0
$\begingroup$

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?

$\endgroup$
3
  • 1
    $\begingroup$ Try using general results about subspaces. Is the intersection of two subspaces a subspace? $\endgroup$ – Elchanan Solomon Mar 27 '12 at 1:55
  • $\begingroup$ 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. $\endgroup$ – intervade Mar 27 '12 at 2:05
  • $\begingroup$ 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. $\endgroup$ – David Mitra Mar 27 '12 at 2:07
3
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
3
  • 1
    $\begingroup$ Of course, that's only useful if OP already knows that those other sets are subspaces. $\endgroup$ – Gerry Myerson Mar 27 '12 at 2:00
  • $\begingroup$ 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 $\mathbb{R}^4$ ? $\endgroup$ – intervade Mar 27 '12 at 2:09
  • 1
    $\begingroup$ @DaltonConley, One vector is not a subspace, unless it's the zero vector. $\endgroup$ – lhf Mar 27 '12 at 2:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.