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Thread: Orthogonality Preserving Linear Map.

  1. #1
    Junior Member RaisinBread's Avatar
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    Orthogonality Preserving Linear Map.

    I'm having trouble with the following problem:

    Let $\displaystyle V$ be a real vector space equipped with an inner product $\displaystyle \langle\cdot,\cdot\rangle$ (and thus with a norm induced by that inner product).
    Let $\displaystyle T:V\to V$ be a bijective linear transformation such that if $\displaystyle \langle u,v\rangle=0$, then $\displaystyle \langle T(u),T(v)\rangle=0$.

    Given an orthonormal basis $\displaystyle \{b_1,...,b_n\}$ is $\displaystyle V$, it clearly is the case that $\displaystyle \{T(b_1),...,T(b_n)\}$ is an orthogonal basis of $\displaystyle V$ and hence $\displaystyle \left\{\frac{T(b_1)}{\|T(b_1)\|},...,\frac{T(b_n)} {\|T(b_n)\|}\right\}$ is an orthonormal basis.

    I now have to show that $\displaystyle \|T(b_1)\|=\cdots=\|T(b_n)\|$, but everything I've tried so far has lead me to a dead end.

    Any hint or tip from someone who knows how to prove this would be greatly appreciated (I'm not looking for someone to post a complete solution).
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  2. #2
    MHF Contributor

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    Re: Orthogonality Preserving Linear Map.

    since our original basis is orthonormal:

    $\displaystyle \|b_1\| = \|b_j\|$ for j = 2,...,n so:

    $\displaystyle \|b_1\|^2 = \|b_j\|^2$

    $\displaystyle \langle b_1,b_1 \rangle - \langle b_j,b_j \rangle = 0$

    $\displaystyle \langle b_1,b_1 \rangle - \langle b_1,b_j \rangle + \langle b_j,b_1 \rangle - \langle b_j,b_j \rangle = 0$

    $\displaystyle \langle b_1,b_1-b_j \rangle + \langle b_j,b_1-b_j \rangle = 0$

    $\displaystyle \langle b_1+b_j,b_1-b_j \rangle = 0$

    $\displaystyle \langle T(b_1+b_j),T(b_1-b_j) \rangle = 0$

    i think it's clear where i'm going with this....
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