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Showing differentiation is a linear map

  1. Mar 8, 2012 #1
    1. The problem statement, all variables and given/known data

    linearmap.png


    3. The attempt at a solution

    For part ii) I wrote it out as a matrix, getting

    \begin{array}{ccccccc}
    0 & 0 & 0 & 0 & ... & 0 \\
    0 & 0 & 2 & 0 & ... & 0 \\
    0 & 0 & 0 & 6 & ... & 0 \\
    . & . & . & . & . & . \\
    0 & 0 & 0 & 0 & ... & N(N-2) \end{array}

    So the rank = N-2 and the nullity = 2

    However for part i) I'm not entirely sure how to go about it. Is it enough to show the map can be written as a matrix form, or do I need to show the additivity and scalar multiplication things hold over differentiation? Or is either ok?

    Cheers.
     
  2. jcsd
  3. Mar 8, 2012 #2
    My guess is that the teacher is expecting you to show the explicit requirements for a linear transformation are fulfilled, for which you don't need to use matrices at all.

    But once you have shown that a matrix can be used (formally proving this is probably pretty hard), the first part is implied.

    The teacher probably doesn't want to you prove from first principles that differentiation is linear (unless this is an upperlevel analysis course in disguise). Just show briefly that the known rules about derivatives imply that they are linear. At least that is what my prof. wanted from a similar question when I took LA
     
  4. Mar 13, 2012 #3
    Thanks.

    I've kinda got another question though. I thought I understood the matrix for part ii, but thinking about it it doesn't really make much sense. Can anyone help me understand it?

    Cheers.
     
  5. Mar 13, 2012 #4
    First of all, the matrix isn't quite right, but I didn't notice that the first time I looked at it. That may be your only issue, but I'll be explicit about everything in case something else is confusing you.

    Does it make sense that you can view polynomials in a vector space? Just like in a vector space, the coefficients of any single term in the polynomial do not effect the other terms. If you have [itex]5x^3 [/itex], the 5 only effects the "[itex]x^3[/itex] direction". When you add up all of the terms, the number of [itex]x^3[/itex] added in will be 5 no matter how many times [itex]x^2[/itex] you also add in. Yes, when you evaluate there may be different values of [itex]x[/itex] such that the polynomial equals the same thing, but we are not dealing with final value when we differentiate. The map for differentiation is,
    [itex]
    D: \mathbb{R}[x]\rightarrow\mathbb{R}[x]
    [/itex] Hopefully that part makes sense. If not, make sure that you are comfortable with the idea that polynomials fit the requirements to be a vector space.

    When you differentiate a single power of [itex]x[/itex], the exponent is multiplied by the coefficient and then that value (coeff x exponent) is moved into the position as coefficient for one power down of [itex]x[/itex]. In calc you think about this as subtracting one from the exponent, but here you are being asked to imagine that all (up to N) of the possible values of [itex]x^n[/itex] exist, but may be multiplied by zero the same way that a vector point down the [itex]x-[/itex]axis doesn't mean that [itex]y\mbox{ and } z[/itex] disappear, they just have zeros for coefficients.

    So our N+1 dimensional vector [itex][c_0,c_1,c_2,\ldots,c_N]^T[/itex] of the coefficients of the polynomial [itex]c_0x^0+c_1x^1+c_2x^2+\ldots+c_Nx^N[/itex] can be written as a column vector with [itex]c_0[/itex] at the top and [itex]c_N[/itex] at the bottom. When you write the matrix for [itex]D[/itex], it puts [itex]c_1[/itex] in the [itex]c_0[/itex] place and gets rid of [itex]c_0[/itex]. That is just the normal rule from calculus. Then [itex]c_n*n\rightarrow c_{n-1}[/itex] follows in general so the matrix will have 1,2,3,4... just off the diagonal

    [itex]
    \begin{array}{cccccc}
    0&1&0&0&\ldots&0\\
    0&0&2&0&\ldots&0\\
    0&0&0&3&\ldots&0\\
    \ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\
    0&0&0&0&\ldots&N\\
    0&0&0&0&\ldots&0\\
    \end{array}
    [/itex]

    When you square this matrix, you will get


    [itex]
    \begin{array}{cccccc}
    0&0&2&0&\ldots&0\\
    0&0&0&2*3&\ldots&0\\
    \ldots&\ldots&\ldots&\ldots&\ldots&\ldots\\
    0&0&0&0&\ldots&N(N-1)\\
    0&0&0&0&\ldots&0\\
    0&0&0&0&\ldots&0\\
    \end{array}
    [/itex]

    Which is the same as taking the derivative twice. If you are not convinced, try an easy polynomial like [itex]1+x+x^2+x^3+x^4+x^5[/itex] using your normal calc rules and then do it with a matrix. Then you can put the [itex]c_n[/itex] in as coefficients to finally remove all doubt.
     
    Last edited: Mar 13, 2012
  6. Mar 13, 2012 #5
    Yeah, that's what I got. Except I got N(N-1) rather then (N-1)(N-2).

    Thanks a lot for the help.
     
  7. Mar 13, 2012 #6
    Yes, that is a typo. sorry.
     
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