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Math Help - Compute the matrices for the given linear transformations

  1. #1
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    Compute the matrices for the given linear transformations

    The format that this question is presented in is chiefly what has me stumped. Either I'm unfamiliar with it or I just can't remember how to do it in this format.

    Compute the matrices for the given linear transformations T:F^n\rightarrow F^n with respect to the standard basis for F^n.

    a) T(a_1,...,a_n)=(a_1,...,a_1), (a_1,...,a_n)\in F^n

    b) T(a_1,...,a_n)=(a_n,a_{n-1},...,a_1), (a_1,...,a_n)\in F^n

    c) T(a_1,...,a_n)=(a_1+a_2,...,a_{n-1}+a_n,0), (a_1,...,a_n)\in F^n
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  2. #2
    Member HappyJoe's Avatar
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    Let me consider part b).

    The notation just means that T maps a typical vector (a_1,...,a_n) in F^n to the vector (a_n,...,a_1), also in F^n.

    The matrix of this transformation with respect to the standard basis for F^n (which is (1,0,0,....,0), (0,1,0,....,0), ...., (0,0,...,0,1)) is the matrix, whose first column is the coordinates of T applied to the first basis vector, and similarly for the other columns. To be concrete, T applied to the first basis vector is:

    T(1,0,...,0) = (0,0,...,1).

    Hence the first column of the matrix is

    0
    0
    .
    .
    .
    0
    1

    Similarly for the other columns, so the matrix will look something like this:

    0 0 . . . 0 1
    0 0 . . . 1 0
    0 0 . . . 0 0
    . . . . . . . .
    . . . . . . . .
    . . . . . . . .
    0 1 . . . 0 0
    1 0 . . . 0 0
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  3. #3
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    It might help to look at simple examples first.

    Suppose n= 3. Then
    a) T(a, b, c)= (a, a, a) so, looking at the standard basis, T(1, 0, 0)= (1, 1, 1) , T(0, 1, 0)= (0, 0, 0), T(0, 0, 1)= (0, 0, 0) and those are the columns of the matrix representation of T: \begin{bmatrix}1 & 0 & 0 \\1 & 0 & 0 \\ 1 & 0 & 0\end{bmatrix}.

    b) T(a, b, c)= (c, b, a) so, looking at the standard basis, T(1, 0, 0)= (0, 0, 1), T(0, 1, 0)= (0, 1, 0), T(0, 0, 1)= (1, 0, 0) and those are the columns of the matrixs representation of T: \begin{bmatrix}0 & 0 & 1 \\ 0 & 1 & 0\\ 1 & 0 & 0\end{bmatrix}

    c) T(a, b, c)= (a+ b, b+ c, 0) so, looking at the standard basis, T(1, 0, 0)= (1, 0, 0), T(0, 1, 0)= (1, 1, 0), T(0, 0, 1)= (, 1, 0) and those are the columns of the matrixs representation of T: \begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 0\\ 0 & 0 & 0\end{bmatrix}
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  4. #4
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    I think your a) is wrong i'm pretty sure the answer is the transopse of what you said.
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  5. #5
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    If you are refering to my solution (in 3 dimensions) of \begin{bmatrix}1 & 0 & 0 \\ 1 & 0 & 0 \\1 & 0 & 0\end{bmatrix}, then it is sufficient to note that
    \begin{bmatrix}1 & 0 & 0 \\ 1 & 0 & 0 \\1 & 0 & 0\end{bmatrix}\begin{bmatrix}a \\ b\\ c\end{bmatrix}= \begin{bmatrix} a \\ a\\ a\end{bmatrix}

    while the transpose \begin{bmatrix}1 & 1 & 1 \\ 0 & 0 & 0 \\0 & 0 & 0\end{bmatrix} gives
    \begin{bmatrix}1 & 1 & 1 \\ 0 & 0 & 0 \\0 & 0 & 0\end{bmatrix}\begin{bmatrix}a \\ b\\ c\end{bmatrix}= \begin{bmatrix}a+ b+ c \\ 0 \\ 0\end{bmatrix}

    Or are you using a convention of writing T(a, b, c) as \begin{bmatrix}a & b & c\end{bmatrix}\begin{bmatrix}1 & 1 & 1 \\ 0 & 0 & 0 \\0 & 0 & 0\end{bmatrix}= \begin{bmatrix} a & a & a\end{bmatrix}?

    If so, then you are correct- but we are using different conventions.
    Last edited by HallsofIvy; October 9th 2010 at 08:15 AM.
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