Question

Find the projection of the vector v onto the subspace S.

S = span (0,1,1), (1,1,0),    v = (3,4,2)

proj_s v = ______

Answer 1

Given that ,

$S=Span \{ (0,1,1),(1,1,0)\}$ and $v=(3,4,2)$

let $u_1=(0,1,1) \ and \ u_2=(1,1,0)$

Then $S=Span\{u_1,u_2\}$ .

Since $u_1$ and $u_2$ are not orthogonal ,first apply the Gram-Schmidt algorithm to find an orthogonal basis for $S$ .

Set $w_1=u_1=(0,1,1)$

Then find $w_2=u_2-\frac{<u_2,w_1>}{<w_1,w_1>}w_1$ $=(1,1,0)-\frac{1}{2}(0,1,1)$

$=(1,\frac{1}{2},-\frac{1}{2})$

Where $w_1\ and \ w_2$ are orthogonal basis of $U$ .

Now we have to calculate the Fourier coefficient of $v$ with respect to $u_i$ i,e,

$c_1=\frac{<v,w_1>}{<w_1,w_1>}=\frac{6}{2}=3$

$c_2=\frac{<v,w_2>}{<w_2,w_2>}=\frac{3+2-1}{1+\frac{1}{4}+\frac{1}{4}}=\frac{4}{(\frac{3}{2})}$ $=\frac{8}{3}$

Then projection $Proj(v,U)$ $=c_1w_1+c_2w_2$

$=3(0,1,1)+\frac{8}{3}(1,\frac{1}{2},-\frac{1}{2})$

$=(0,3,3)+(\frac{8}{3},\frac{4}{3},-\frac{4}{3})$

$=(\frac{8}{3},\frac{13}{3},\frac{5}{3})$