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Polynomial Subspaces?

  1. Sep 18, 2012 #1

    ElijahRockers

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    1. The problem statement, all variables and given/known data

    Determine whether the following are subspaces of P4:

    a) The set of polynomials in P4 of even degree
    b) The set of all polynomials of degree 3
    c) The set of all polynomials p(x) in P4 such that p(0) = 0
    d) The set of all polynomials in P4 having at least one real root

    3. The attempt at a solution

    The book defines the vector space Pn as being all polynomials of degree n-1.

    So I'm assuming that if R is a polynomial in P4,
    then

    [itex] R = r_3 x^3 + r_2 x^2 + r_1 x + r_0[/itex]

    where [itex] r_n [/itex] is real.

    For a set to be a subspace of P4 then, it must consist of polynomials of degree 3 or less, and according to the stipulations given in a), must be even degrees. Now I am not quite sure how to interpret this part. Does that mean the polynomial must contain elements of degree two or less? Or does this mean it must contain elements of only degree 2 and degree 0?

    Regardless, when I test scalar multiplication and vector addition it seems to work fine in both cases, but the book says that only part c) is actually a subspace, and the rest are not.

    [itex] \alpha R_2 = \alpha (r_2 x^2 + r_1 x + r_0) [/itex]
    [itex] \alpha R_2 = \alpha r_2 x^2 + \alpha r_1 x + \alpha r_0 [/itex]

    [itex] R_2 + Q_2 = (r_2 x^2 + r_1 x + r_0) + (q_2 x^2 + q_1 x + q_0) [/itex]
    [itex] R_2 + Q_2 = ((r_2+q_2)x^2 + (r_1+q_1) x + (r_0+q_0))[/itex]

    This resulting polynomial is still even degree, right? And even if I interpret the question the other way, meaning it does not contain r_1*x^1, this would still be true.

    Is it because r2 can be anything? I could see that if r2 = 0, then the polynomial would be of odd degree in my initial interpretation (though not in my second interpretation) and would fail, but I'm not sure if that's what they are trying to get at.

    Any advice would be appreciated!
     
    Last edited: Sep 18, 2012
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  3. Sep 18, 2012 #2

    jbunniii

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    I think you are misunderstanding what the degree of a polynomial is. The degree is equal to the highest power that appears (with a nonzero coefficient) in the polynomial.

    Thus

    * the degree of x^3 + x^2 + x + 1 is 3
    * the degree of x^2 + x + 1 is 2
    * the degree of x + 1 is 1
    * the degree of 1 is 0
    * the degree of 0 is usually taken to be [itex]-\infty[/itex]

    So in general, if

    p(x) = r_3 x^3 + r_2 x^2 + r_1 x + r_0

    then the degree of p(x) is equal to the largest n such that r_n is nonzero. (Or it's [itex]-\infty[/itex] for the zero polynomial.)
     
  4. Sep 18, 2012 #3

    ElijahRockers

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    Ok, so that was my first interpretation of the question. So a polynomial in P4 (according to my text) could have up to degree 3, as I thought.

    The set of even degree polynomials in P4 then, would be what I had. A polynomial either with degree 2 as it's highest degree, or degree 0 with it's highest degree (I'm not sure that's considered even, but that's kind of trivial anyway). Correct?
     
  5. Sep 18, 2012 #4

    jbunniii

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    Yes, that's correct. So suppose you have two such polynomials, and add them together. Is the result guaranteed to have even degree?
     
  6. Sep 18, 2012 #5

    ElijahRockers

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    Well it all depends on the coefficients. If r2 = -q2 then adding the polynomials together would annihilate the degree 2 term resulting in a degree 1 polynomial. Is that what the question is getting at?

    So I can find which values of coefficients that will prove that those three sets are not subspaces of P4 and just say "well if this coefficient equals such-and-such, then this set will not be a subspace"? That's really what I was asking, I guess I just wanted to provide as much info as possible just incase.
     
  7. Sep 18, 2012 #6

    jbunniii

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    Yes, exactly.

    To be more precise, "if I choose these coefficients, then I have found a counterexample showing that the set is not closed under addition, so it is not a subspace."
     
  8. Sep 18, 2012 #7

    ElijahRockers

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    Couldn't I have proven the same thing using scalar multiplication if α≠0? For any polynomial with r2 = 0, r1 ≠ 0, that would have made the same point, right?
     
  9. Sep 18, 2012 #8

    jbunniii

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    I bet you can find a counterexample for (b) pretty easily now.
     
  10. Sep 18, 2012 #9

    jbunniii

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    No, because that polynomial would have degree 1, not 2, so it doesn't have even degree.
     
  11. Sep 18, 2012 #10

    ElijahRockers

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    That's what I was trying to do. Come up with a counter example. :P
     
  12. Sep 18, 2012 #11

    ElijahRockers

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    And yes, now I can see b) pretty easily, and d) is obviously disproved by finding some combination of the coefficients that results in no real roots.

    But what about c)? The book says that it's a subspace of P4, but if I chose r0 ≠ 0, then R(0) ≠ 0, thus is wouldn't be a subspace, right? Or am I thinking about that wrong?
     
  13. Sep 18, 2012 #12

    jbunniii

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    Right, but in order to find a counterexample violating the scalar multiplication property, you would need a polynomial [itex]p(x)[/itex] that has even degree, but such that [itex]ap(x)[/itex] does not have even degree, for some nonzero scalar [itex]a[/itex].

    By the way, I'm not sure how the degree of the zero polynomial is defined in your book/course, but if it's defined to have degree [itex]-\infty[/itex], then the zero polynomial doesn't have even degree, so that's another reason why the set of polynomials with even degree does not form a subspace. Any subspace must contain zero!
     
  14. Sep 18, 2012 #13

    jbunniii

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    It's true that if [itex]r_0 \neq 0[/itex], then [itex]R(0) \neq 0[/itex]. But all that shows is that this polynomial isn't contained in the set. The set is defined to be only those polynomials that DO satisfy R(0) = 0. The task is to check whether that set is closed under addition and scalar multiplication.
     
  15. Sep 18, 2012 #14

    ElijahRockers

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    I am not very satisfied with my book. The only thing it says about the zero polynomial is a sort of 'by the way' while describing the Pn vector space.

    "In this case, the zero vector is the zero polynomial,

    [itex]z(x) = 0x^{n-1} + 0x^{n-2} + ... + 0x + 0 [/itex].

    And that is the only mention of the zero polynomial I can find in the entire book, even after looking through the index. Needless to say I'm glad I'm borrowing the book from a classmate and didn't have to buy it.
     
  16. Sep 18, 2012 #15

    ElijahRockers

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    Ok that clears things up a bit! Thank you so much!
     
  17. Sep 18, 2012 #16

    jbunniii

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    OK, in that case I wouldn't make any assumptions about the degree of the zero polynomial. Perhaps the author has implicitly decided to assign it degree zero. I don't think there's universal agreement among authors regarding this. You may want to check with your instructor (if you have one) to find out what definition you should use.
     
  18. Sep 18, 2012 #17
    I haven't read the rest of this thread but if the question is what is the degree of the zero polynomial, I've seen it defined formally as -∞. The degree of any other constant polynomial is 0. So the degree of P(x) = 47 is zero; but the degree of P(x) = 0 is -∞.
     
  19. Sep 18, 2012 #18

    jbunniii

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    Yes, I've seen the same.

    The reason the question came up is because Elijah's book doesn't define it explicitly, and at least one of his candidate subspaces (the set of all polynomials of even degree) wouldn't contain the zero polynomial if its degree is defined to be [itex]-\infty[/itex], but it would if its degree is defined to be 0. Fortunately that set can be disqualified as a subspace on other grounds, because it's not closed under addition.
     
  20. Sep 19, 2012 #19

    ElijahRockers

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    I've been doing some thinking on the degree of the zero polynomial and it makes the most sense to have it defined as degree -∞, like you were saying.

    I wish the teacher didn't pull the homework questions directly from the book. They are not bad questions, but the book doesn't clearly explain a lot of the concepts we're working with. (Unless it does, and I just missed it)

    But the homework is all done in time for class this morning, so thank you for your help!
     
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