Proof concerning inner products

Prove that dimV=dimU^$\displaystyle \perp$+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^$\displaystyle \perp$,

so dim nullT=dimU^$\displaystyle \perp$=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^$\displaystyle \perp$+dimU=m+k.

ami rite?