setting up for basis of vector

Question

I don't know how to get started with this Find a basis for the subspace $V$ of $\mathbb{R}^4$ consisting of all vectors of the form $(a,b,c,a+b+c)$, and state its dimension. Another way or writing this subspace is $V=\{(a,b,c,d):\ d=a+b+c\}$ I know how to get the dimension and how to work it but I just don't know how to set it up. What do I do with $a, b, c$, and $d$ to get me started?

Answer 1

IMO, the best strategy is to split the vector using the free parameters $a,b,c$: \begin{align*} v\in V &\Longleftrightarrow v= (a,b,c,a+b+c)\\ &\Longleftrightarrow v = (a,0,0,a) + (0,b,0,b) + (0,0,c,c) \\ &\Longleftrightarrow v = a(1,0,0,1) + b(0,1,0,1) + c(0,0,1,1) \\ &\Longleftrightarrow v \in \left\langle \ (1,0,0,1),\ (0,1,0,1),\ (0,0,1,1) \ \right\rangle \\ \end{align*}

Thus, $V=\left\langle \ (1,0,0,1),\ (0,1,0,1),\ (0,0,1,1) \ \right\rangle$, so $\lbrace \ (1,0,0,1),\ (0,1,0,1),\ (0,0,1,1) \ \rbrace $ generates $V$. Moreover, $$(a,b,c,a+b+c) = (0,0,0,0) \Longleftrightarrow a=b=c=0$$ which means that $\lbrace \ (1,0,0,1),\ (0,1,0,1),\ (0,0,1,1) \ \rbrace $ is also linearly independent.

Answer 2

You know that all the vectors in this subspace look like $(a,b,c,a+b+c)$. So you want to break this up into a linear combination of vectors, each specified by one parameter ($a$, $b$, or $c$). So let $V = (a,b,c,a+b+c)$. We want vectors X,Y,Z such that $V = aX + bY + cZ$. As the answer above suggests, $X=(1,0,0,1)$, $Y=(0,1,0,1)$, $Z = (0,0,1,1)$ does the trick. You would then just need to argue that $X,Y, Z$ are linearly independent.