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Vector Space Basis

  • #1

Homework Statement



Find a basis for (1, a, a^2) (1, b, b^2) (1, c, c^2)

Homework Equations





The Attempt at a Solution



M(1, a, a^2) + N(1, b, b^2) + K(1, c, c^2) = (0, 0, 0)

M + N + K = 0
Ma + Nb + Kc = 0
Ma^2 + Nb^2 + Kc^2 = 0

This is as far as I got. I tried monkeying around with these 3 equations to no avail.
 
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Answers and Replies

  • #2
662
1
I don't understand the question: you are asked to find a basis for a collection of 3 vectors?
 
  • #3
33,632
5,287
I don't understand the question, either. Please give us the exact wording of the problem. Also, tell us what kinds of things a, b, and c are.
 
  • #4
Show that for any three distinct values a,b,c, the vectors (1,a,a2), (1,b,b2), (1,c,c2) are linearly independent.
 
  • #5
THanks for all your help
 
  • #6
33,632
5,287
Show that for any three distinct values a,b,c, the vectors (1,a,a2), (1,b,b2), (1,c,c2) are linearly independent.
Now, the work you showed in the OP begins to make sense. Your problem statement (shown below) is completely misleading. The objective is not to find a basis at all.

Homework Statement



Find a basis for (1, a, a^2) (1, b, b^2) (1, c, c^2)

Homework Equations





The Attempt at a Solution



M(1, a, a^2) + N(1, b, b^2) + K(1, c, c^2) = (0, 0, 0)

M + N + K = 0
Ma + Nb + Kc = 0
Ma^2 + Nb^2 + Kc^2 = 0

This is as far as I got. I tried monkeying around with these 3 equations to no avail.
So here you have a system of three equations in three unknowns, M, N, and K. You must have some experience in solving systems like this.
 
  • #7
Yea sorry that was a different problem...

I don't know how to solve a system like that above^.
 
  • #8
33,632
5,287
I suspect that you actually do.
Can you solve a system like this?
x + 2y + z = 3
2x - y + 3z = 4
x - 3y + z = 2
I'm not asking you to solve this system - I'm just asking if you know how.
For this system, the variables are x, y, and z. For your system, the variables are M, N, and K.

If you really don't know how to solve a system of equations, then I don't see how you are going to show that the given vectors are linearly independent.
 
  • #9
Yeah I can do that,^ but normally I just monkey around with them, so in that case I would do something like multiply the 1st by -2 and add that to the second and then substitute, etc...

I tried employing that method to this problem, and it didn't work.
 
  • #11
Can someone please just show me how to solve it. I don't know how to solve it when the variables are general and not actual numbers. Normally I just get rid of one variable by mult/subtracting equations... but it won't work in this case. Please help!
 
  • #13
OMG someone please!! I'm like crying right now
 
  • #14
33,632
5,287
Can someone please just show me how to solve it. I don't know how to solve it when the variables are general and not actual numbers. Normally I just get rid of one variable by mult/subtracting equations... but it won't work in this case. Please help!
Yes, it will work in this case.

M + N + K = 0
Ma + Nb + Kc = 0
Ma^2 + Nb^2 + Kc^2 = 0

Remember, the variables are M, N, and K.

To get rid of the first variable in the 2nd equation, add -a times the first equation to the second.

M + N + K = 0
(b-a)N +(c-a)K = 0

Do something similar to eliminate the M term from the 3rd equation.

After this, add some multiple of the 2nd equation to the 3rd equation, to get rid of the N term from the third equation. At this point, your 3rd equation will involve only K, so you can solve for it.

Knowing K, you can substitute into the 2nd equation and solve for N.

Finally, knowing K and N, you can substitute into the 1st equation to get M.
 

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