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[Linear Algebra] Kernel and range

  1. Jul 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Let P2 be the vector space of all polynomials of a degree at most 2 with real coefficients. Let T: P2→ℝ be the functioned defined by:

    ##T(p(t)) = p(2) - p(1)##

    a) Find a non-zero element of the Kernel of T. (I think I figured this one out, but I'm not too sure).
    b) Find a specific element of p(t) for which T(p(t)) = π. Describe the image of T as an explicit subset of ℝ. ("Image" and "Range" mean the same.)

    2. Relevant equations
    ##p(t) = a_0 + a_1t + a_2t^2##

    3. The attempt at a solution

    a) What I understood by the "non-zero element of ker(T)" part of the question was that they didn't want me to just write f(x) = 0. So, this is what I wrote down:

    ##ker(T) = \{f(x) ∈ P^2 : f(2) = f(1)\}##

    b) I'll write down my attempt at a solution:

    If we're look for a ##p(t)## such that ##T(p(t)) = π##, then we're going to need an equation that follows the requirements of ##p(2) - p(1) = π##. I was thinking solving the following linear system:

    ##a_0 + 2a_1 + 4a_2t = π + 2##
    ##a_0 + a_1 + a_2 = 2##

    But I feel like I'm overcomplicating things, because the margin given to me in this question doesn't permit the space required to solve this system of linear equations. There must be a clever way of doing this, no? I was thinking about substituting ##a_0 = 0##, since I was asked for a specific solution but I don't know… It seems as if I'm over-complicating things.
     
  2. jcsd
  3. Jul 23, 2016 #2

    fresh_42

    Staff: Mentor

    You shouldn't switch between notations, like ##P_2## and ##P^2##.
    With respect to a): I understood it as to give an example of such a vector ##f##. So which polynomial ##f(t) \neq 0## would satisfy ##f(2) = f(1)##?
    It might help to write the conditions on kernel and range in terms of your unknowns: ##a_0 \, , \, a_1\, , \, a_2 \,.##

    As to b). The same advice here. What is ##T(p(t))## written out? If you do so, it should be easy to find values, such that ##T(p(t)) = \pi.##
    (And you have a "t" too many in your equations.)
     
  4. Jul 24, 2016 #3

    pasmith

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    Homework Helper

    Hint for part (2): What is T(p) for p(t) = at?
     
  5. Jul 24, 2016 #4

    vela

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    Staff Emeritus
    Science Advisor
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    ker(T) is a set, which is what you wrote down. The question asks you to find a non-zero element of this set.

    By the way, note that the term is spelled kernel; you misspelled it (with an a) in the thread title. I corrected it.
     
  6. Jul 24, 2016 #5
    Excuse the typos… So if f(2) = f(1), then: ##a_0 + a_1 + a_2 = a_0 + 2a_1 + 4a_2##… Cancel out the ##a_0##s, and you'd get: ##a_1 + a_2 = 2a_1 + 4a_2##

    With a little bit of isolation, you'd get:
    ##a_1 = 2a_1##
    ##a_2 = 4a_2##

    So, ##ker T= \{f(x) ∈ P_2 : a_1 = 2a_1; a_2 = 4a_2; a_0 = a_0\}## ?

    EDIT: Forgot to add the fact that ##a_0 = a_0##
     
  7. Jul 24, 2016 #6

    fresh_42

    Staff: Mentor

    You must not separate the ##a_i## because your codomain ##\mathbb{R}## is one-dimensional. Thus you can only say ##3a_2 + a_1 = 0## for polynomials ##p(t) = a_2 t^2 +a_1t +a_0 \in \ker(T)##. With that you can easily find a ##p(t) \in \ker(T) - \{0\}##.
    The range can be treated similar.
     
  8. Jul 24, 2016 #7
    Ok so ##T(p(t)) = p(2) - p(1)## and if we allow ##p(t) = at## then: ##T(p(t)) = 2a - a## If we want it to equal π, then we would set ##2a - a = π##. Obviously, ##a = π## in this example.

    I'm just gonna type up my attempt at extending what you said up to the actual question:

    ##T(p(t)) = 4a_2 + 2a_1 + a_0 - a_2 - a_1 - a_0 = 3a_2 + a_1 = π##

    So, ##img T= \{f(x) ∈ P_2 : 3a_2 + a_1 = π\}##?
     
  9. Jul 24, 2016 #8
    I'm sorry, I don't understand what you mean (we haven't taken codomain.) After solving the image part (hopefully I did so correctly?) I can understand where I took a weird route in isolating the variables, but why isn't that a valid move? Anyway, here is how I'd solve it using the method that (I think) you're describing:

    ##4a_2 + 2a_1 + a_0 = a_0 + a_1 + a_2##
    ##3a_2 - a_1 = 0##

    Thus: ##ker T = \{f(x) ∈ P_2 : 3a_2 - a_1 = 0\}##
     
  10. Jul 24, 2016 #9

    fresh_42

    Staff: Mentor

    No. ##img \; T## denotes the entire image, range of ##T##, not only the ones which end up in ##\pi##. It is not a set of polynomials anymore since you evaluated them at points ##2## and ##1## and got a real number out of it. So the image is a set of real numbers.
    Which are the real numbers, that can be written as ## f(2) - f(1) = 3a_2 + a_1##?
    Are there real numbers which can't be written this way?

    For the example with ##\pi## you have to find numbers ##a_2## and ##a_1## such that ##3a_2 + a_1 = \pi##.
    pasmith's hint meant: you can choose ##a_2=0## and only bother with ##a=a_1##.


    It has to be ##+a_1##, but yes. And it's better to say what the ##a_i## mean in this context, so
    ##ker \, T = \{ f(x) \in P_2 : f(x) = a_2x^2 + a_1 x + a_0 \wedge 3a_2 +a_1 = 0\}##

    codomain is the english term for the set which you map to. In your case ##\mathbb{R}##.
     
  11. Jul 24, 2016 #10

    Aaah! Silly me, of course the img isn't what I stated! So since the question is asking for a specific element, I just need to formulate a polynomial… right? So I'll just let ##a_2 = π## and ##a_1 = -2π##. So, the specific element would be: ##πt^2 - 2πt = p(t)##

    And yeah the minus was an accident in the kernel. thanks for the help, I think I got it!
     
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