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I have a homework exercise that I am having a difficult time understanding/visualizing:

Choose $x = (x_1, x_2, x_3, x_4)$ in $\mathbb{R}^4$. It has $24$ rearrangements like $(x_2, x_1, x_3, x_4)$ and $(x_4, x_3, x_1, x_2)$. Those $24$ vectors including $x$ itself span a subspace $S$. Find specific vectors $x$ so that dimension of $S$ is $0$, $1$, $3$, $4$.

I appreciate any help that can be given. Thanks! Source: 3.5.42, P183, Intro to Lin Alg, by G Strang

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1 Answer 1

up vote 1 down vote accepted

$0$ is easy, not much choice.

As to $1$, think of a nonzero vector that doesn't change when rearranged.

As to $4$, think of the standard basis.

As to $3$, think of the subspace $x_1+x_2+x_3+x_4= 0$.

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I think I have an understanding of when S=0,1 and 4, but I am not sure what to do for S=3 –  user42864 Feb 21 '13 at 2:52
    
@user42864, pick a nonzero element in that subspace, and see what you can do with the vectors you obtain by rearranging it. Please let me know if you can do it. –  Andreas Caranti Feb 21 '13 at 9:21
    
When I visualize this problem, I feel that I should be working with the column spaces of x1, x2, etc in R4. Am I correct? –  user42864 Feb 21 '13 at 13:52
    
Never mind. I've got it. –  user42864 Feb 21 '13 at 15:49
    
@user42864: Could it please be elucidated, in Andreas Caranti's answer and not as a comment, whether the dimension of $S$ can be $2$? I'd appreciate it ! –  Law Area 51 Proposal - Commit Nov 26 '13 at 8:14

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