Show that W is a subspace of R3

  • Thread starter jhamm11
  • Start date
  • #1
3
0
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


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 Ive 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?
 

Answers and Replies

  • #2
3
0
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?
 
  • #3
33,730
5,422
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.

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 Ive 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?
 
  • #4
3
0
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.
 
  • #5
33,730
5,422
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.
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.


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

Related Threads on Show that W is a subspace of R3

  • Last Post
Replies
15
Views
5K
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
5
Views
10K
  • Last Post
Replies
6
Views
44K
  • Last Post
Replies
7
Views
13K
  • Last Post
Replies
9
Views
16K
Replies
3
Views
9K
  • Last Post
Replies
17
Views
2K
  • Last Post
Replies
14
Views
1K
Replies
6
Views
633
Top