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

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

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

    \|b_1\|^2 = \|b_j\|^2

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

    \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

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

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

    \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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