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

**Mattpd** · June 22nd 2010, 07:15 PM

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 $\mathbb{R}^{4}$ a vector subspace?

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

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

**enderwiggin** · June 22nd 2010, 07:33 PM

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

Finally, show that if $(u_1,u_2,u_3,u_4)$ and $(v_1,v_2,v_3,v_4)$ are in U, the their sum, $(u_1+v_1,u_2+v_2,u_3+v_3,u_4+v_4)$ 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.

**maddas** · June 22nd 2010, 08:43 PM

For (a), note that the set is the intersection of the kernels of the maps $(u_1,u_2,u_3,u_4)\mapsto u_2 - 2u_3 - u_4$ and $(u_1,u_2,u_3,u_4)\mapsto 3u_1-u_4$ , which are obviously linear. Since the kernel of linear maps are subspaces and the intersection of subspaces are subspaces, (a) is a subspace.

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