Extending a linearly independent set to a basis by adding a vector Q) Let $V=R^4$. Consider the subspace $U=[{(a_1,a_2,a_3,a_4) \in R^4|a_1+a_2+a_3=0}]$ of $V$. Consider the elements $u_1=(0,0,0,1)$ and $u_2=(5,-2,-3,0)$ of $U$. Find another element $u_3 \in U$ such that ${u_1,u_2,u_3}$ is a a basis of U.
A) So far I have found $u_3=(1,-1,0,0) \in U$. I have shown that $u_1, u_2,$ and $u_3$ are linearly independent using a matrix. However, I don't know how to show that they span U, which would then prove that the set ${u_1,u_2,u_3}$ is a basis of U. Do I also need to prove that $U$ is a subspace of $R^4$? Also, if there is a better method to solve these sort of questions that would be very helpful!
 A: You presumably don't need to prove that $U$ is a subspace, since the question refers to "the subspace $U$" (so you can assume you already know it is a subspace).
To show your set spans $U$, you could directly take an arbitrary element of $U$ and show how to write it as a linear combination of your vectors.  As an example of how to get started on this, given $(a,b,c,d)\in U$, you can say that $d$ will need to be the coefficient of $u_1$, since that's the only way to get the last coordinate right, and $-c/3$ will need to be the coefficient of $u_2$ to get the third coordinate right.  You can then solve for what coefficient of $u_3$ to use, and use the fact that $a+b+c=0$ to show that your linear combination really does give all four coordinates correctly.
If you want to avoid this work, you can take some shortcuts using a little dimension theory.  You have three linearly independent vectors, so they will automatically be a basis of $U$ if you can prove that $U$ is 3-dimensional.  You could prove this in several ways.  You could try to find another, simpler basis of $U$ which has 3 elements.  You could consider the fact that $U$ is the kernel of the linear map $T:V\to\mathbb{R}$ given by $T(a,b,c,d)=a+b+c$.  We then have $\dim V=\dim \ker(T)+\dim \operatorname{im}(T)$, and you can use this to conclude that $\dim \ker(T)=3$.
