0$ such that $|\langle Tx,y\rangle|\leq c$ for all $x\in D(T)$ with $\|x\|=1$. Being bounded, the functional extends to all of $H$. Then
$$
\sum_k|\lambda_k|^2\,|y_k|^2=\sup\Big\{\Big|\sum_kx_k\,\lambda_ky_k\Big|:\ \sum_k|x_k|^2=1\Big\}=\sup\{|\langle Tx,y\rangle|:\ \|x\|=1\}\leq c. 
$$
Thus
$$
D(T^*)=\{y:\ \sum_k|\lambda_k|^2\,|y_k|^2<\infty\}
$$
and
$$
T^*y=\sum_k\lambda_ky_ke_k. 
$$
Now suppose that $x\in D(T^*)$ and $(T^*+i)x=0$. That is, $T^*x=-ix$. The fact that $x\in D(T^*)$ guarantees that $\sum_k\lambda_kx_ke_k\in H$. The equality $T^*x=-ix$ is
$$
\sum_k\lambda_kx_ke_k=-i\sum_kx_ke_k. 
$$
Comparing coordinate-wise, this gives us $\lambda_kx_k=-ix_k$. Because $\lambda_k$ is real, this equality can only happen when $x_k=0$. Being the case for all $k$, we get that $x=0$ and so $T^*+i$ is injective.
A: You can also show that $(x,T(y))= (T(x),y)$ , in fact : $$(\sum  x_i e_i,\sum \lambda_i y_i e_i)= (\sum \lambda_i x_i e_i, \sum y_i e_i)= \sum_{i,j} \lambda_i x_i y_j(e_i,e_j)$$
