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Prove a subset

  • #1

Homework Statement


Show that W ={(x,y,z) : x +2y +3z =0} is a subspace of R3. By finding a subset S of W such that span(S) = W.


Homework Equations


Ab = X?


The Attempt at a Solution



I don't have an attempt because i'm completely lost where to start . Can someone point me in the right way please.
 
Last edited:

Answers and Replies

  • #2
radou
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Start with the definition of a subspace. Obviously every element of W is in R^3, so W is a subset (set!) of R^3. Now, what is the easiest subspace "test"?
 
  • #3
Start with the definition of a subspace. Obviously every element of W is in R^3, so W is a subset (set!) of R^3. Now, what is the easiest subspace "test"?
So prove that is its

Independent
Closed Under Addition
Closed Under Scalar Multiplication?

How to I find S?
 
  • #4
Fredrik
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The equation defines a plane, so you need to find two linearly independent vectors in that plane.
 
  • #5
Is this system correct?

(x, y, z) = y(-2,1,0) +z(-3,0,1)

So S would be <(-2,1,0),(-3,0,1)>
 
  • #6
Fredrik
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Yes, that appears to be correct. But I don't like the <,> notation. Does your book or teacher write the span that way?
 
  • #7
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Is this system correct?

(x, y, z) = y(-2,1,0) +z(-3,0,1)

So S would be <(-2,1,0),(-3,0,1)>
Those vectors work. I would write it as S = {<-2, 1, 0>, <-3, 0, 1>}. This set of vectors spans W. Now what you need to do is to show that W is a subspace of R3.

Something you said earlier in this thread deserves comment:
judahs_lion said:
So prove that is its

Independent
Closed Under Addition
Closed Under Scalar Multiplication?
The first one you listed is not a test for showing that a set is a subspace of some vector space. The vectors in a basis have to linearly independent, and have to span some subspace.

What is the test you missed?
 
  • #8
Those vectors work. I would write it as S = {<-2, 1, 0>, <-3, 0, 1>}. This set of vectors spans W. Now what you need to do is to show that W is a subspace of R3.

Something you said earlier in this thread deserves comment:


The first one you listed is not a test for showing that a set is a subspace of some vector space. The vectors in a basis have to linearly independent, and have to span some subspace.

What is the test you missed?
Contains the zero vector?
 
  • #9
How do I test for the zero vector?
 
  • #10
33,503
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Does W contain the zero vector? It's a very easy test.
 
  • #11
Does W contain the zero vector? It's a very easy test.
I was think of testing S for zero vector.

So for (x, y, x) =(0,0,0)

0 + 2(0) +3(0) = 0 Checks out.

Proper test? So for being so needy, math is a weak point of mine.
 
  • #12
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Sure, that's another way to do it.

Your set W represents a plane in three-dimensional space. This plane contains the origin, which you can tell by noticing that (0, 0, 0) is a solution of the equation x + 2y + 3z = 0.

Set S contains two vectors that lie in the plane. It doesn't contain the zero vector. You want to test set W to see whether it (W) contains the zero vector.

Since there are two vectors, and neither is a multiple of the other, they are linearly independent. Since the plane is two dimensional, and you have two linearly independent vectors, the two vectors span the plane, which means that every vector in the plane is a linear combination of the vectors you found. Instead of noticing that (0, 0, 0) is a point on this plane, you instead found that 0*<-2, 1, 0> + 0*<-3, 0, 1> = <0, 0, 0>, which says that the zero vector is in W (or equivalently, that (0,0,0) is on the plane).
 
  • #13
Thank you
 
  • #14
the matrix of { 1+x^2, x +x^3} = [1,0;0,1;1,0;0,1] ?
 
  • #15
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the matrix of { 1+x^2, x +x^3} = [1,0;0,1;1,0;0,1] ?
What's the actual question?
 
  • #16
Fredrik
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If a set is closed under linear combinations, it automatically contains the zero vector. So checking if it contains 0 is only useful in some cases when you're trying to prove that a set isn't a subspace.
 

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