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Linear transformation between bases

  1. Sep 6, 2009 #1
    Hi !

    I am a little bit confused with notation in the following:

    Let A=

    [tex]\begin{bmatrix}
    2 & 3 & 4 \\
    8 & 5 & 1 \\
    \end{bmatrix}[/tex]

    and consider A as a linear transformation mapping [tex]\mathbb{R}^3[/tex] to [tex]\mathbb{R}^2[/tex]. Find the matix representation of A with respect to the bases

    [tex]\begin{bmatrix}
    1\\
    1\\
    0\\
    \end{bmatrix} , [/tex] [tex]\begin{bmatrix}
    0\\
    1\\
    1\\
    \end{bmatrix} , [/tex] [tex]\begin{bmatrix}
    1\\
    0\\
    1\\
    \end{bmatrix} [/tex] of [tex]\mathbb{R}^3[/tex] and

    [tex]\begin{bmatrix}
    3\\
    1\\
    \end{bmatrix} , [/tex] [tex]\begin{bmatrix}
    2\\
    1\\
    \end{bmatrix} [/tex] of [tex]\mathbb{R}^2[/tex]

    It seems to be a lot of A´s in here with different meanings, and I suppose it is what confuses me :(. Anyway I solved it as follows:

    [tex]\begin{bmatrix}
    3 & 2\\
    1 & 1\\
    \end{bmatrix}^{-1} * [/tex] [tex]\begin{bmatrix}
    2 & 3 & 4\\
    8 & 5 & 1\\
    \end{bmatrix} * [/tex] [tex]\begin{bmatrix}
    1 & 0 & 1\\
    1 & 1 & 0\\
    0 & 1 & 1\\
    \end{bmatrix} = [/tex] [tex]\begin{bmatrix}
    -21 & -5 & -12\\
    34 & 11 & 21\\
    \end{bmatrix} [/tex]
    I am still not sure that I have not confused myself with all the different A´s :( Am I on the right track or completely lost ?
     
  2. jcsd
  3. Sep 6, 2009 #2
    I think you are right but no guarantees. Just as a vector has different representations (components) in different bases, a matrix has different representations in different bases.
     
  4. Sep 6, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    In general, to find the matrix representation of A, from U to V, with [itex]\{u_1, u_2, ..., u_n\}[/itex] a basis for U and [itex]\{v_1, v_2, ..., v_m\}[/itex] a basis for V:

    Apply A to each of [itex]\{u_1, u_2, ..., u_n\}[/itex] in turn and write the result as a linear combination of the [itex]\{v_1, v_2, ..., v_m\}[/itex]. The coefficients form the columns of the matrix.

    for example, here
    [tex]u_1= \begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}[/tex]
    so
    [tex]Au_1= \begin{bmatrix}2 & 3 & 4 \\8 & 5 & 1 \\\end{bmatrix}\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}= \begin{bmatrix}5 \\ 13\end{bmatrix}[/tex]

    Now, [5, 13]= a[3, 1]+ b[2, 1] gives 3a+ 2b= 5 and a+ b= 13. Multiplying the second equation by 2, 2a+ 2b= 26 and, subtracting that from the first equation a= -21. -21+ b= 13 gives b= 34. The first column of the matrix you want is
    [tex]\begin{bmatrix}-21 \\ 34\end{bmatrix}[/tex]
     
    Last edited: Sep 6, 2009
  5. Sep 6, 2009 #4
    Thx for the replies. Thank you for a very clear and good explanation (HallsofIvy), the book I am reading is very compact and does not give very good explanations. There was a typo in the end of your reply:

    [tex] \begin{bmatrix}-21 \\ 13\end{bmatrix} [/tex]

    should be

    [tex] \begin{bmatrix}-21 \\ 34\end{bmatrix} [/tex]
     
  6. Sep 6, 2009 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Right! Thanks.

    No, I'll go back and edit my post so I can claim I never made that silly mistake!
     
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