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Math Help - Standard matrix of T

  1. #1
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    Standard matrix of T

    Assume that R2>R2 is a linear transformation such that
    T|1| =|1|
    |2| |-3|

    and
    T|3| =|2|
    |-1| |1|
    Find the standard matrix of T.

    Ok, so I get that |1,2|=e1+2e2 and that |3,-1|=3e1-e2, but what do I do from there on? It shows in the solutions that
    e1=1/7|1,2| + 2/7|3, -1|
    e2=3/7|1,2|-1/7|3,-1|

    I understand that the are using the same thing from e1+2e2 and 3e1-e2, but where is the divisor of 7 coming from?

    Sorry if it's formatted badly, I can scan the original page? Thanks
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  2. #2
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    Okay so you know that

    T(e_1+2e_2)=e_1-3e_2 and
    T(3e_1-e_2)=2e_1+e_2

    Now use the linearity of the operator to get

    T(e_1+2e_2)=T(e_1)+2T(e_2)=e_1-3e_2 and
    T(3e_1-e_2)=3T(e_1)-T(e_2)=2e_1+e2

    This gives a system of equations that you can solve for T(e_1),T(e_2)

    After you solve for them you should get

    T(e_1)=\frac{1}{7}(5e_1-e_2); T(e_2)=\frac{1}{7}(e_1-10e_2)

    These are the columns of the matrix representation of the transform.

    \begin{bmatrix} \frac{5}{7} & \frac{-1}{7} \\ \frac{1}{7} & \frac{-10}{7}\end{bmatrix}=\frac{1}{7}\begin{bmatrix} 5 & -1\\ 1 & -10\end{bmatrix}
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  3. #3
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    Thanks for the reply but I'm still confused. I don't get where you said it gives a system of equations for T(e1) and T(e2). Where would I get that from?

    is it just
    1 -1
    2 1
    Which I'm pretty sure is wrong... could you maybe put in the step of where you got the system of equations, and where from?

    Thanks a lot
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  4. #4
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    Quote Originally Posted by shibble View Post
    Assume that R2>R2 is a linear transformation such that
    T|1| =|1|
    |2| |-3|

    and
    T|3| =|2|
    |-1| |1|
    Find the standard matrix of T.

    Ok, so I get that |1,2|=e1+2e2 and that |3,-1|=3e1-e2, but what do I do from there on? It shows in the solutions that
    e1=1/7|1,2| + 2/7|3, -1|
    e2=3/7|1,2|-1/7|3,-1|

    I understand that the are using the same thing from e1+2e2 and 3e1-e2, but where is the divisor of 7 coming from?

    Sorry if it's formatted badly, I can scan the original page? Thanks
    Let's take the following as given:
    T is completely determined once the images of any basis for R^2 are known. *

    With * in mind, I think I would read the initial statement, i.e.,
    that T(1,2) = (1,-3) and T(3,-1) = (2,1), as telling me that
    with respect to the basis B = {(1,2), (3,-1)} T is uniquely determined.

    In this context the matrix  \begin{bmatrix}5/7 & 1/7 \\ -1/7 & -10/7\end{bmatrix} is the matrix representation of T
    relative to basis B. Call it M'.

    But, the problem asks you to find the "standard matrix" of T.
    I read that as meaning the matrix representation of T relative to
    the standard (or natural) basis E = {(1,0),(0,1)}.

    So I think there's more work to do.

    You already have M'.
    Can you say what the matrix  \begin{bmatrix}1/7 & 3/7 \\ 2/7 & -1/7\end{bmatrix} might represent?
    Maybe the matrix of transition from E to B?
    What about  \begin{bmatrix}1 & 3 \\ 2 & -1\end{bmatrix} ?
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  5. #5
    Behold, the power of SARDINES!
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    Quote Originally Posted by shibble View Post
    Thanks for the reply but I'm still confused. I don't get where you said it gives a system of equations for T(e1) and T(e2). Where would I get that from?

    is it just
    1 -1
    2 1
    Which I'm pretty sure is wrong... could you maybe put in the step of where you got the system of equations, and where from?

    Thanks a lot
     T(e_1+2e_2)=T(e_1)+2T(e_2)=e_1-3e_2 and
    T(3e_1-e_2)=3T(e_1)-T(e_2)=2e_1+e2

    This is the system of equations or more explicitly

    T(e_1)+2T(e_2)=e_1-3e_2 and
    3T(e_1)-T(e_2)=2e_1+e2

    Solving this gives you the transform of the standard basis.
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  6. #6
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    Quote Originally Posted by TheEmptySet View Post
     T(e_1+2e_2)=T(e_1)+2T(e_2)=e_1-3e_2 and
    T(3e_1-e_2)=3T(e_1)-T(e_2)=2e_1+e2

    This is the system of equations or more explicitly

    T(e_1)+2T(e_2)=e_1-3e_2 and
    3T(e_1)-T(e_2)=2e_1+e2

    Solving this gives you the transform of the standard basis.
    Hi, I don't know if its because it's late, or because I'm just really stupid, but I'm still really lost.

    To solve, do you mean in this fashion:

    T(e_1)+2T(e_2)=e_1-3e_2
    3T(e_1)-T(e_2)=2e_1+e2

     \begin{bmatrix} 1 & -3 \\ 2 & 1\end{bmatrix}

    >>
     \begin{bmatrix} 1 & -3 \\ 0 & 7\end{bmatrix}

    so

    T(e_1)+2T(e_2)=-3
    3T(e_1)-T(e_2)=7

    Just a wild guess here

    I completely get this:

    T(e_1)+2T(e_2)=e_1-3e_2 and
    3T(e_1)-T(e_2)=2e_1+e2

    and I also get that you have to use

    T(e_1)+2T(e_2) and
    3T(e_1)-T(e_2)
    where the e1 and e2 are (1,2) and (3,-1) respectively, but how do you find out what T is for each?
    From what I'm looking at, the multiple stays the same, but there is a divisor for each T, in this case 7.
    How do I solve for that T? Hopefully I'm being clear enough


    EDIT I GOT IT, I guess I just needed some sleep Thanks guys, makes sense now
    Last edited by shibble; August 2nd 2010 at 07:44 AM.
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