Need a help in understanding a solution of a problem in Israel Gohberg. The question and its solution is given in the following pictures:


By (2.a) he means what is given in the following picture:

All the solution is clear for me except the last 2 lines:
1- it is not clear for me why he did not name $x_{j}$ by $(\xi_{j})$ instead.Could anyone explain this for me please?
2- The idea that he uses in the last 2 lines is not clear for me in general, could anyone clarify it for me please?
Thanks! 
 A: The notational issue is with sequences of vectors in $\ell^2$, which is in turn a space of sequences of complex numbers.
$$x_j = (\underbrace{0, 0, \ldots, 0}_{j-1}, 1, 0, \ldots)$$ is an element of $\ell^2$, so $(x_j)_{j=1}^\infty$ is a sequence of vectors in $\ell^2$. These particular vectors $x_j$ are usually denoted by $e_j$.
Author in this context uses the notation $(\xi_j)_{j=1}^\infty$ to represent a sequence $(\xi_1, \xi_2, \ldots, ) \in \ell^2$ of complex numbers. It would be confusing to use the same notation $(\xi_j)_{j=1}^\infty$ to represent a sequence of vectors in $\ell^2$.

The idea used here is that for any bounded linear map $A : H \to H$ on a Hilbert space $H$ we have the decomposition
$$H = \operatorname{Ker} A \oplus \overline{\operatorname{Im} A^*} = \operatorname{Ker} A^* \oplus \overline{\operatorname{Im} A}$$
In particular for $D_\omega : \ell^2 \to \ell^2$ we have:
$$\ell^2 = \operatorname{Ker} D_\omega \oplus \overline{\operatorname{Im} D_\omega^*} = \operatorname{Ker} D_\omega^* \oplus \overline{\operatorname{Im} D_\omega}$$
First we determined by direct calculation that
$$\operatorname{Ker} D_\omega = \big\{(\xi_j)_{j=1}^\infty\in\ell^2 : \omega_j\ne 0 \implies \xi_j = 0 \big\}$$
Then we noticed that the adjoint operator is given by $$D_\omega^*\left((\xi_j)_{j=1}^\infty\right) = \left(\overline{\omega_j}\xi_j\right)_{j=1}^\infty$$
and we calcualated:
$$\operatorname{Ker} D_\omega^* = \big\{(\xi_j)_{j=1}^\infty\in\ell^2 : \overline{\omega_j}\ne 0 \implies \xi_j = 0 \big\}= \operatorname{Ker} D_\omega$$
because $\overline{\omega_j} = 0$ if and only if $\omega_j = 0$.
So now we simply have:
$$\overline{\operatorname{Im} D_\omega} = \left(\operatorname{Ker} D_\omega^*\right)^\perp = \left(\operatorname{Ker} D_\omega\right)^\perp$$
It remains to calculate the orthogonal complement of the subspace $\operatorname{Ker} D_\omega$, i.e. the space of all vectors orthogonal to $\operatorname{Ker} D_\omega$. We claim that:
$$\left(\operatorname{Ker} D_\omega\right)^\perp = \big\{(\xi_j)_{j=1}^\infty\in\ell^2 :\omega_j= 0 \implies \xi_j = 0 \big\}$$
Indeed, for any sequence $(\xi_j)_{j=1}^\infty\in\ell^2$ such that $\omega_j= 0 \implies \xi_j = 0$ and for any sequence $(\eta_j)_{j=1}^\infty \in \operatorname{Ker} D_\omega$ we have
$$\left\langle(\xi_j)_{j=1}^\infty, (\eta_j)_{j=1}^\infty \right\rangle = \sum_{j=1}^\infty \xi_j \overline{\eta_j} = 0$$
since if $\omega_j = 0$ then $\xi_j = 0$, and if $\omega_j \ne 0$ then $\eta_j = 0$ (because $(\eta_j)_{j=1}^\infty \in \operatorname{Ker} D_\omega$).
Therefore, $(\xi_j)_{j=1}^\infty\in\ell^2 \perp \operatorname{Ker} D_\omega$. We can conclude $$\{(\xi_j)_{j=1}^\infty\in\ell^2 :\omega_j= 0 \implies \xi_j = 0 \big\} \subseteq \left(\operatorname{Ker} D_\omega\right)^\perp$$
Conversely, take $(\xi_j)_{j=1}^\infty\in\ell^2 \in \left(\operatorname{Ker} D_\omega\right)^\perp$. Notice that the vector $$x_j = e_j = x_j = (\underbrace{0, 0, \ldots, 0}_{j-1}, 1, 0, \ldots)$$
is in $\operatorname{Ker} D_\omega$ for any $j \in \mathbb{N}$ such that $\omega_j = 0$. That is because for any $k \in \mathbb{N}$ such that $\omega_k \ne 0$ we have that $k$-th coordinate of $x_j$ is zero, which is precisely the condition for a sequence to be in $\operatorname{Ker} D_\omega$.
Therefore, for every $k \in \mathbb{N}$ such that $\omega_k = 0$ we have:
$$0 = \left\langle \underbrace{x_k}_{\in \operatorname{Ker} D_\omega}, \underbrace{(\xi_j)_{j=1}^\infty}_{\in \left(\operatorname{Ker} D_\omega\right)^\perp}\right\rangle = \xi_k$$
Therefore, $\omega_k = 0 \implies \xi_k = 0$ so $(\xi_j)_{j=1}^\infty \in \big\{(\xi_j)_{j=1}^\infty\in\ell^2 :\omega_j= 0 \implies \xi_j = 0 \big\}$. Hence, $$\left(\operatorname{Ker} D_\omega\right)^\perp \subseteq \big\{(\xi_j)_{j=1}^\infty\in\ell^2 :\omega_j= 0 \implies \xi_j = 0 \big\}$$
Therefore, we conclude $$\left(\operatorname{Ker} D_\omega\right)^\perp = \big\{(\xi_j)_{j=1}^\infty\in\ell^2 :\omega_j= 0 \implies \xi_j = 0 \big\}$$
so $$\overline{\operatorname{Im} D_\omega} = \big\{(\xi_j)_{j=1}^\infty\in\ell^2 :\omega_j= 0 \implies \xi_j = 0 \big\}$$
