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Math Help - Finding T*x given T and x

  1. #1
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    Finding T*x given T and x

    The problem: V= R^2 , T(a,b)= (2a+b, a-3b), x= (3,5). Evaluate T* at the given vector in V.
    Thoughts:
    I solved this by putting T in matrix form ( \left[ \begin{array}{cc} 2 & 1 \\ 1 &-3 \end{array} \right] ) finding T^* ( \left[ \begin{array}{cc} 2 & 1 \\ 1 &-3 \end{array} \right] ) and computing  T^*x which equals \left[ \begin{array}{c} 11 \\ -12 \end{array} \right] .
    I believe this method to be correct (though I wouldn't really be shocked if you told me it wasn't), however, I'd like to be able to solve this problem using the property <T(X), y> = <x, T^*(y)> because it is my understanding that this is the only way to solve some of these problems. But I can't figure out how this is possible despite at least an hours attempt. *bangs head against wall*
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  2. #2
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    Quote Originally Posted by kaelbu View Post
    The problem: V= R^2 , T(a,b)= (2a+b, a-3b), x= (3,5). Evaluate T* at the given vector in V.
    Thoughts:
    I solved this by putting T in matrix form ( \left[ \begin{array}{cc} 2 & 1 \\ 1 &-3 \end{array} \right] ) finding T^* ( \left[ \begin{array}{cc} 2 & 1 \\ 1 &-3 \end{array} \right] ) and computing  T^*x which equals \left[ \begin{array}{c} 11 \\ -12 \end{array} \right] .


    Uh?? And how do you actually find what T^{*} is?? This is the problem in this question!

    Yet if we remember that T^{*} is easily representable by means of T when we choose an orthonormal basis for our vector space then the problem is easy.

    Tonio


    I believe this method to be correct (though I wouldn't really be shocked if you told me it wasn't), however, I'd like to be able to solve this problem using the property <T(X), y> = <x, T^*(y)> because it is my understanding that this is the only way to solve some of these problems. But I can't figure out how this is possible despite at least an hours attempt. *bangs head against wall*
    .
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  3. #3
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    I used the formula T*_{ij} = (-1)^{i+j} det [j ,i] , I think ... on second thought (this actually gives  \left[ \begin{array}{cc} 2 & -1 \\ -1 & -3 \end{array} \right]  , I think ) that may have not been what I used, but it's what I MEANT to use. But that really is beside the point, because I most certainly do not know how to use this mysterious formula for my next problem.
    I tried to use the orthonormal basis  \beta = { e_1 , e_2 } And do something involving T(x) = < x, y> where y= T(e_1) e_1 + T(e_2)e_2  which = -1. But I had no idea where to go from there or if that was anywhere near what I supposed to do.
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