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# Vector Space help needed watch

1. I've attached the relevant question. I'm not really sure what the steps I need to take are. Do the standard basis vectors work?
Attached Images

2. (Original post by pineapplechemist)
I've attached the relevant question. I'm not really sure what the steps I need to take are. Do the standard basis vectors work?
The standard basis does not work: (1,0,0) doesn't satisfy the equation, sadly.

Could you have a guess at the dimension (with reasoning if you can) before I attempt to help? Don't worry about the basis for now.
3. (Original post by Smaug123)
The standard basis does not work: (1,0,0) doesn't satisfy the equation, sadly.

Could you have a guess at the dimension (with reasoning if you can) before I attempt to help? Don't worry about the basis for now.
I'd guess the dimension needs to be three but I don't really understand why I'm afraid.
4. Geometrically, do you know what type of set the equation

a + 2b - 2c = 0

describes?
5. (Original post by Smaug123)
The standard basis does not work: (1,0,0) doesn't satisfy the equation, sadly.

Could you have a guess at the dimension (with reasoning if you can) before I attempt to help? Don't worry about the basis for now.
Would (0,1,1), (1,0,0.5) and (0,-1,-1) form a basis because they satisfy the equation?
6. (Original post by pineapplechemist)
Would (0,1,1), (1,0,0.5) and (0,-1,-1) form a basis because they satisfy the equation?
(0, 1, 1) + (0, -1, -1) = 0, so these two vectors are not linearly independent.

The fact is, the vector space you are trying to find a basis for does NOT have dimension 3, so you aren't going to be able to find a basis with 3 elements.

It might help to look at it like this:

If you have a vector (a, b, c) in the vector space, then we know that

a+2b-2c = 0.

So suppose you are given a, b. What must c equal? (*)

So we know that (a, b, c) = (a, b, Aa+Bb) where (A, B) are constants you should have found in (*).

Rewrite it as where A, B are constant vectors.

At this point, what can we say about A, B?

Do they span our vector space? (If so, why, if no, why)?
Are they linearly independent?

So...
7. (Original post by DFranklin)
(0, 1, 1) + (0, -1, -1) = 0, so these two vectors are not linearly independent.

The fact is, the vector space you are trying to find a basis for does NOT have dimension 3, so you aren't going to be able to find a basis with 3 elements.

It might help to look at it like this:

If you have a vector (a, b, c) in the vector space, then we know that

a+2b-2c = 0.

So suppose you are given a, b. What must c equal? (*)

So we know that (a, b, c) = (a, b, Aa+Bb) where (A, B) are constants you should have found in (*).

Rewrite it as where A, B are constant vectors.

At this point, what can we say about A, B?

Do they span our vector space? (If so, why, if no, why)?
Are they linearly independent?

So...
Thanks, my solution is attached. Is it ok?
Attached Images

8. No. You haven't correctly written (a, b, c) in the form aA + bB
9. (Original post by DFranklin)
No. You haven't correctly written (a, b, c) in the form aA + bB
This any better? not one hundred percent sure I understand. Thanks again.
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10. Still wrong. You need to write it in the for aA + bB where A, B are constant vectors. (i.e. don't involve a or b).

E.g. if you had (a, b, c) = (a, b, 7a + 9b), then you would need to rewrite this as a (1, 0, 7) + b (0, 1, 9).
11. (Original post by DFranklin)
Still wrong. You need to write it in the for aA + bB where A, B are constant vectors. (i.e. don't involve a or b).

E.g. if you had (a, b, c) = (a, b, 7a + 9b), then you would need to rewrite this as a (1, 0, 7) + b (0, 1, 9).
Thanks, the exxample helped big style. This better? Also the dimension will be two right?
Attached Images

12. Yes.
13. (Original post by DFranklin)
Yes.
Thank you
14. Should you not say:
A basis of V is:
linearly independent
spans V

And where you've written a(1,0,1/2) and b(0,1,1) should you not write they form a basis of V.

I think you've gotten a bit confused about R^3 and V. V is a subspace of R^3. Everything else seems right though.

(Although I wouldn't trust this post until someone else confirms...)

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