Prove that \(f_1(x)=1,f_2 (x)=x,f_3 (x)=((3x^2-1))/2\) are orthogonal over \((-1,1)\) using matrices
@OOPS
I answered. :) and I am sure you got it. hihihi
you have deleted it before reading it completely
i know onlhy linear algebra, never took abstract... so im not good with fields but i can read wiki for keywords if you show me the matrix method :)
It's not hard, friend, that from theoretical linear algebra, not abstract any polynomial is in the form a + bx + cx^2 + dx^3+.......
okay...
so that {1, x, x^2, x^3,......} is the basis of polynomial set Consider P_2, we have basis of P_2 is {1, x,x^2}
please wait, so basis is a set of linear independent vectors that span the entire vector space, right ?
it seems like you're defining a basis whose dimension is infinite or something
so that the first one is 1 + 0x + 0x^2 the second is 0+1x +0x^2 the third is -1/2 +0 x+ 3/2 x^2 and Yes for your question
P_2 is subset of P, and Wolg P is a vector space
from that I can construct the matrix as shown.
I see...
And it 's quite easy to prove they are independent vectors and then orthogonal.
this would be the matrix \[\large \left[\begin{array}\\ 1&0&0\\0&1&0\\-1/2&0&3/2 \end{array}\right]\] ?
yes
we need to show determinant is 0 for independence
what about orthogonality ?
*not 0
using inner product
If you show that rank of the matrix =3 it should solve the problem.
f1 dot f2 =0 and so on...
orthogonal \(AA^T =I\)
Sweet :3
this looks more elegant than the integrals xD
Yes, there are many ways, but I choose the simplest way for my simple brain
just a final noob question : we need to take inner product of pairs of \(column\) vectors right ?
I use row, not column
It's the same but always go with the easiest.
row space is orthogonal <==> column space is orthogonal ?
The reason I choose row is (1,0,0) is the first vector as f(1), so that it makes more sense than picking (1,0,1/2) where I don't know what it stands for.
and I need prove f(1) perpendicular to f(2) and f(3) . That 's it
If you take it as a 3*3 matrix yes, you can use columns or rows.
Oh I see... i think i get it :) thanks !
:)
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