Question

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]

Answer 1

Answer:

Hence, we have:

$proj_W(v)=[\dfrac{464}{21},\dfrac{167}{21},\dfrac{71}{21},\dfrac{-131}{7}]$

Step-by-step 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:

$proj_W(v)=(\dfrac{v\cdot w}{w\cdot w})w+(\dfrac{v\cdot w'}{w'\cdot w'})w'$

Here we have:

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

Now,

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

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

and

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

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

Hence, we have:

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