# Finding basis of subspace of $\mathbb{R}^4$ such that $x_1 + x_2 + x_3 + x_4 = 1$

I want to find the basis of the following subspace

$$W = \{(x_1, x_2, x_3, x_4)^T \in \mathbb{R}^4 : x_1 + x_2 + x_3 + x_4 = 1\}$$

We clearly see that, from the constraint, we get the condition

$$x_4 = 1 - x_1 - x_2 - x_3$$

so these are the vectors of the form

$$(x_1, x_2, x_3, 1-x_1-x_2-x_3)^T$$

and this space should have dimension $$3$$, but how do I proceed to find it explicitly?

• Hint: start with $\vec {v_1}=(1,0,0,-1)\in W$. Can you find a vector in $W$ which is linearly independent from that? And so on. – lulu Dec 30 '19 at 13:50
• @lulu, your $\vec v_1\not\in W$ – Martund Dec 30 '19 at 13:51
• $W$ is not a subspace of $\mathbb{R^{4}}.$ – math Dec 30 '19 at 13:51
• @Martund you are correct! I read the condition as requiring that $\sum x_i=0$. As stated, $W$ isn't a subspace. – lulu Dec 30 '19 at 13:52

As already remarked in the comments, $$W$$ is not a linear subspace of $$R^4$$. What you can do is the following: the condition $$x_4=1−x_1−x_2−x_3$$ allows you to find a parametrization for $$W$$. By choosing the first three $$x_i$$ as free variables, e.g. $$x_1 = r, x_2 = s, x_3 = t$$ you can write $$(x_1,x_2,x_3,x_4) = (r,s,t,1-r-s-t) = (0,0,0,1) + r (1,0,0,-1) + s (0,1,0,-1) + t (0,0,1,-1).$$ Your set $$W$$ is the subspace spanned by the latter three vectors translated over $$(0,0,0,1)$$.