1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Linear algebra: Vector subspaces

  1. Apr 10, 2012 #1
    1. The problem statement, all variables and given/known data
    Is the subset of
    P= {a0 + a1x + a2x2 + ... + anxn}
    formed only by the polynomials that satisfy the condition:
    a vector subspace?

    2. Relevant equations
    I already proved the subset is not closed under addition so I know it's not a vector subspace, however, the answer my teacher marked as correct reads: "No, it's closed under the product but not under addition."

    3. The attempt at a solution
    Let Pb and Pc be elements of a subset of P= {a0 + a1x + a2x2 + ... + annn} with their coefficients given by any series in function of n (this is an assumption I'm not sure I can make but I find no other way to tackle the product step without making it).

    Pb + Pc = {b0+c0 + (b1+c1)x + ... + (b3+c3)x3 + ... + (bn+cn)xn}
    (b1+c1)≤0 and (b3+c3)≥0
    (b1+c1)≥0 and (b3+c3)≤0

    Therefore, it is not closed under addition.

    Pb * Pc = {b0c0 + (b1c1)x + ... + (b3c3)x3 + ... + (bncn)xn}
    (b1c1)= 0
    (b3c3)= 0
    (because of their positions in the series).

    Therefore, it is not closed under the product.

    Was the assumption mistaken? If so, can you give me a clue how to prove that it is/isn't closed under the product?
  2. jcsd
  3. Apr 10, 2012 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I don't get what your logic is here.

    You want to show that subset P has to be closed under scalar multiplication. That is, if p ∈ P and r ∈ R, then (rp) ∈ P.
  4. Apr 10, 2012 #3
    Thank you for replying. I know, I realized after two hours that I needed to use scalar product and not whatever I was doing, lol. How would you do the first part, though? I suck at this type of problems :/.
  5. Apr 10, 2012 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Since you're trying to disprove a statement, all you need to do is find a single counterexample. So first assume
    p_b &= b_0 + b_1x + b_2x^2 + \cdots + b_n x^n \in P \\
    p_c &= c_0 + c_1x + c_2x^2 + \cdots + c_n x^n \in P \\
    \end{align*} which means ##b_1b_3 \le 0## and ##c_1c_3 \le 0##. Then form the sum
    $$ p = p_b + p_c = (b_0+c_0) + (b_1+c_1)x + (b_2+c_2)x^2 + \cdots + (b_n+c_n)x^n.$$ You want to show there exists a case where ##(b_1+c_1)(b_3+c_3) \gt 0## which would imply that p is not an element of P. In other words, find specific numbers for b1, b3, c1, and c3 that satisfy those three conditions.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook