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Find the basis of a kernel and the dimension of the image

  1. Jan 15, 2017 #1
    1. The problem statement, all variables and given/known data
    Let ##n>1\in\, \mathbb{N}##. A map ##A:\mathbb{R}_{n}[x]\to\mathbb{R}_{n}[x]## is given with the rule ##(Ap)(x)=(x^n+1)p(1)+p^{'''}(x)##
    a)Proof that this map is linear
    b)Find some basis of the kernel
    b)Find the dimension of the image
    2. Relevant equations
    ##\mathbb{R}_{n}[x]##
    is defined as the set of all polynomial with real coeficient that have the power less or equal to n
    ##kerA=\{x;Ax=0\}##
    ##imA=\{Ax,x\in\mathbb{R}_{n}[x]\}##
    ##p(x)=a_{n}x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots+a_{2}x^2+a_{1}x^1+a_{0}x^0##

    3. The attempt at a solution
    a)Prove that this map is linear
    ##p,q\in A \\
    (A(p+q))(x)=(x^n+1)(p+q)(1)+(p+q)^{'''}(x)=((x^n+1)(p)(1)+(p)^{'''}(x)+(x^n+1)(q)(1)+(q)^{'''}(x)=Ap(x)+Aq(x)
    \\
    \text{ }
    \\
    A(\theta p)(x)=\theta((x^n+1)(p)(1)+(p)^{'''}(x))=\theta(Ap)(x)
    ##
    c)
    I started doing this by writing out the possible polynomial
    ##(Ap)(x)=(x^n+1)p(1)+p^{'''}(x)\\
    p(1)=a_{n}+a_{n-1}+a_{n-2}+a_{n-3}+\ldots+a_{2}+a_{1}+a_{0}\\
    p^{'''}(x)=6a_{n}\binom{n}{3}x^{n-3}+6a_{n-1}\binom{n-1}{3}x^{n-4}+\ldots+6a_{4}\binom{4}{3}x+6a_{3}
    ##
    now I placed all of this together
    ##
    (Ap)(x)=(x^n+1)\displaystyle\sum_{i=0}^{n}a_{i}+6a_{n}\binom{n}{3}x^{n-3}+6a_{n-1}\binom{n-1}{3}x^{n-4}+\ldots+6a_{4}\binom{4}{3}x+6a_{3}
    ##
    Then I paired all of the same coeficients together and got
    ##
    a_{n}(x^{n}+1+6\binom{n}{3}*x^{n-3}) \\
    a_{n-1}(x^{n}+1+6\binom{n-1}{3}*x^{n-4}) \\
    a_{n-2}(x^{n}+1+6\binom{n-2}{3}*x^{n-5}) \\
    \vdots\\
    a_{3}(x^n+7) \\
    a_{2}(x^n+1) \\
    a_{1}(x^n+1) \\
    a_{0}(x^n+1) \\
    ##
    Here I noticed that the bottom 3 functions are linearly dependent, which means that If I want to find the basis or dimension I should take ##a_{0}\, and\, a_{1}## out. Then I also noticed that all of the above (an to a3) are also linearly dependent on a2 so I subtracted them and got
    ##
    a_{n}(x^{n-3}) \\
    a_{n-1}(x^{n-4}) \\
    a_{n-2}(x^{n-5}) \\
    \vdots\\
    a_{3}(6) \\
    a_{2}(x^n+1) \\
    ##
    as my basis for the image of A therefore the ##dimension(imA)=n-1##
    This is as far as I have gotten. I've tried solving b) by setting the whole polynomial I got equal to 0 however I have no idea how to continue from there.
    thank you
     
    Last edited: Jan 15, 2017
  2. jcsd
  3. Jan 15, 2017 #2

    Math_QED

    User Avatar
    Homework Helper

    Find n + 1 polynomials that are in the kernel and are linear independent and you are done.

    Also, if you would have made the b question first, you could have deduced the dimension of the kernel, and using the second dimension theorem ##dim(Ker(A)) + dim(Im(A)) = dim(\mathbb{R}_n[X])## you would immediately know the dimension of the image of the map without finding a basis.
     
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