Turn on thread page Beta
    • Thread Starter
    Offline

    3
    ReputationRep:
    Hello,

    as the title states:

    {1 + 2x + x^2, 1 - x- 4x^2, x - x^2, 1 + 3x} in P2

    Now i know it does not span as dimP2 = 3 and there are four vectors (I could be wrong). For a basis, do you check for independence?

    Thanks!
    Offline

    11
    ReputationRep:
    (Original post by mathsRus)
    Hello,

    as the title states:

    {1 + 2x + x^2, 1 - x- 4x^2, x - x^2, 1 + 3x} in P2

    Now i know it does not span as dimP2 = 3 and there are four vectors (I could be wrong). For a basis, do you check for independence?

    Thanks!
    What is the actual question that you are trying to answer?

    First of all, a vector does not have a basis. A vector space has a basis.

    Secondly having "too many vectors" does not prevent a set of vectors from being a spanning set.

    Third, yes, a basis is a linearly independent spanning set.

    I think that's all correct. Not done any linear algebra for ages.
    Offline

    18
    ReputationRep:
    (Original post by BuryMathsTutor)
    What is the actual question that you are trying to answer?
    Hope it's OK to piggy back on your answer to emphasize/clarify...

    First of all, a vector does not have a basis. A vector space has a basis.
    True. From context I think it's clear that he's looking at the vector space of polynomials of degree <=2 and he has a set composed of four elements (therefore vectors in this context) of the vector space.

    Secondly having "too many vectors" does not prevent a set of vectors from being a spanning set.

    Third, yes, a basis is a linearly independent spanning set.
    Both correct.

    Note that having "too many vectors" does prevent a set of vectors from being linearly independent - any set with more vectors than the dim of the vector space cannot be a lin indep set.

    What's still not very clear is what he's actually supposed to do.

    Most likely seems to be "find a subset of this set that forms a basis for P2". There are various ways of doing this, but I would probably proceed as follows (calling the vectors v1, v2, v3, v4).

    Start with the empty set as your "basis in progress". Repeatedly:

    Take the next vector in the list. Check if it can be written as a lin. comb. of your "basis in progress". If it can, discard it, otherwise add it to the list.

    If you can add 3 vectors to the list, you have 3 lin indep vectors over a vector space of dimension 3 and you're done.

    If you can't, then your vectors don't span P2 and you can't choose a basis from them.
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: October 20, 2015
Poll
Do you think parents should charge rent?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.