Representations of $SU(2)$ and the complex subspace of its weight spaces I'm following these notes on Representation Theory and Quantum Mechanics by Noah Miller. In chapter 12, the author is classifying all of the irreducible representations of $SU(2)$, and I'm running into some confusion with a particular claim that's made. I'll quote it here and point out my confusion,

Say that $\pi: SU(2) \rightarrow GL(V)$ is a finite dimensional irreducible representation of $SU(2)$ where $V$ is a complex vector space. There will be a highest weight vector $v_{k_H} \in V$. I claim that the following subspace $W \in V$ is closed under the action of all Lie algebra representation elements $\pi'(X)$:
$$W \equiv \mathrm{span}_{\mathbb{C}}\big\{v_{k_H}, \pi'(S_-) v_{k_H}, \pi'(S_-)^2v_{k_H}, \pi'(S_-)^3 v_{k_H},...\big\}$$
(By "closed" I mean that for all $w \in W$ and $X \in \mathfrak{su}(2)$, we have $\pi'(X)w \in W$.)

As for the notation here, the skew-adjoint generators of $SU(2)$ have been defined as $X_1, X_2, X_3$, where as the self-adjoint matrices $S_j \in \mathfrak{su}(2)_\mathbb{C}$ are defined as,
$$S_j \equiv i X_j$$
Now the proof of this subspace $W$ being closed begins like so,

Proof: As $W$ is a complex vector space, all vectors $w \in W$ and Lie algebra representation elements $\pi'(X)$ satisfy $\pi'(X)w \in W$ if and only if $i \pi'(X) w \in W$. Therefore, if we show that $W$ is closed under the action of the complexified Lie algebra $\mathfrak{su}(2)_{\mathbb{C}}$, we are done.

I don't understand this argument.
 A: Since $W$ is a complex vector space, for any $w\in W$, $zw \in W$ where $z \in \mathbb{C}$. This is as per the definition of a complex vector space, that it be closed under multiplication by complex scalars such as $z$. An example of this would be the fact that $iw \in W$. Suppose then we define $w' = iw$, and make the claim in the opposite direction, that if $iw = w' \in W$ then this implies that $w\in W$ also. Well this is simply another multiplication by a complex scalar, namely by $-i$. See that $-i(w') = -i(iw) = w$ and so we must conclude that $w \in W$.
Let's repeat this but define $w' = \pi'(X)w$. Then by the same logic, if we can show that $iw' = i\pi'(X)w \in W$ for all $X$ and $w$, then by definition we will have shown that all $w' = \pi'(X)w \in W$ as well. So it's just a matter of demonstrating that $W$ is closed on the complexified representation of the $\mathfrak{su}(2)$ Lie algebra, all the $i\pi'(X)$, and we'll have implicitly shown that it's closed under all $\pi'(x)$.
