Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    Determine if the given set is a subspace of P_n for an appropriate value of n. Justify your answer.

    All polynomials of the form p(t) = a + t^2, where a is in R.

    My answer:

    The given set is NOT a subspae of P_n because...


    It is NOT closed under addition: If there are two polynomials, u(t) = b + t^2 and v(t) = c+t^2, then... u(t)+v(t) = b + t^2 + c + t^2 = (c+b) + (2)t^2 ? p(c+b)

    Another way to show that p(t) is not a subspace is to show how the zero polynomial is not included: Because "a" can be any real number, it can equal to zero, but the coefficient next to t^2 is equal to 1 and therefore cannot equal zero.

    Is my answer correct? Is there anything wrong with it?

    Thank you in advance.
    • PS Helper
    Offline

    14
    PS Helper
    Your answer is fine. Saying that the zero polynomial isn't in it is enough, and so is saying that it's not closed under addition; you don't need to say both.

    Although I can't see where the "P_n for an appropriate value of n" bit comes into this; surely they're all in P_2?
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by nuodai)
    Your answer is fine. Saying that the zero polynomial isn't in it is enough, and so is saying that it's not closed under addition; you don't need to say both.

    Although I can't see where the "P_n for an appropriate value of n" bit comes into this; surely they're all in P_2?
    Thank you...is it ok if you also check the way I show that it's not closed under addition?

    "u(t)+v(t) = b + t^2 + c + t^2 = (c+b) + (2)t^2 which is not equal to p(c+b)
    Offline

    14
    ReputationRep:
    (Original post by Artus)
    Thank you...is it ok if you also check the way I show that it's not closed under addition?

    "u(t)+v(t) = b + t^2 + c + t^2 = (c+b) + (2)t^2 which is not equal to p(c+b)
    Not quite.
    This much is right:
    "u(t)+v(t) = b + t^2 + c + t^2 = (c+b) + (2)t^2"
    Then all you need is that this isn't in the space, so is not of the form a+t^2(which it isn't). p(c+b) would be something entirely different, and unrelated.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Slumpy)
    Not quite.
    This much is right:
    "u(t)+v(t) = b + t^2 + c + t^2 = (c+b) + (2)t^2"
    Then all you need is that this isn't in the space, so is not of the form a+t^2(which it isn't). p(c+b) would be something entirely different, and unrelated.
    Thank you for answering...I wanted to give you a thumbs up but accidentally clicked on the thumbs down...is there any way I can change it? I'm so sorry.
    Offline

    17
    ReputationRep:
    (Original post by Artus)
    Thank you for answering...I wanted to give you a thumbs up but accidentally clicked on the thumbs down...is there any way I can change it? I'm so sorry.
    I've given a counteracting +ve (and since "mine's bigger than yours", he should now be quids-in on the transaction!).
 
 
 
Poll
Who is your favourite TV detective?
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.