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Thread: Proof concerning inner products

  1. #1
    Jul 2009

    Proof concerning inner products

    Prove that dimV=dimU^ \perp+dimU

    Let T project from V---->U. Assume V=nullT+rangeT.
    let dim V=n, dim rangeT=m, and dim nullT=k.
    Extend the basis of V by {, w1...wk} where wi is in nullT
    and uj is in rangeT, for 1<=i<=k and 1<=j<=m. Since T(uj)=uj and T(wi)=0, (for wi=/=0)
    <T(uj), T(wi)>=0, thus wi is 0 in rangeT, and is orthogonal to uj, by the inner
    product being 0, so wi is in both nullT and orthU. Therefore, nullT=U^ \perp,
    so dim nullT=dimU^ \perp=k. Since T maps from V---->U and V-nullT=rangeT,
    knowing that nullT is not in U means that rangeT=U, thus dim rangeT=dimU=m, and dimV=dimnullT+dimrangeT=dimU^ \perp+dimU=m+k.

    ami rite?
    Last edited by evilpostingmong; Jul 5th 2009 at 05:53 AM.
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