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Matrix of. Linear operator question

  1. Aug 27, 2012 #1
    I am trying to figure out what the matrix of this linear operator would be:
    T:M →AMB where A, M, B are all 2X2 matrices with respect to the standard bases of a 2x2 matrix viz. e11, e12 e21 and e22. Any ideas? Il know it should be 2X8 matrix. I am trying to teach myself Abstract algebra using Artin's book and this is listed in the problem section in one of the chapters.

  2. jcsd
  3. Aug 27, 2012 #2


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    Hey frowdow and welcome to the forums.

    Just to clarify what are the domain and range (or codomain) of the actual mappings?
  4. Aug 27, 2012 #3


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    What are the images of the standard basis elements under T?
  5. Aug 27, 2012 #4


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    In general, one can find the matrix representing a given linear transformation, in a given basis, by applying the linear transformation to each basis "vector" in turn, writing the result as a linear combination of the basis vectors. The coefficients of the linear combination form the columns of the matrix.

    [tex]e_{11}= \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}[/tex]
    If, say,
    [tex]A= \begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix}[/tex]
    [tex]B= \begin{bmatrix}b_{11} & b_{12} \\ b_{21} & b_{22}\end{bmatrix}[/tex]
    [tex]Ae_{11}B= \begin{bmatrix}a_{11} & a_{12} \\ a_{21} & a_{22}\end{bmatrix} \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}\begin{bmatrix}b_{11} & b_{12} \\ b_{21} & b_{22}\end{bmatrix}[/tex]
    [tex]= \begin{bmatrix}a_{11} & 0 \\ a_{21} & 0\end{bmatrix}\begin{bmatrix}b_{11} & b_{12} \\ b_{21} & b_{22}\end{bmatrix}= \begin{bmatrix}a_{11}b_{11} & a_{11}b_{21} \\ a_{21}b_{12} & a_{21}b_{22}\end{bmatrix}[/tex]
    [tex]= a_{11}b_{11}e_{11}+ a_{11}b_{21}e_{12}+ a_{21}b{21}e_{21}+ a_{21}b_{22}e_{22}[/tex]
    so the first 'column' consists of those four matrices. That is, the matrix is 2 by 2 but each entry is a 2 by 2 matrix so, expanded, it has 4 rows and 4 columns.
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