Back
to top
Vector Space help needed
Watch
Announcements
Page 1 of 1
Go to first unread
Skip to page:
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?
0
reply
Report
#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?
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?
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.
0
reply
(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.
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.
0
reply
Report
#4
Geometrically, do you know what type of set the equation
a + 2b - 2c = 0
describes?
a + 2b - 2c = 0
describes?
0
reply
(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.
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.
0
reply
Report
#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?
Would (0,1,1), (1,0,0.5) and (0,-1,-1) form a basis because they satisfy the equation?
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...
0
reply
(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...
(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...
0
reply
(Original post by DFranklin)
No. You haven't correctly written (a, b, c) in the form aA + bB
No. You haven't correctly written (a, b, c) in the form aA + bB
0
reply
Report
#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).
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).
0
reply
(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).
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).
0
reply
Report
#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...)
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...)
0
reply
X
Page 1 of 1
Go to first unread
Skip to page:
Quick Reply
Today on TSR
-
Have your say on 2021 assessmentOfqual and DfE want students' thoughts
- Applying to uni? Chat to others here
- Struggling with lockdown uni work
- Win a year of Headspace
- Housemate's BF just wont leave