Not quite sure how to start with this one



Any help would be appreciated

Sorry also this one lol

the inner product space is a generalization of the dot product into vector spaces, hopefully this makes it so you can figure it out yourself

Question 3:
Proof in <-- direction first.
If u and v are dependent, it means that v can be written as u = cv where is a scalar not equal to 0.

|<u,v>| = |<cv,v>|

= |c||<v,v>|
= |c| ||v||^2
= |c| ||v|| ||v||
= ||cv|| ||v||
= ||u|| ||v||

now proof --> direction
let |<u,v>| = ||u|| ||v||
if u or v = 0 then they are linearly dependent.( everything is dependent to 0)
therefore u,v =/ 0

let w = u - <u,v>/<v,v> · v
you will probably notice that the equation above is u - (projection u on v) which is orthogonal to v

solving for u
u = w + <u,v>/<v,v> · v

now sub it in

|<u,v>| = |<w + <u,v>/<v,v> · v , v>|

we get
|<u,v>| = ||w + <u,v>/<v,v> · v|| ||v||
from the fact that we let |<u,v>| = ||u|| ||v||

now square both sides
|<u,v>|^2 = ||w + <u,v>/<v,v> · v||^2 ||v||^2
|<u,v>|^2 = ||w||^2 + |<u,v>/<v,v>|^2 · ||v||^2 ||v||^2
|<u,v>|^2 = ||w||^2 + ||u||^2 ||v||^2

which mean ||w||^2 = 0
so w=0

therefore we get
u = 0 + <u,v>/<v,v> · v
this shows u,v are linearly dependent

Question 5: let p(x) = x-1 and q(x) = x^2
using the Gram-Schmidt Process

u = x-1
v = x^2 - projection q(x) on p(x)

projection q(x) on p(x) = <p(x), q(x)>/<p(x),p(x)>· p(x)
now compute the inner product defined by:
<p,q> = p(-1)q(-1) + p(0)q(0) + p(1)q(1)

projection q(x) on p(x) = -2/5 · p(x)

v = x^2 +2/5 · p(x)
v = x^2 + 2/5 (x-1)
v = x^2 +2/5x - 2/5

therefore your orthogonal complement to U is {x-1, x^2+2/5x-2/5}

Wow I will read over this in more detail when I finish working on the other questions but thank you so much!!

no problem. I know it can be difficult/confusing to read in the current format, so feel free to ask if you don't understand

|<u,v>|^2 = ||w||^2 + |<u,v>/<v,v>|^2 · ||v||^2 ||v||^2
|<u,v>|^2 = ||w||^2 + ||u||^2 ||v||^2

Can you explain how you went from the top line to the bottom line?

recall <v,v> = ||v||^2
so you have
|<u,v>/<v,v>|^2 · ||v||^2 ||v||^2
expanding |<u,v>/<v,v>|^2
you get |<u,v>|^2 / ||v||^4 · ||v||^2 ||v||^2

||v||^4 · ||v||^2 ||v||^2 cancel
which leaves you with
|<u,v>|^2

we let |<u,v>| = ||u|| ||v||
so then
|<u,v>|^2 = ||u||^2 ||v||^2

edit: i left the ||w||^2 out because its the same in both lines

This post was edited by cdexswzaq on Sep 30 2014 09:32pm

ooh thats clever okay thanks!

isnt this part wrong?

|<u,v>|^2 = ||w + <u,v>/<v,v> · v||^2 ||v||^2
|<u,v>|^2 = ||w||^2 + |<u,v>/<v,v>|^2 · ||v||^2 ||v||^2

arent you missing a few components?