Show that W is a subspace of R3

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SUMMARY

The discussion centers on proving that the set W = {(x,y,z) | z = y - 2x} is a subspace of R3. To establish this, three conditions must be satisfied: (1) the zero vector (0,0,0) must be in W, (2) the sum of any two vectors in W must also be in W, and (3) any scalar multiple of a vector in W must remain in W. The participants initially attempted to use specific vectors and constants, but the correct approach involves using arbitrary vectors and demonstrating that they satisfy the equation defining W.

PREREQUISITES
  • Understanding of vector spaces and subspace criteria
  • Familiarity with the equation of a plane in R3
  • Knowledge of vector addition and scalar multiplication
  • Basic algebraic manipulation skills
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  • Learn how to determine if a vector satisfies a plane equation in R3
  • Explore examples of subspaces in R3 beyond the given equation
  • Practice proving subspace criteria with various vector sets
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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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