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Linear Algebra: Linear indepency of a set of Polynomials

  1. Feb 24, 2013 #1
    1. The problem statement, all variables and given/known data

    Let {p, q} be linearly independent polynomials. Show that {p, q, pq} is linearly independent if and only if deg p ≥ 1 and deg q ≥ 1.

    2. Relevant equations

    λ1p + λ2q = 0 ⇔ λ1 = λ2 = 0

    3. The attempt at a solution

    λ1p + λ2q + λ3pq = 0

    I know if λ3 = 0, then the coefficients of and q must be zero, but if λ3 ≠ 0, how do I show the degrees must be greater than or equal to 1?
     
  2. jcsd
  3. Feb 25, 2013 #2
    Assuming {p, q} is linearly independent.
    {p, q, pq} is linearly independent [itex]\Leftrightarrow [/itex] deg p ≥ 1 and deg q ≥ 1.

    {p, q, pq} is linearly independent [itex]\Rightarrow [/itex] deg p ≥ 1 and deg q ≥ 1. (1)
    {p, q, pq} is linearly independent [itex]\Leftarrow [/itex] deg p ≥ 1 and deg q ≥ 1. (2)

    You can easily prove (1) by the transposition (~q => ~p):
    If deg p =0 or deg q =0 then {p, q, pq} is not linearly independent

    And (2), if deg p ≥ 1 and deg q ≥ 1 .What happens with deg pq?
     
    Last edited: Feb 25, 2013
  4. Feb 25, 2013 #3
    Thanks for the reply! Today I was able to prove (1) by contraposition but I thought that meant I had to prove (2) by contraposition as well. Can I just prove the positive statement for (2)?

    Edit: Wait, taking statement A as {p, q, pq} is linearly independent and statement B as deg p ≥ 1 and deg q ≥ 1, is the contrapositive of (1) ~A => ~B or ~B => ~A? I used the first one, but I'm beginning to think its the second one.
     
    Last edited: Feb 25, 2013
  5. Feb 25, 2013 #4
    (1)
    {p, q, pq} is linearly independent as P
    deg p ≥ 1 and deg q ≥ 1 as Q
    ~P is {p, q, pq} is linearly dependent
    ~Q is p < 1 or deg q < 1
    Then ~Q => ~P is:
    ~Q is p < 1 or deg q < 1 =>~P is {p, q, pq} is linearly dependent

    PD: Is contraposition the right word? I searched in wiki to know how it's named in English, in Spanish is "contrarecripoco" and I thought it translation was "transposition"

    (2) I thought something like:
    If deg p ≥ 1 and deg q ≥ 1, then deg (pq) ≥ 2, so λ_{1}p + λ_{2}q + λ_{3}pq = 0 only if λ_{3}=0, and as p and q are linearly independent...
    I don't remember exactly how to show that degres must be lesser than 1, but if you write both polynomials as sums, then the product of them will have a x with a power greater than one and it thought it was trivial enough to say that you can't make λ_{1}*(0)+λ_{2}*(0)+λ_{3}*aX^{n}=0 (with a≠0) without making λ_{3}=0 (because the polynomial sum to be equal to zero needs to be every coefficient of a power of X be equal to zero).
     
    Last edited: Feb 25, 2013
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