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Thread: Inner product space

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    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 $\displaystyle T$ be singular and $\displaystyle S$ its twin transformation. Since $\displaystyle T$ is singular, there exists an $\displaystyle x\in V, x\neq 0$ with $\displaystyle Tx=0$. Thus we have $\displaystyle \langle Tx, Sx\rangle = \langle 0,Sx\rangle = 0$, and we also have that $\displaystyle \langle Tx, Sx\rangle = \langle x, x\rangle\neq 0$, a contradiction.

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