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Math Help - Linear Transformation help !!

  1. #1
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    Linear Transformation help !!

    Let T: M2x2(R)------>P3(R) be a linear transformation defined by
    T ([a b]) = (a-b) + (a-d)x +(b-c)x2 +(c-d)x3
    [c d]


    Consider the bases v ={[1 0] , [0 1] ,[1 0] ,[ 0 0]} of M2x2
    - - - -------- -- -- ---- [1 0 ] [0 1] [0 1] [1 1]

    and u = {x, x-x2,x-x3,x-1} of P3(R)


    a) find [T]uv

    b) use [T]uv to find a basis for the kernel of T

    c) use [T]uv to find a basis for the image of T
    d) State the nullity and rank of T. Is T injective? surjective?
    Last edited by dave52; February 2nd 2013 at 12:44 PM.
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Linear Transformation help !!

    let me just be clear.  [T]_{uv} is the transformation that goes from Vector Space U to V ?
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  3. #3
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    Re: Linear Transformation help !!

    u is a basis for M(2x2) and v is a basis for P(3)
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  4. #4
    Senior Member jakncoke's Avatar
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    Re: Linear Transformation help !!

    so since any vector in  M_{2x2}(\mathbb{R}) can be written as a linear combination of basis vectors  \{b_1,b_2,b_3,b_4\}
    c_1b_1 + c_2b_2 + c_3b_3 + c_4b_4 is any vector in V. since  L(c_1b_1 + c_2b_2 + c_3b_3 + c_4b_4) = c_1L(b_1)+...+c_4L(b_4) we need to find out the coordinates of the linear transformation of the basis vectors of V with respect to basis of U.

    let b_1 = \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix} , b_2 = \begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix} , b_3 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} , b_4 = \begin{bmatrix} 0 & 0 \\ 1 & 1 \end{bmatrix}
    so L(b_1) = 1 + x - x^2 + x^3 , which written in terms of the basis of U is \begin{bmatrix} x & x-x^2 & x-x^3 & x-1 \end{bmatrix}* \begin{bmatrix} 2 \\ 1 \\ -1 \\ -1 \end{bmatrix}  = 2x + x - x^2 -x + x^3 -x + 1  = x - x^2 + x^3 + 1 = L(b_1)  . I did this for all basis vectors of V.
    and i got the following Linear Transformation Matrix from V to U

     \begin{bmatrix}2 & -2 & 0 & -2 \\ 1 & -1 & 0 & 1 \\ -1 & 1 & 1 & 0 \\ -1 & 1 & -1 & 0 \end{bmatrix} which has only 3 linearly indepdent vectors.
    So  Dim(Img(T)) = 3 so Dim(Ker(T)) = 1. Which is as so since only the matrix consisting of all entries identical  \begin{bmatrix}a & a \\ a & a \end{bmatrix} a \in \mathbb{R} maps to the 0 matrix under this Linear Transformation.

    Nullity is the Dim of the Kernel which is 1
    Rank if the Dim of the image which is 3.

    So the basis for the kernel is just  \begin{bmatrix}1 & 1 \\ 1 & 1 \end{bmatrix}

    The basis for the image of the Linear Transformation are the following coordinates in P  \begin{bmatrix}2  & 0 & -2 \\ 1  & 0 & 1 \\ -1  & 1 & 0 \\ -1  & -1 & 0 \end{bmatrix}

    Since the Ker contains more than just the 0 vector, it is certainly not injective. Since the Rank of T is 3 (Dim of codomain is 4), it is also not surjective.
    Last edited by jakncoke; February 2nd 2013 at 02:13 PM.
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