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The Matrix Of A Linear Transformation

  1. Nov 23, 2005 #1
    I saw a similar post to this one, but i just got lost in the mess of the whole thing. So i just started a new thread.

    A question reads:

    Let T: Pn ---> Pn be defined by T[P(x)] = p(x) + xp'(x), where p'(x) denotes the derivative. Show that T is an isomorphism by finding Mbb(T) when B = {1, x, x^2, ... , x^n}

    From doing the other question and problems in the text book, i know how to find Mdb(T). I suppose that finding Mbb(T) would be very similar.

    I did it like this:

    Mbb(T) = [ CbT(1) CbT(x) CbT(x^2) ... CbT(X^n) ]
    Mbb(T) = [ Cb(1) Cb(2x) Cb(3x^2) .... Cb((n+1)X^n]

    and it gives this nxn matrix:

    [1 0 0 ......... 0]
    [0 2 0 ......... 0]
    [0 0 3 ......... 0]
    [0 0 0 ... (n+1)]

    now, an isomorphism means that the linear transformation is both one-to-one and onto.

    How do you tell that its an isomorphism by just looking at the matrix?
     
  2. jcsd
  3. Nov 23, 2005 #2

    Physics Monkey

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    In this case, your transformation is isomorphic if the matrix is invertible. Why?
     
  4. Nov 23, 2005 #3
    Their is a theorm in my text book.

    "Let A be an mXn matrix, and let TA : R^n ---> R^m be the linear transformation inudced by A, that is, Ta(X) = AX for all X in R^n.

    1. Ta is onto if and only if rank A = m
    2. Ta is one-to-one if and only if A = n

    oh wait.. that dosn't work in this problem...

    ill brb
     
  5. Nov 23, 2005 #4
    Their is an example in the textbook...

    If U is any invertable mxm matrix, the map T: Mmn ----> Mmn given by T(X) = UX is an isomorphism by Example 6 Section 8.2

    This example is too long to type out... but it dosn't show any matrix. It just uses the transformation to show it.
     
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