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Math Help - Inner product space

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Inner product space

    Let V be Inner product space(real or complex).
    For this exercise we define "twin transformations" S,T:V-->V if for all u,w in V : <Tu,Sw>=<u,w>.

    1. Prove that if T is singular,so T not have "twin transformations".
    2. Prove that if T nonsingular, so there is only one "twin transformations".
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Let V be Inner product space(real or complex).
    For this exercise we define "twin transformations" S,T:V-->V if for all u,w in V : <Tu,Sw>=<u,w>.

    1. Prove that if T is singular,so T not have "twin transformations".
    Indirect proof: Let T be singular and S its twin transformation. Since T is singular, there exists an x\in V, x\neq 0 with Tx=0. Thus we have \langle Tx, Sx\rangle = \langle 0,Sx\rangle = 0, and we also have that \langle Tx, Sx\rangle = \langle x, x\rangle\neq 0, a contradiction.

    2. Prove that if T nonsingular, so there is only one "twin transformations".
    Since T is non-singular we have that for every basis vector e_i there exists a uniquely determined x_i\in V, such that T x_i = e_i.
    Therefore the component of the image S e_j of the basis vector e_j under S in the direction of e_i is given by \langle e_i, S e_j\rangle = \langle T x_i, S e_j\rangle = \langle x_i, S e_j\rangle, and thus is uniquely determined by T.
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