Let R^{4 }have the Euclidean inner product. Expressw= (-1, 2, 6, 0) in the formw=w_{1 }+w_{2}, wherew_{1}is in the space W spanned byu_{1}= (-1, 0, 1, 2) andu_{2}= (0, 1, 0, 1), andw_{2 }is orthogonal to W

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- Jan 17th 2013, 05:30 PMrocker71Inner Product Space question
Let R

^{4 }have the Euclidean inner product. Express**w**= (-1, 2, 6, 0) in the form**w**=**w**_{1 }+**w**_{2}, where**w**_{1}is in the space W spanned by**u**_{1}= (-1, 0, 1, 2) and**u**_{2}= (0, 1, 0, 1), and**w**_{2 }is orthogonal to W - Jan 17th 2013, 06:03 PMBariothRe: Inner Product Space question
Hi the euclidean inner produit, I guess that is the basic scalar product as $\displaystyle <u,v> = u^tv$?

ps. I'm from french school so just wana make sure! - Jan 17th 2013, 06:25 PMDevenoRe: Inner Product Space question
do you know how to do an orthogonal projection onto W?

- Jan 18th 2013, 04:55 AMrocker71Re: Inner Product Space question
I honestly have no idea how to approach this question.

- Jan 18th 2013, 06:41 AMBariothRe: Inner Product Space question
every vector in W are of the form $\displaystyle a_1u_1+a_2u_2=(-a_1, 0 , a_1,2a_1)+(0,a_2,0,a_2)=(-a_1,a_2,a_1,[2a_1+a_2])=m$

where $\displaystyle a_1$ and $\displaystyle a_2$ are in $\displaystyle R^1$

since dimW1+dimW2=4

we know that dimW2 = 2. There are 2 vector x1 and x2 as <m,x1>=<m,x2> =0, and x1 and x2 are linearly independant.

let x1=(b,c,d,e)

$\displaystyle <m,x1>=-a_1b+ca_2+da_1+e(2a_1+a_2)=a_1(d-b+2e)+a_2(c+e)$ so we can take x1=(1,0,1,0)

doing the same for x2 one that would work is x2=(0,1,2,-1)

there we've our 2 basic for w2 and w1 we could now try to find out to write the vector w given. - Jan 18th 2013, 03:14 PMDevenoRe: Inner Product Space question
what we want is a basis for the orthogonal complement of W, W

^{⊥}(we're already given the basis for W).

since every vector in W^{⊥}is orthogonal to all of W, this means:

(-1,0,1,2).(x,y,z,w) = 0

(0,1,0,1).(x,y,z,w) = 0

for any (x,y,z,w) in W^{⊥}.

from the second equation, we get: y+w = 0.

from the first, we get: -x+z+2w = 0.

since dim(W^{⊥}) = 2, we will have "two free parameters". we know that:

y = -w

x = z + 2w.

so if we select z and w, this will determine x and y.

so let's try z = 1, w = 0 for our first choice. this leads to (1,0,1,0). next, try z = 0, w = 1, this leads to (2,-1,0,1).

this gives us the basis {(-1,0,1,2),(0,1,0,1),(1,0,1,0),(2,-1,0,1)} for R^{4}.

since this IS a basis, we can write:

(-1,2,6,0) = a(-1,0,1,2) + b(0,1,0,1) + c(1,0,1,0) + d(2,-1,0,1), for some UNIQUE real numbers a,b,c,d.

that is:

(-1,2,6,0) = (-a+c+2d,b-d,a+c,2a+b+d)

or:

-a+c+2d = -1

b-d = 2

a+c = 6

2a+b+d = 0

we can solve this equation for a,b,c,d like we do any system of linear equations (form an augmented matrix, row-reduce, or any other method). since i'm lazy, and i hate row reduction with a passion, i will do this, instead:

b = d+2

a = 6-c

(c-6)+c+2d = -1

2(6-c)+d+2+d = 0

2c+2d = 5

-2c+2d = -14

4d = -9, so:

d = -9/4

c = 19/4

b = -1/4

a = 5/4

substituting back in, we have:

(-1,2,6,0) = (5/4)(-1,0,1,2) - (1/4)(0,1,0,1) + (19/4)(1,0,1,0) - (9/4)(2,-1,0,1), which you can verify.

thus w_{1}= (5/4)(-1,0,1,2) - (1/4)(0,1,0,1) = (-5/4,-1/4,5/4,9/4) <--this is in W,

and w_{2}= (19/4)(1,0,1,0) - (9/4)(2,-1,0,1) = (1/4,9/4,19/4,-9/4),

furthermore, w_{2}is orthogonal to W:

(-1,0,1,2).(1/4,9/4,19/4,-9/4) = -1/4 + 19/4 - 18/4 = 0

(0,1,0,1).(1/4,9/4,19/4,-9/4) = 9/4 - 9/4 = 0

since it is orthogonal to the basis vectors of W, and thus any linear combination of them (including w_{1}).