There is a natural map $U : \ell^2_{a}\rightarrow \ell^2$ given by
$$
U(x_0,x_1,x_2,\cdots) = (a^{0/2}x_0,a^{1/2}x_1,a^{2/2}x_2,a^{3/2}x_3,\cdots)
$$
$U$ is a unitary map between the two Hilbert spaces. If $S_a$ is the shift operator on $\ell^2_a$, then $S_a$ may be pulled back to $\ell^2$ by looking at $T : \ell^2 \rightarrow \ell^2$ defined by
$$
T = US_aU^{-1},
$$
which is described by
\begin{align}
(x_0,x_1,x_2,\cdots) & \mapsto (x_0,a^{-1/2}x_1,a^{-2/2}x_2,\cdots) \\
& \mapsto (0,x_0,a^{-1/2}x_1,a^{-2/2}x_2,\cdots) \\
& \mapsto (0,a^{1/2}x_0,a^{2/2}a^{-1/2}x_1,a^{3/2}a^{-2/2}x_2,\cdots) \\
& = \sqrt{a}(0,x_0,x_1,x_2,\cdots).
\end{align}
That is, $T=\sqrt{a}S$, where $S$ is the ordinary shift operator on $\ell^2$.
Since $\dim \ker T<\infty$ and $\mathrm{ran}\,T$ is closed, it results that $T$ is semi-Fredholm. Hence $T^*$ is semi-Fredholm with $\mathrm{codim}\,\mathrm{ran}\,T^*<\infty$ (and $\mathrm{ran}\,T^*$ is closed). Furthermore we have
$$(1) \quad \mathrm{ind}\,T=-\mathrm{ind}\,T^*.$$
Now let $K:= T -T^*$. Since $K$ is compact, $T-K$ is semi- Fredholm with
$$(2) \quad \mathrm{ind}\,T=\mathrm{ind}\,(T-K).$$
But we have $T^*=T-K$, hence
$$(3) \quad \mathrm{ind}\,T=\mathrm{ind}\,T^*.$$
From $(1)$ and $(3)$ we now see, that $ \mathrm{ind}\,T=\mathrm{ind}\,T^*$ is finite $=0$ and therefore $\mathrm{codim}\,\mathrm{ran}\,T<\infty.$
Best Answer
The operator $T$ restricted to $(\ker T)^\perp$ is injective. If it is surjective at the same time the spaces $(\ker T)^\perp$ and $K^n$ are isomorphic. Therefore they have the same dimension.