# Direct sum of subspaces of the three dimensional space

$\newcommand{\span}[0]{\mathrm{span}}$I got stuck showing the following problem:

If $\mathbb{R}^3 = W\oplus U$ where $W=\span\{e_1\}$ then $U = \span\{e_2,e_3\}$

I tried this way:
Since $\mathbb{R}^3 = W\oplus U$ Then $\dim(W)+\dim(U) = \dim(\mathbb{R}^3) = 3$ Which implies that $\dim(U)=2$ now we get that $U = \span\{u_1,u_2\}$ where $u_1$ and $u_2$ are vectors in $\mathbb{R}^3$
We must show that $U \subseteq \span\{e_2,e_3\}$ and that $U \supseteq \span\{e_2,e_3\}$ but I failed to show it.

EDIT: actually this question rose from this question :
Linear Transformation defined by a Matrix and Invariant Subspaces

• Wait, you mean $e_1, e_2, e_3$ is a fixed basis, such as $e_1 = (1,0,0)$ etc? If so, think of the case of $\mathbb{R}^{2}$ first. – Andreas Caranti Jan 9 '14 at 9:58
• The $e_i$s are standard vectors of $\mathbb{R}^3$ – MathNerd Jan 9 '14 at 10:06
• OK, then think of the ways in which you can, starting with $e_1$, find two vectors $v_2, v_3$ such that $e_1, v_2, v_3$ is a basis for $\mathbb{R}^{3}$. – Andreas Caranti Jan 9 '14 at 10:09
• Another clarification, with $\oplus$ do you just mean a direct sum, or something else? – Andreas Caranti Jan 9 '14 at 10:12
• I do not understand the problem. What does the “if” do at the beginning of the sentence? – Carsten S Jan 9 '14 at 10:12

This has been going on for too long.

Any $U$ of the form $$U = \langle a_1 e_1 + a_2 e_2 + a_3 e_3, b_1 e_1 + b_2 e_2 + b_3 e_3 \rangle$$ with $$\det \begin{bmatrix} a_2 & a_3 \\ b_2 & b_3\end{bmatrix} \ne 0$$ would satisfy $\mathbb{R}^{3} = W \oplus U$.

Equivalently, $$U = \langle c e_1 + e_2, d e_1 + e_3 \rangle.$$

Ok so take $v\in\mathbb{R}^3$, then $v=a_1 e_1 + a_2e_2 + a_3e_3$ for unique $a_1,a_2,a_3\in\mathbb{R}$.

Then $v = (a_1 e_1) + (a_2 e_2 + a_3 e_3)$ tells you that anything of the form $ae_2 + be_3$ is in $U$.

As you already noted $U$ is $2$-dim and so we are done, since $e_1,e_2$ are a basis for $\{ae_2 + be_3\,|\,a,b\in\mathbb{R}\}$.

Alternatively note that since $(1,0,1), (1,1,0)\in\mathbb{R}^3$ and $W$ measures the first component so that $e_2,e_3\in U$ hence the span of them (which is $2$-dimensional) must lie in $U$.