Show that W is a subspace of R3

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Homework Help Overview

The problem involves determining whether the set W = {(x,y,z)|z=y-2x} is a subspace of R³. Participants are exploring the necessary conditions for a subset to qualify as a subspace, specifically focusing on vector addition and scalar multiplication properties.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to verify that the zero vector is in W, and whether arbitrary vectors can be tested against the defining equation. There is also a focus on the correct application of the subspace criteria, including the need for generality in selecting vectors and constants.

Discussion Status

Some participants have attempted to apply specific vectors and constants to demonstrate the subspace properties, while others have pointed out the necessity of a more general approach. There is ongoing clarification about the conditions that must be satisfied for W to be considered a subspace.

Contextual Notes

Participants are grappling with the definitions and requirements for subspaces, particularly in relation to the specific equation provided. The discussion reflects uncertainty about how to properly demonstrate the necessary conditions without relying on specific examples.

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Homework Statement



Let W = R3 be the set W = {(x,y,z)|z=y-2x}. Show that W is a subspace of R3

Homework Equations


From the equations my teacher gave me I know that if W is to be a subspace it needs to follow:

1.) U,VεW then (U + V)εW,
2.)UεW and K(random constant)εW then KUεW


The Attempt at a Solution


So I haven't really figured out how to attempt this problem but from what I know is that you need to assign U and V to something. From what I've done so far I have U = (1,-2,-1) and V = (2,-4,-2). I got those numbers from the z=y-2x equation. Then U+V = (3,-6,-3), which I believe froms the first equation I gave.
Then I believe I pick a random constant for K, so ill choose 3. So 3U = (3,-6,-3) which proves the second equation.

Are these the proper steps to solving this problem?
 
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Also I just realized that I mixed up x and y in when I placed them in U and V, but are the steps still correct?
 
jhamm11 said:
ε

Homework Statement



Let W = R3 be the set W = {(x,y,z)|z=y-2x}. Show that W is a subspace of R3

Homework Equations


From the equations my teacher gave me I know that if W is to be a subspace it needs to follow:

1.) U,VεW then (U + V)εW,
2.)UεW and K(random constant)εW then KUεW
It must also be true that 0 ε W.
jhamm11 said:

The Attempt at a Solution


So I haven't really figured out how to attempt this problem but from what I know is that you need to assign U and V to something. From what I've done so far I have U = (1,-2,-1) and V = (2,-4,-2). I got those numbers from the z=y-2x equation. Then U+V = (3,-6,-3), which I believe froms the first equation I gave.
Then I believe I pick a random constant for K, so ill choose 3. So 3U = (3,-6,-3) which proves the second equation.

Are these the proper steps to solving this problem?

No, they aren't. You can't pick any particular vectors in W and you can't pick a particular constant. You need to show that
1) 0 ε W.
2) For any vectors u and v in W, u + v ε W.
3) For any constant k and any vector u in W, ku ε W.

If you are given an arbitrary vector u = <u1, u2, u3> in R3, how can you tell whether that vector is in W?
 
So to show that 0 εW, do i plug (0,0,0) into z=y-2x?

And if I was given any arbitrary vector I should just be able to plug it into z=y-2x correct?
I am not given any vectors though, just what is shown. I cannot quite wrap my head around this.
 
jhamm11 said:
So to show that 0 εW, do i plug (0,0,0) into z=y-2x?
Sort of. A better way to show this is to show that (0, 0, 0) satisfies the equation z = y - 2x.
jhamm11 said:
And if I was given any arbitrary vector I should just be able to plug it into z=y-2x correct?
Any vector u in W must satisfy this equation.

Here's how to approach this problem:

Let u = <u1, u2, u3> be an arbitrary vector in W. From the definition of set W, it must be true that u3 = u2 - 2u1.

Now take another arbitrary vector v in W. Show that u + v ##\in## W.

For the third part, show that for any arbitrary real number k, and any vector u ##\in## W, then ku ##\in## W.


jhamm11 said:
I am not given any vectors though, just what is shown. I cannot quite wrap my head around this.
 

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