I know if I have the following equation
$$ \mathbb{R}^3=\text{span} \left\{ \begin{bmatrix}1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 0 \end{bmatrix} \right\} \oplus W $$
Then $W=\text{span} \left\{ \begin{bmatrix}0 \\ 0 \\ 1 \end{bmatrix}\right\}$ as it has to span over the last entry.
How do I find a suitable space for $W$ in the following
$$ \mathbb{R}^3=\text{span} \left\{ \begin{bmatrix}0\\ \frac{-2}{\sqrt{5}} \\ \frac{1}{\sqrt{5}} \end{bmatrix}, \begin{bmatrix}\frac{5}{3\sqrt{5}}\\ \frac{2}{3\sqrt{5}} \\ \frac{4}{3\sqrt{5}} \end{bmatrix} \right\} \oplus W $$
would it be $\text{span} \left\{ \begin{bmatrix}0 \\ \frac{5}{3\sqrt{5}} \\ 0 \end{bmatrix}\right\}$ so that the sum of the entries is $\frac{5}{3\sqrt{5}}$?
In the second case, calling the span vectors $b_1$ an $b_2$, respectively, notice that
$$b_1+3b_2 = \begin{bmatrix}\sqrt{5} \\ 0 \\ \sqrt{5} \end{bmatrix}$$
and
$$3b_2 -4b_1 = \begin{bmatrix}\sqrt{5} \\ 2\sqrt{5} \\ 0 \end{bmatrix}$$
thus the vector $\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}$ is not in the span of $b_1$ and $b_2$. So one choice of $W$ you can make is
$$W = \operatorname{span}\left\{\begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix}\right\}$$