Vector subspace questions

**acevipa** · September 9th 2010, 03:06 AM

1) Suppose that $\mathbf{v}$ is a vector in $\mathbb{R}^n$ . Show that the line segment defined by:

$S=\{\mathbf{x}\in\mathbb{R}^n:\mathbf{x}=\lambda\m athbf{v}, \mbox{for} \ 0\leq\lambda\leq10\}$

is not a subspace of $\mathbb{R}^n$

Since its a line through the origin, the $\mathbf{0}$ vector is in $S$ .

$\mathbf{x_1}=\lambda \mathbf{v_1} \ \ \mbox{and} \ \ \mathbf{x_2}=\lambda \mathbf{v_2}$ such that $\mathbf{x_1, x_2}\in \mathbf{x}$

$\mathbf{x_1}+\mathbf{x_2}=\lambda\mathbf{v_1}+\lam bda\mathbf{v_2}$

$=\mathbf{x_1}+\mathbf{x_2}=\lambda(\mathbf{v_1}+\m athbf{v_2})$

$\mathbf{v_1}+\mathbf{v_2}\in \mathbf{v}\Longrightarrow \mathbf{x_1}+\mathbf{x_2}\in \mathbf{x}$

Therefore, its closed under scalar multiplication

What would you do next for this question?

2) Show that the set:

$S=\{\mathbf{x}\in\mathbb{R}^n:2x_1+3x_2-4x_3=4x_1-2x_2+3x_3=0\}$

is a subspace of $\mathbb{R}^3$

For this question, do you only need to consider say $2x_1+3x_2-4x_3=0$ and show that is a subspace?

**Ackbeet** · September 9th 2010, 04:50 AM

For question 1, I don't think you're supposed to subscript the v's. That is, all the vectors in S look like scalar multiples of a single, given vector v, just with different multiples in the interval [0,10]. And here, perhaps, you can see what fails in terms of the vector space axioms.

For question 2, I would think of the problem geometrically. You have two equations there. Geometrically, what shape does each equation represent separately? And if they both have to be true, what sort of shape is described by the simultaneous equations?

**acevipa** · September 9th 2010, 04:58 AM

Originally Posted by Ackbeet

For question 1, I don't think you're supposed to subscript the v's. That is, all the vectors in S look like scalar multiples of a single, given vector v, just with different multiples in the interval [0,10]. And here, perhaps, you can see what fails in terms of the vector space axioms.

For question 2, I would think of the problem geometrically. You have two equations there. Geometrically, what shape does each equation represent separately? And if they both have to be true, what sort of shape is described by the simultaneous equations?

For question 1, how would you show closure under multiplication and addition.

For question 2. Would they both be planes through the origin?

**Ackbeet** · September 9th 2010, 05:03 AM

For question 1, how would you show closure under multiplication and addition.

I wouldn't. I would show that it's not closed under either operation. Try this: let $\mathbf{x}_{2}=\mathbf{x}_{1}=8\mathbf{v}$ . What can you do with those two vectors?

For question 2. Would they both be planes through the origin?

Correct. And the two planes are not the same plane through the origin. Since they are both through the origin, their intersection is nonempty. Question: if two planes intersect, and they are not the same plane, what is the nature of the intersection? What shape is it?

**acevipa** · September 9th 2010, 05:11 AM

Originally Posted by Ackbeet

I wouldn't. I would show that it's not closed under either operation. Try this: let $\mathbf{x}_{2}=\mathbf{x}_{1}=8\mathbf{v}$ . What can you do with those two vectors?

Correct. And the two planes are not the same plane through the origin. Since they are both through the origin, their intersection is nonempty. Question: if two planes intersect, and they are not the same plane, what is the nature of the intersection? What shape is it?

Okay so for question 1. When you add $\mathbf{x_1}+\mathbf{x_2}=16\mathbf{v}$

Since 16 is not possible, its not closed under vector addition.

For question 2. Wouldn't the nature of their intersection be a line?

**Ackbeet** · September 9th 2010, 05:23 AM

For question 1, I'd say you're done.

For question 2: you're correct. So now you're down to showing that a line through the origin is a vector space. Since you're thinking about a subspace of 3D space, all you have to show is that the space is nonempty (which we've already done), and closed under scalar multiplication and vector addition. How do you propose to go about showing the last two things?

**acevipa** · September 9th 2010, 05:27 AM

Originally Posted by Ackbeet

For question 1, I'd say you're done.

For question 2: you're correct. So now you're down to showing that a line through the origin is a vector space. Since you're thinking about a subspace of 3D space, all you have to show is that the space is nonempty (which we've already done), and closed under scalar multiplication and vector addition. How do you propose to go about showing the last two things?

Couldn't I just show that only one of the planes is satisfied by vector addition and scalar multiplication?

**acevipa** · September 9th 2010, 05:30 AM

Or could I find the line of intersection of the two planes?

**Ackbeet** · September 9th 2010, 05:34 AM

Couldn't I just show that only one of the planes is satisfied by vector addition and scalar multiplication?

Logically speaking, you sort of can. Suppose you show that vector addition and scalar multiplication are closed with respect to one plane equation. Then you could argue that similarly, the other plane equation (since they both go through the origin) is also closed under those operations. Therefore, if you start out with the set satisfying both equations, it will be closed under both operations.

**Ackbeet** · September 9th 2010, 05:35 AM

Reply to Post # 8:

That is also a valid approach.

**acevipa** · September 9th 2010, 05:37 AM

Originally Posted by Ackbeet

Reply to Post # 8:

That is also a valid approach.

What's the best way to approach the question?

**Ackbeet** · September 9th 2010, 05:42 AM

What's the best way to approach the question?

I honestly don't know. Looking at the two approaches, I'd probably go with the one outlined in post # 9, because then you don't have to row reduce a matrix. Incidentally, you might be able to save yourself some effort by noting that both of those planes can be written as $\hat{n}\cdot\mathbf{x}=0$ for some fixed vector $\hat{n}$ . That might simplify the work, since you know something about how dot products behave.

**acevipa** · September 9th 2010, 05:46 AM

Originally Posted by Ackbeet

I honestly don't know. Looking at the two approaches, I'd probably go with the one outlined in post # 9, because then you don't have to row reduce a matrix. Incidentally, you might be able to save yourself some effort by noting that both of those planes can be written as $\hat{n}\cdot\mathbf{x}=0$ for some fixed vector $\hat{n}$ . That might simplify the work, since you know something about how dot products behave.

Okay so say I wanted to do it the long way. Could I just show that $2x_1+3x_2-4x_3=0$ and $4x_1-2x_2+3x_3=0$ are both vector spaces and hence the set is a subspace.

**acevipa** · September 9th 2010, 05:49 AM

Just to clarify a few things. We're showing that the line of intersection of the two planes is a subspace of $\mathbb{R}^3$ .

**Ackbeet** · September 9th 2010, 05:57 AM

Reply to Post # 13:

Yes, essentially, if you've proven that a nonempty intersection of two vector spaces using the same arithmetic is a vector space (which it is). If you haven't gone through that exercise, I would recommend it to you.

Reply to Post # 14:

That's correct as far as it goes. Technically, it only works because the line of intersection includes the origin as a point on the line.

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