# Vectors, subsets, and subspaces... Can someone explain this question more simply?

• Jun 22nd 2010, 06:15 PM
Mattpd
Vectors, subsets, and subspaces... Can someone explain this question more simply?
I am having trouble understanding the question, could anyone break it down a little? I learned today about 3 cases that need to be checked, but my notes aren't helping too much.

Is the given subset U of $\displaystyle \mathbb{R}^{4}$ a vector subspace?

a) The set U of all vectors u in $\displaystyle \mathbb{R}^{4}$ such that $\displaystyle u_{2}-2u_{3} - u_{4}=0$ and $\displaystyle 3u_{1}+u_{4}=0$

b) The set U of all vectors u in $\displaystyle \mathbb{R}^{4}$ such that $\displaystyle u_{2}-2u_{3} - u_{4}=0$ and $\displaystyle 3u_{1}+u_{4}\leq 0$
• Jun 22nd 2010, 06:33 PM
enderwiggin
First, you have to prove that the 0 element, aka (0,0,0,0) is in the subspace. This is trivial.

Second, show that if $\displaystyle (u_1,u_2,u_3,u_4)$ is in the subspace, then $\displaystyle \alpha (u_1,u_2,u_3,u_4)$ is in the subspace. You can try this by substitution. (Hint - seperate out the cases where $\displaystyle \alpha$ is positive and negative)

Finally, show that if $\displaystyle (u_1,u_2,u_3,u_4)$ and $\displaystyle (v_1,v_2,v_3,v_4)$ are in U, the their sum, $\displaystyle (u_1+v_1,u_2+v_2,u_3+v_3,u_4+v_4)$ is in U. This can also be done by substitution.

If my intuition is correct, then the first one will be a subspace, and the second will not.
• Jun 22nd 2010, 07:43 PM
For (a), note that the set is the intersection of the kernels of the maps $\displaystyle (u_1,u_2,u_3,u_4)\mapsto u_2 - 2u_3 - u_4$ and $\displaystyle (u_1,u_2,u_3,u_4)\mapsto 3u_1-u_4$, which are obviously linear. Since the kernel of linear maps are subspaces and the intersection of subspaces are subspaces, (a) is a subspace.