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Polynomial vector space

  1. Sep 7, 2008 #1
    [/b]1. The problem statement, all variables and given/known data[/b]
    Let P_2 be the set of all real polynomials of degree no greater than 2.
    Show that both B:={1, t, t^2} and B':= {1, 1-t, 1-t-t^2} are bases for P_2.

    If we regard a polynomial p as defining a function R --> R, x |--> p(x), then p is differentiable, and
    D: P_2 --> P_2, p |--> p' = dp/dx - defines a linear transformation.
    Find the matrix of D with respect to the bases
    (i) B in both the domain and co-domain
    (ii) B in the domain and B' in the codomain.
    (iii) B' in the domain and B in the codomain
    (iv) B' in both the domain and codomain.

    [/b]2. The attempt at a solution[/b]
    I used 3x3 matrices and row reduction for B' to show that they are both bases, i.e. 3 pivot variables in reduced row form. B was just the identity matrix.
    Its the second part that I don't understand. For me, the domains of B and B' are all real numbers, or have I already found the matrices of D in both domains by showing with row echelon matrices that B and B' are bases.
    This is very confusing and my mind is now doing many laps around the same circuit.
    Any clarification would be great.
     
  2. jcsd
  3. Sep 7, 2008 #2

    HallsofIvy

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    Suppose T is a Linear transformation from vector space U with basis Bu to vector space V with basis Bv. To find the matrix representing T using basis Bu in the domain and basis Bv in the codomain, do the following:

    1) Apply T to the first vector in basis Bu. The result, of course, will be in V and can be written as a linear combination of the vectors in basis Bv. The coefficients in that linear combination are the numbers in the first column of the matrix.

    2) Apply T to the second vector in basis Bu. The result, of course, will be in V and can be written as a linear combination of the vectors in basis Bv. The coefficients in that linear combination are the numbers in the second column of the matrix.

    Continue applying T to each basis vector in turn to get each column of the matrix.

    For example, the first vector in B is "1". Differentiating that gives 0. That, written in the B basis is (0)1+ (0)x+ (0)x2. The first column consists entirely of 0's. The second vector in B is "x". Differentiating that gives 1. That, written in the B basis is (1)(1)+ (0)x+ (0)x2. The second column is <1 0 0>. The third vector in B is "x2". Differentiating that gives 2x. That, written in the B basis is (0)1+ (2)x+ (0)x2. The third column is <0 2 0>. (i) is
    [tex]\left[\begin{array}{ccc}0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0\end{array}\right][/tex]

    Notice that matrix has determinant 0 and is not invertible. That is because differentiation is not one-to-one and so is not invertible.
     
  4. Sep 7, 2008 #3
    so....
    B matrix is [tex]
    \left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]
    [/tex]
    is the domain?
    codomain of B is
    [tex]
    \left[\begin{array}{ccc}0 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 0\end{array}\right]
    [/tex]
    as you said
    B' domain is
    [tex]
    \left[\begin{array}{ccc}1 & 1 & 1 \\ 0 & -1 & -1 \\ 0 & 0 & -1\end{array}\right]
    [/tex]
    codomain of B'
    [tex]
    \left[\begin{array}{ccc}0 & -1 & -1 \\ 0 & 0 & -2 \\ 0 & 0 & 0\end{array}\right]
    [/tex]
    Are these ok??

    To find the matrices D with respect to the 4 part question, is it just a matter of solving the augmented matrices formed in each part, so that part i) would be the codomain matrix of B and part ii)the codomain matrix of B', given the identity natrix is the domain of B?

    Why this is scary is because of the magnitude of how wrong I could be lol.
     
    Last edited: Sep 7, 2008
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