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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:
    a1a3≤0
    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).

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

    Therefore, it is not closed under addition.

    Product:
    Pb * Pc = {b0c0 + (b1c1)x + ... + (b3c3)x3 + ... + (bncn)xn}
    where
    (b1c1)(b3c3)≤0
    if
    (b1c1)= 0
    and/or
    (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

    vela

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    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

    vela

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    Since you're trying to disprove a statement, all you need to do is find a single counterexample. So first assume
    \begin{align*}
    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.
     
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