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find the orthogonal projection of v= [19,12,14,-17] onto the subspace W spanned by [ [ -4,-1,-1,3] ,[ 1,-4,4,3] ] proj w (v) = [answer,answer,answer,answer]

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2 votes

Answer:

Hence, we have:


proj_W(v)=[(464)/(21),(167)/(21),(71)/(21),(-131)/(7)]

Explanation:

By the orthogonal decomposition theorem we have:

The orthogonal projection of a vector v onto the subspace W=span{w,w'} is given by:


proj_W(v)=((v\cdot w)/(w\cdot w))w+((v\cdot w')/(w'\cdot w'))w'

Here we have:


v=[19,12,14,-17]\\\\w=[-4,-1,-1,3]\\\\w'=[1,-4,4,3]

Now,


v\cdot w=[19,12,14,-17]\cdot [-4,-1,-1,3]\\\\i.e.\\\\v\cdot w=19* -4+12* -1+14* -1+-17* 3\\\\i.e.\\\\v\cdot w=-76-12-14-51=-153


w\cdot w=[-4,-1,-1,3]\cdot [-4,-1,-1,3]\\\\i.e.\\\\w\cdot w=(-4)^2+(-1)^2+(-1)^2+3^2\\\\i.e.\\\\w\cdot w=16+1+1+9\\\\i.e.\\\\w\cdot w=27

and


v\cdot w'=[19,12,14,-17]\cdot [1,-4,4,3]\\\\i.e.\\\\v\cdot w'=19* 1+12* (-4)+14* 4+(-17)* 3\\\\i.e.\\\\v\cdot w'=19-48+56-51\\\\i.e.\\\\v\cdot w'=-24


w'\cdot w'=[1,-4,4,3]\cdot [1,-4,4,3]\\\\i.e.\\\\w'\cdot w'=(1)^2+(-4)^2+(4)^2+(3)^2\\\\i.e.\\\\w'\cdot w'=1+16+16+9\\\\i.e.\\\\w'\cdot w'=42

Hence, we have:


proj_W(v)=((-153)/(27))[-4,-1,-1,3]+((-24)/(42))[1,-4,4,3]\\\\i.e.\\\\proj_W(v)=(-17)/(3)[-4,-1,-1,3]+((-4)/(7))[1,-4,4,3]\\\\i.e.\\\\proj_W(v)=[(68)/(3),(17)/(3),(17)/(3),-17]+[(-4)/(7),(16)/(7),(-16)/(7),(-12)/(7)]\\\\i.e.\\\\proj_W(v)=[(464)/(21),(167)/(21),(71)/(21),(-131)/(7)]

User Mgronber
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