# Basis for subspace in $\mathbb{R}^4$

How would I start to answer this:

Show that the vectors $(1,0,0,1)$, $(0,1,0,1)$, and $(0,0,1,1)$ form a basis for the subspace $V$ of $\mathbb{R}^4$ which is defined by the equation $x_1+x_2+x_3-x_4=0$.

I understand that to be a basis the vectors must be linearly independent and generate $V$, however, I do not know how to set this up.

Also, all of the examples I have encountered have the same number of elements and dimensions and this one has $3$ elements and $4$ dimensions. Does this make a difference?

Since $$\{ (x_{1},x_{2},x_{3},x_{4}) \in \mathbb{R}^{4} \mid x_{1}+x_{2}+x_{3} - x_{4} = 0\} = \{ (s,t,u,s+t+u) \mid s,t,u \in \mathbb{R} \},$$ since $(s,t,u,s+t+u) = s(1,0,0,1) + t(0,1,0,1) + u(0,0,1,1)$ for all $s,t,u \in \mathbb{R}$, and since $(1,0,0,1), (0,1,0,1), (0,0,1,1)$ are linearly independent in $\mathbb{R}^{4}$, we are done!

• Better than my answer. Well done! – Moya Aug 30 '15 at 3:41
• What happens to the x sub 4? I'm sorry, I'm very new to this. Oh wait, is s+t+u considered the x sub 4 because it is the commonality between the vectors? Also, why does it change to R^3? – Aksel'sRose Aug 30 '15 at 4:02
• It is just a simple parametrization. Note that we have $x_{1}+x_{2}+x_{3} = x_{4}$ by hypothesis. – Megadeth Aug 30 '15 at 4:14
• In my opinion, given the level of the question of OP, you should be as explicit as possible in your manipulations, and thus, avoid relabelling stuff like that without an explanation. – YoTengoUnLCD Aug 30 '15 at 4:19
• @YoTengoUnLCD: That someone posts an elementary question in SOME field does not imply that he is new to ALL branches, for sure. So it is better to write a "real" answer. Every learner should convince himself what is left before him so that he can really learn something out of this process. – Megadeth Aug 30 '15 at 4:34

Hint: Consider the vector $v = (x_1,x_2,x_3,x_4)$ and the matrix $A$ whose columns are the vectors given. Then $Av = 0$ implies what about $x_1,x_2,x_3,$ and $x_4$?

Let $W$ be the subspace spanned by these vectors.
Well, first you should determine that these are linearly independent (which is simple). Then, verify that each of these vectors satisfy the equation, so $W$ will be a subspace of $V$. From here, you can verify that $V$ has $3$ dimensions (your previous proof showed that $\dim V\geq 3$, so it suffices to show $V\not=\mathbb{R}^4$). Then the result will follow since $W\subset V$ is a subspace with the same dimension as the whole space, so $W=V$.