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Thread: Image & Kernal

  1. #1
    Member iPod's Avatar
    Jul 2009

    Image & Kernal

    I have attached the problem:

    We are dealing with polynomials and we have a linear transformation $\displaystyle F:P_2 \rightarrow P_3$ , meaning we are transforming a polynomial of degree 2 to a polynomial of degree 3.

    An intellectual guess for the basis of the Image of the linear transformation would be $\displaystyle 1,x,x^2,x^3$, however I'm not too sure.

    I would like some guidance on how to encounter questions like these about Images and Kernals.
    Thank you.
    Attached Thumbnails Attached Thumbnails Image & Kernal-lin-alg-prob.png  
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  2. #2
    MHF Contributor

    Mar 2011

    Re: Image & Kernal

    well, a little thought should convince you that we can't "transform" a smaller space into a bigger one (in a linear way), we can only (at best) place a "copy" of our smaller space inside the bigger one.

    the number-one tool to use in these situations is the rank-nullity theorem, so dim(P2) = dim(ker(F)) + dim(im(F)).

    since dim(P2) = 3, even if dim(ker(F)) = 0 (the smallest possible value), the biggest dim(im(F)) could possibly be is 3. this means that {1,x,x2,x3} couldn't possibly be a basis, as that set has 4 elements (4 is bigger than 3, see? this vector space stuff is easy).

    since it's not really clear how big a space we get by taking that integral, let's look at the kernel first. at least this will tell us how many basis vectors we have to come up with for the image.

    now ker(F) consists of all the polynomials of degree 2 or less (plus the 0-polynomial, which doesn't really have a degree) that F takes to the 0-polynomial in P3.

    so if f(x) is in ker(F):

    F(f)(x) = 0 (= 0 + 0x + 0x2 + 0x3, NOT "the real number 0").

    so we're going to have to bite the big one, and actually do some integration, here (i hope i can remember HOW). let's take a "typical polynomial in P2", say:

    f(x) = a + bx + cx2, and just "see what happens". so

    $\displaystyle F(f)(x) = \int_{x-1}^{3x+1} f(t)\ dt = \int_{x-1}^{3x+1} a + bt + ct^2\ dt$

    $\displaystyle = at + \frac{b}{2}t^2 + \frac{c}{3}t^3 \bigg|_{x-1}^{3x+1}$

    $\displaystyle = a(3x+1) + \frac{b}{2}(3x+1)^2 + \frac{c}{3}(3x+1)^3 - a(x-1) - \frac{b}{2}(x-1)^2 - \frac{c}{3}(x-1)^3$

    $\displaystyle = 3ax + a + \frac{9b}{2}x^2 + 3bx + \frac{b}{2} + 9cx^3 + 9cx^2 + 3cx + \frac{c}{3} - ax + a - \frac{b}{2}x^2 + bx - \frac{b}{2} - \frac{c}{3}x^3 + cx^2 - cx + \frac{c}{3}$

    $\displaystyle = \frac{26c}{3}x^3 + (4b + 10c)x^2 + (2a + 4b + 2c)x + (2a + \frac{2c}{3})$

    (whew! i hope i did that right. you might want to double-check it).

    since this is identically 0, this gives us the four equations:

    26c/3 = 0
    4b + 10c = 0
    2a + 4b + 2c = 0
    2a + 2c/3 = 0

    from which we see that c = 0, and thus a and b must also be 0. hence dim(ker(F)) = 0, so F is injective. this tells us that we can form a basis for im(F), by taking the images of any basis for P2. and we happen to have one handy: {1,x,x2}. so it suffices to calculate F(1)(x), F(x)(x), and F(x2)(x) (those look kind of funny, don't they? perhaps we should have chosen another variable to use for P2. too late, we're forging ahead).


    $\displaystyle F(1)(x) = \int_{x-1}^{3x+1} dt = 3x+1 - (x-1) = 2x + 2$

    $\displaystyle F(x)(x) = \int_{x-1}^{3x+1} t\ dt = \frac{1}{2}((3x+1)^2 - (x-1)^2)$

    $\displaystyle = \frac{1}{2}(9x^2 + 6x + 1 - x^2 + 2x - 1) = 4x^2 + 4x$

    $\displaystyle F(x^2)(x) = \int_{x-1}^{3x+1} t^2\ dt = \frac{1}{3}((3x+1)^3 - (x-1)^3)$

    $\displaystyle = \frac{1}{3}(27x^3 + 27x^2 + 9x + 1 - x^3 + 3x^2 - 3x + 1) = \frac{26}{3}x^3 + 10x^2 + 2x + \frac{2}{3}$

    so {2x+2,4x2+4x,(26/3)x3+10x2+2x+(2/3)} is a basis for im(F).
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