Image and pseduo-inverse of an operator Let $\mathcal{H}$ be a Hilbert space and $(e_n)_{n\in \Bbb N}$ be an orthonormal basis for $\mathcal{H}$. 
Define the surjective operator $T\in B(\mathcal{H})$ such that $Te_{2n-1}=\frac{1}{2^n}e_1$ and $Te_{2n}= e_{n+1}$ for each $n\in\Bbb N$.
There are several questions: 
$\bullet$ What is the pseudo-inverse $T^\dagger$ of $T$? 
 How is $T^\dagger$  define? 
$\bullet$Decomposing $e_{2n-1}$ to $ f_{n,1}\oplus f_{n,2}$ where $f_{n,1} \in R(T^*)$ and $f_{n,2}\in \ker T$ for each $n$. Could we conclude $f_{n,1}, f_{n,2}$? ($R(T^*)$ is the image of $T^*$ and $\ker T$ is the kernel of $T$) 

The answer to the previous version of the question, which was is $e_{2k-1} \in R(T^*)$, is no.
We have
\begin{align}
T^*x &= \sum_{n=1}^\infty \langle T^*x, e_n\rangle e_n \\
&= \sum_{n=1}^\infty \langle x, Te_n\rangle e_n \\
&= \sum_{n=1}^\infty \langle x, Te_{2n-1}\rangle e_{2n-1} + \sum_{n=1}^\infty \langle x, Te_{2n-1}\rangle e_{2n}\\
&= \langle x, e_1\rangle\sum_{n=1}^\infty \frac1{2^n} e_{2n-1} + \sum_{n=1}^\infty \langle x, e_{n+1}\rangle e_{2n}
\end{align}
so if $T^*x = e_{2k-1}$, we must have $$1 = \langle T^*x, e_{2k-1}\rangle = \langle x, Te_1\rangle = \frac1{2^k}\langle x, e_1\rangle$$ so $\langle x, e_1\rangle = 2^k$ but then
$$T^*x = 2^k\sum_{n=1}^\infty \frac1{2^n} e_{2n-1} + \sum_{n=1}^\infty \langle x, e_{n+1}\rangle e_{2n} \ne e_{2k-1}$$
