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 {u1...um, 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.

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