Linear Algebra: Determine if set forms vector space

[0specificity]

Homework Statement

Determine whether the following set forms a vector space:
{(x1, x2, x3) E R^3 | x1 + 2x2 - x3 = 0 and x1x2 = 0}

Homework Equations

The axioms!

The Attempt at a Solution

I know that the first equation in the set fulfills the axioms for a vector space, since it is of form {x| Ax = 0}. However, I am unsure about the second equation, and how the two must relate in order to form a vector space. Any help would be greatly appreciated!

 

[Robert1986]

Homework Statement

Determine whether the following set forms a vector space:
{(x1, x2, x3) E R^3 | x1 + 2x2 - x3 = 0 and x1x2 = 0}

Homework Equations

The axioms!

The Attempt at a Solution

I know that the first equation in the set fulfills the axioms for a vector space, since it is of form {x| Ax = 0}. However, I am unsure about the second equation, and how the two must relate in order to form a vector space. Any help would be greatly appreciated!

OK, so you do not need to run through all the vector space axioms to do this. All you need to do is to show that this set is closed under linear combinations. That is, let a and b be scalars, and let x and y be vectors in your set. Then you need to show that ax + by is also in your set.

Also, you seem confused on what {(x1, x2, x3) E R^3 | x1 + 2x2 - x3 = 0 and x1x2 = 0} means. I'm not sure what you mean by {x| Ax = 0}, what is A supposed to be? What {(x1, x2, x3) E R^3 | x1 + 2x2 - x3 = 0 and x1x2 = 0} means is that if you take a vector in R^3, then the first coordinate plus twice the second minus the third is equal to 0 AND the product of the first and second coordinates is equal to 0.

 

[0specificity]

Thank you so much for your help! But I'm still incredibly confused as to how to prove this (I'm sorry I feel really idiotic). I know that for the first equation, if vectors x and y are in that equation, the sum of x+y is also fulfills that equation. In addition, I know that a scalar b times any vector x for that equation implies that bx also fulfills that equation. However, I do not know how to go about using this to see if the whole set is a vector space.

If x1x2 = 0 and y1y2 = 0, x1x2 + y1y2 = 0. However, the vector (x+y) plugged into the equation comes out as x1x2 + y1y2 + x1y2 + x2y1 = 0. Can I assume that (x1y2 + x2y1) = 0?

 

[Robert1986]

Thank you so much for your help! But I'm still incredibly confused as to how to prove this (I'm sorry I feel really idiotic). I know that for the first equation, if vectors x and y are in that equation, the sum of x+y is also fulfills that equation. In addition, I know that a scalar b times any vector x for that equation implies that bx also fulfills that equation. However, I do not know how to go about using this to see if the whole set is a vector space.

If x1x2 = 0 and y1y2 = 0, x1x2 + y1y2 = 0. However, the vector (x+y) plugged into the equation comes out as x1x2 + y1y2 + x1y2 + x2y1 = 0. Can I assume that (x1y2 + x2y1) = 0?

Ok, so first of all, don't feel idiotic. Linear Algebra is the start of something that is really uncomfortable to most of us, but once you get the hang of it, you'll love it.

So, here is how you prove that your set is closed under x1 + 2x2 - x3 = 0.

Let a and b be scalars, let x=(x_1,x_2,x_3) and y=(y_1,y_2,y_3) be vectors in your set S.

ax + by = a(x_1,x_2,x_3) + b(y_1,y_2,y_3) = (ax_1 + by_1, ax_2 + by_2, ax_3 + by_3).
Now, ax_1 + by_1 -2(ax_2 + by_2) - ax_3 + by_3 = a(x_1) + a2(x_2) -a(x_3) + b(y_1) + b2(y_2) -b(y_3) = (a)(x_1 +2x_2 -x_3)+(b)(y_1 +2y_2 -y_3) = a0+b0 = 0. So, this linear combination satisfies the equation x1 + 2x2 - x3 = 0 and thus your set is closed under that eqation. Now do the same for for other condition.

 

[0specificity]

Thank you so much!

 

[0specificity]

Sorry one more question! I followed the procedure you showed me, and I'm ending up with x_2y_1 + x_1y_2 = 0. Since it was never established that either of the terms are equal to zero, is this set thus not a closed space and not a vector space?

 

[Robert1986]

You are correct that it was never established that x_2y_1 + x_1y_2 = 0. And you are correct that this set is not a vector space. However, there is one more thing that you need to do to complete the proof that this isn't a vector space. You need to prove that there are some vectors such that x_2y_1 + x_1y_2 =! 0 (=! is not equals). The best way to do this is to come up with a counter example. Once you do that, you're done.