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

  1. #1
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    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 \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
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  2. #2
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    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.
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  3. #3
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    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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