Linear transformation and it's properties Let M be a p-1-dimensional subspace

cyrsvab 2022-02-14 Answered
Linear transformation and its
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venenolundicoofw
Answered 2022-02-15 Author has 12 answers

First, the statement is false if t=0 in RpM because in this case, t is the class of an element t′ of M and M=kerBBt=01.
Now, if t0, then tM and the line L=t satisfies ML=Rp. Thus if π is the projection on L along M (i.e., if x=xM+xL, then π(x)=xL) and if ψ:LR is defined by ψ(λt)=λ,λR, we see that B=ψπ satisfies kerB=M and Bt=1.
Edit: 1. B is linear: indeed, if x=xM+λtRp,y=yM+μtRp and aR then ax+y=(axM+yM)+(aλ+μ)t so
B(ax+y)=ψ(π(ax+y))=ψ((aλ+μ)t)=aλ+μ
and aB(x)+B(y)=aψ(π(x))+ψ(π(y))=aψ(λt)+ψ(μt)=aλ+μ
By definition of π,Mkerπ thus MkerB. Now, if kerBM then kerB is a subspace of Rp containing strictly a (p1)-subspace, hence kerB would be all Rp. This is false because Bt=1. Thus kerB=M.

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