Show that W is a subspace of R3

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In summary, to show that W is a subspace of R3, you need to prove that 0 ε W, and for any vectors u and v in W, u + v ε W, and for any constant k and any vector u in W, ku ε W. To do this, you can start by taking an arbitrary vector u = <u1, u2, u3> in W and showing that it satisfies the equation z = y - 2x. Then, take another arbitrary vector v in W and show that u + v also satisfies the equation. Finally, for any constant k, show that ku also satisfies the equation. These steps will prove that W satisfies the properties of being a subspace, therefore
  • #1
jhamm11
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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 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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  • #2
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
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?
 
  • #4
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
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.
 

What is a subspace?

A subspace is a subset of a vector space that contains the zero vector and is closed under vector addition and scalar multiplication. It has all the properties of a vector space, but it is smaller in dimension.

What is R3?

R3, or three-dimensional Euclidean space, is a vector space that represents all possible 3-dimensional points. It consists of all ordered triples (x, y, z) where x, y, and z are real numbers.

How do you show that W is a subspace of R3?

In order to show that W is a subspace of R3, we need to prove that it satisfies the three properties of a subspace: closure under vector addition, closure under scalar multiplication, and containing the zero vector. This can be done by showing that any linear combination of vectors in W is also in W, and that the zero vector is in W.

What does it mean for W to be closed under vector addition?

Being closed under vector addition means that if we take any two vectors in W and add them together, the resulting vector will also be in W. This is an important property of subspaces because it ensures that the space remains within the original vector space.

Why is it necessary for W to contain the zero vector?

The zero vector is necessary for W to be a subspace because it is a requirement for all vector spaces. The zero vector is the identity element for vector addition, and without it, the space would not be closed under addition. It also ensures that W has a non-empty set, which is necessary for a subspace.

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