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Thread: Is this onto or 1 to 1 ?

  1. #1
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    Is this onto or 1 to 1 ?

    T: R^3 ----> W , where W is the vector space of all symmetric 2x2 matrices

    T [ a b c ] =
    [ a+b+c b-2c ]
    [ b-2c a-c ]

    I'm not sure how to show it is 1 to 1. After being transformed, I see that a = c and b = 2c and 2a = -b.

    But I think I know it is not "onto" because T of R^3 cannot possibly cover all of the range of M22. A side-question on a little notation problem, is the space of 2x2 matrices M22 ?
    Last edited by crystal-maiden; Mar 21st 2009 at 04:45 PM. Reason: typo
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  2. #2
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    [quote=crystal-maiden;285899]T: R^3 ----> W , where W is the vector space of all symmetric 2x2 matrices

    T [ a b c ] =
    [ a+b+c b-2c ]
    [ b-2c a-c ]

    I am new to abstract algebra, but I have a few thoughts/questions.

    But I think I know it is not "onto" because T of R^3 cannot possibly cover all of the range of M22. A side-question on a little notation problem, is the space of 2x2 matrices M22 ?
    1) A basis for the vector space of symmetric matrices is
    $\displaystyle \left( \begin{array}{cc}
    1 & 0 \\
    0 & 0 \end{array} \right)$,$\displaystyle \left( \begin{array}{cc}
    0 & 1 \\
    1 & 0 \end{array} \right)$,$\displaystyle \left( \begin{array}{cc}
    0 & 0 \\
    0 & 1 \end{array} \right)$
    Therefore it has dimension 3, so it is possible to create a 1-to-1 and onto map, T. For instance, $\displaystyle T([a b c]^T) = a*\left( \begin{array}{cc}
    1 & 0 \\
    0 & 0 \end{array} \right)+b*\left( \begin{array}{cc}
    0 & 1 \\
    1 & 0 \end{array} \right)+c*\left( \begin{array}{cc}
    0 & 0 \\
    0 & 1 \end{array} \right) = \left( \begin{array}{cc}
    a & b \\
    b & c \end{array} \right)$
    This T will hit every symmetric matrix (onto), and every element in $\displaystyle R^3$ maps to a unique symmetric matrix.
    This shows it is possible for a T. But your question is asking about your T not any T.

    I'm not sure how to show it is 1 to 1. After being transformed, I see that a = c and b = 2c and 2a = -b.
    2) I do not know how you concluded your values above.


    3) Consider the symmetric matrix:
    $\displaystyle \left( \begin{array}{cc}
    x & y \\
    y & z \end{array} \right)$

    Your transformation is going to take $\displaystyle [a b c]^T$ into the symmetric matrix:
    $\displaystyle \left( \begin{array}{cc}
    a+b+c & b-2c \\
    b-2c & a-c \end{array} \right)$

    This gives us three equations, namely:
    $\displaystyle x = a+b+c$
    $\displaystyle y = b-2c$
    $\displaystyle z = a-c$
    we can express this in terms of matrices
    $\displaystyle \left( \begin{array}{ccc}
    1 & 1 & 1 \\
    0 & 1 & -2 \\
    1 & 0 & -1\end{array} \right)\left( \begin{array}{c}
    a\\
    b\\
    c \end{array} \right)=\left( \begin{array}{c}
    x\\
    y\\
    z\end{array} \right)$
    if you row-reduce this last matrix you will see that it has full rank.
    First, this means that for any $\displaystyle (x,y,z)^T$ (any symmetric matrix) there is a unique solution for $\displaystyle (a,b,c)^T$ (an element in $\displaystyle R^3$). Thus, T is onto.
    Second, the full rank shows that each $\displaystyle (a,b,c)^T$ in $\displaystyle R^3$ maps to a unique $\displaystyle (x,y,z)^T$ in the 2x2 symmetric matrix space. Thus, T is one-to-one.

    Like I said, i am new to abstract algebra, I hope this at least gets you started.
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  3. #3
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    Question

    interesting approach, that would work.
    I was trying to find the kernal of the transformation such that a+b+c=0 and b-2c=0 and a-c = 0
    and if I can prove the kernal is indeed zero, then I know for sure it is 1 to 1.
    *edit* ooo.. I figured it out thanks so much!
    Last edited by crystal-maiden; Mar 22nd 2009 at 06:48 AM. Reason: Figured it out!
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