02-07-2005, 11:32 AM
|
#1 (permalink) |
|
Will power
 
Join Date: Nov 2002
Location: Hanoi and Munich
Posts: 401
 |
Dimension of a vector space
Let V be the vector space, under usual operations, of real polynomials that are of degree at most 3. Let W be the subspace of all polynomials p(x) in V such that p(0) = p(1) = p(-1) = 0. Then dim V + dim W is
A. 4 B. 5 C. 6 D. 7 E. 8 Answer key: SPOILER: B The dimension of V is four (corresponding to four coefficients). A polinomial in W has the form: ax^3 - ax. Hence the dimension of W is one. => dim V + dim W = 5. I'm not sure about my reasoning though. What do you think? Br, vvaann
_ _ _ _ SIG _ _ _ _
Go to TWE section and learn writing with me. Correct my essays and I will participate in reviewing yours! |
|
       
|
|
02-07-2005, 02:15 PM
|
#3 (permalink) |
|
Will power
Join Date: Nov 2002
Location: Hanoi and Munich
Posts: 401
|
Re: Dimension of a vector space
I'm doubtful whether the dimension of the polynomial ax^3 - ax is one.
matroid, I'm going to take the test at the end of this month. It won't be exactly a GRE mathematics test, but it'll be similar to that. How about you?
_ _ _ _ SIG _ _ _ _
Go to TWE section and learn writing with me. Correct my essays and I will participate in reviewing yours! |
|
|
|
|
02-07-2005, 03:52 PM
|
#4 (permalink) |
|
Trying to make mom and pop proud
Join Date: Feb 2005
Posts: 18
|
Re: Dimension of a vector space
vvaann,
dim W=1 because W is all linear combinations of x^3 -x. It might help to think of it in vector notation. W=a{1,0,-1,0} where a is a constant. Since the basis of W consists of only one vetor (not unique, {-1,0,1,0}, etc. would all work), and the dimension of a vector space is defined as the number of vectors in its basis, dimW=1. Thus, the answer is dim V + dim W = 4 + 1 = 5. |
|
|
|
|
02-07-2005, 04:33 PM
|
#5 (permalink) |
|
Will power
Join Date: Nov 2002
Location: Hanoi and Munich
Posts: 401
|
Re: Dimension of a vector space
It's clear now. Thanks matroid and astraltourist!
_ _ _ _ SIG _ _ _ _
Go to TWE section and learn writing with me. Correct my essays and I will participate in reviewing yours! |
|
|
|
|
02-07-2005, 06:31 PM
|
#6 (permalink) | |
|
Within my grasp!
.gif)
Join Date: Nov 2004
Posts: 101
|
Re: Dimension of a vector space
Quote:
I've already taken the subject test with a rather pathetic result.  Waiting for the rejects to come...Best wishes for your test, Md. |
|
|
|
|
|
02-07-2005, 11:41 PM
|
#7 (permalink) |
|
Trying to make mom and pop proud
Join Date: Jan 2005
Posts: 11
|
Re: Dimension of a vector space
Sorry to hijack this thread!
Matroid, I took the subject test in Nov. I wasn't thrilled with my results either:o. Where did you apply? Have you heard from any schools yet? I'm still waiting to hear from the 5 schools I applied to. I think I'll hear back late February or early march. I'm crossing my fingers! |
|
|
|
|
02-13-2005, 06:08 PM
|
#9 (permalink) |
|
Will power
Join Date: Nov 2002
Location: Hanoi and Munich
Posts: 401
|
Re: Dimension of a vector space
ish,
The polynomial is of the form: ax^3 - ax. We can see it as a vector of the form (with elements are coefficients of the polynomial) : (a, 0, 0, -a, 0). The dimension of the subspace is defined as the number of linearly independent vectors which span the subspace. Those linearly independent vectors, when linearly combined, can make up any vector of the subspace. Here, we need only one vector, for example (1, 0, 0, -1, 0), to create, by means of linear combination, any other vector of the subspace. Therefore, its dimension is 1. Br, vvaann
_ _ _ _ SIG _ _ _ _
Go to TWE section and learn writing with me. Correct my essays and I will participate in reviewing yours! |
|
|
|
Contact TestMagic TestMagic Forums Archive
TestMagic Locations
Legal
Privacy
Content Relevant URLs by vBSEO 3.0.0
Copyright © 1998-2008 TestMagic
Ad Management by RedTyger
Linear Mode
