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The question is as follows:

Is the set of all vectors $x = [x_1, x_2, x_3, x_4]^T$ that are linear combinations of $[4, 2, 0, 1]^T$ and $[6, 3, -1, 2]^T$ and in addition satisfy the equation $x_1 = 2x_2$ a subspace of $\mathbb{R}^4$?

Could someone give me some hints on where to start with this problem?

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Try using general results about subspaces. Is the intersection of two subspaces a subspace? – Isaac Solomon Mar 27 '12 at 1:55
Well, I've proven that if $U$ is a subspace of $V$ and $W$ is a subspace of $V$ then $U \cap W$ is a subspace of $V$, but I cannot see how that helps me. I guess I don't understand how $x = [x_1, x_2, x_3, x_4]^T$ could be a linear combination of $[4, 2, 0, 1]^T$ or the other vector. – intervade Mar 27 '12 at 2:05
Why not just verify that it's a subspace directly? (Show that it's closed under scalar multiplication and vector addition, and non-empty.) Note the two vectors satisfy the given equation. – David Mitra Mar 27 '12 at 2:07
up vote 3 down vote accepted

Since the vectors $\vec{x}=[4,2,0,1]^T$ and $\vec{y} = [6,3,-1,2]^T$ both satisfy $x_1 = 2x_2$, it's not hard to see that all linear combinations of $\vec{x}$ and $\vec{y}$ will also satisfy this equation. Hence this equation is redundant. Now the only question that remains is whether the set of linear combinations of $\vec{x}$ and $\vec{y}$ is a subspace. Do you see the answer?

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Here is a roadmap:

  1. Prove that the set of all linear combinations of two vectors is a subspace.

  2. Prove that the set of all vectors whose coordinates satisfy a linear equation is a subspace.

  3. Prove that the intersection of two subspaces is a subspace.

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Of course, that's only useful if OP already knows that those other sets are subspaces. – Gerry Myerson Mar 27 '12 at 2:00
Hmm.. since $[4, 2, 0, 1]^T$ and $[6, 3, -1, 2]^T$ are both subspaces of $\mathbb{R}^4$ (and they both clearly satisfy $x_1 = 2x_2$), then their intersection is a subspace of $\mathbb{R}^4$ ? – intervade Mar 27 '12 at 2:09
@DaltonConley, One vector is not a subspace, unless it's the zero vector. – lhf Mar 27 '12 at 2:13

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