Direct Sum of Three Subspaces Suppose $U = \{(x, y, x+y, x -y, 2x) \in \Bbb F^5 : x, y \in \Bbb F\}$.
Find three subspaces $W_1, W_2, W_3$ of $\Bbb F^5$, none of which equal $\{0\}$ such that $\Bbb F^5 = U \oplus W_1 \oplus W_2 \oplus W_3$.
The answer I got this was :
$W_1 = \mathrm{span}(0, 0, 1, 0, 0)$ 
$W_2 = \mathrm{span}(0, 0, 0, 1, 0)$ 
$W_3 = \mathrm{span}(0, 0, 0, 0, 1)$
My question is first of all, is this right? Secondly, Im a bit confused about how the sum of these subspaces is equal to $\Bbb F^5$. For a vector space to span $\Bbb F^5$ it would have to be something alone the lines of $(a, b, c, d, e) \in \Bbb F^5$ correct? So I chose subspaces that span $\Bbb F^5$ in the sense that they are all linearly independent and that you need n linearly independent vectors in $\Bbb F^n$ in order to span the space. Is that correct thinking and can someone elaborate? I feel like my understanding of this is a bit muddled and I can't find anything to clear it up. Thank you!

Your answer is correct (and one of many possible correct answers). The given space is $U = \mathrm{span}((1,0,1,1,2), (0,1,1,-1,0))$, and those two vectors together with the vectors that span the $W_i$ span all of $\mathbb{R}^5$. There are several ways to see this: you could construct a matrix whose columns or rows are those vectors and see that it is invertible (so that the five vectors are linearly independent and therefore span). You could determine that $(1,0,0,0,0)$ and $(0,1,0,0,0)$ are both in the span of the five given vectors.
