Find a subspace $U \subseteq \mathbb{R}^3$ such that $\mathbb{R}^3 \cong U \oplus W$

Question

Let $W$ be the subspace of $\mathbb{R}^3$ spanned by {($1,1,0$),($1,0,-1$)}. Find a subspace $U \subseteq \mathbb{R}^3$ such that $\mathbb{R}^3 \cong U \oplus W$

I know that $\mathbb{R}^3$ is the vector space containing all vectors of the form {$(x_1,x_2,x_3)$}

I know this means that $\mathbb{R}^3 = U+W$ and their intersection is trivial $U \cap W =$ {$0$} (meaning they do not contain any equal vectors).

So would $U$ be something like $\mathbb{R}^3\setminus$ {$(1,1,0),(1,0,-1)$}?

or perhaps $\mathbb{R}^3 \setminus$ {$(x_1,x_2,0),(x_1,0,-x_3)$}?

How do I go about proving that?

Answer 1

Because of dimensions consideration, it should first be clear that $\;\dim U=1\;$ , so that $\;U=\text{Span}\{v\}\;$ , for some non-zero $\;v\in\Bbb R^3\;$ .

You can then take any vector that complements the given subspace's basis to a basis of the whole $\;\Bbb R^3\;$, for example:

$$U=\text{Span}\left\{\;\begin{pmatrix}0\\1\\0\end{pmatrix}\;\right\}\;,\;\;\text{or}\;\;U=\text{Span}\left\{\;\begin{pmatrix}1\\1\\1\end{pmatrix}\;\right\}$$