Gram-Schmidt Orthonormal Basis & Adjoint for Linear Map in P4([0,1])

  • Thread starter Thread starter rad0786
  • Start date Start date
  • Tags Tags
    Basis
rad0786
Messages
187
Reaction score
0

Homework Statement



Apply the Gram-Schmidt orthogonalization procedure to the canonical basis 1, x, x^2, x^3, x^4 in order to find an orthonormal basis for the space P4([0, 1]) with respect to the inner product <p(x), q(x)> =int(0,1) p(x)q(x) dx

AND USE THIS BASIS TO FIND THE ADJOINT OF THE LINEAR MAP! A(ax^4 + bx^3 +cx^2 + dx + e) = cx^2 (=cx^2 ... no cx^3... latex acts funny?)




Homework Equations





The Attempt at a Solution



Finding the orthogonal basis using the Gram-Schmit algorithm is slimply plugging numbers into a formula, so that is straight forward.

My Basis is:

e1 =1

e2 = x-1/2

e3 = x^2+1/6-x

e4 = x^3-1/20+3/5*x-3/2*x^2

e5 = x^4+1/70-2/7*x+9/7*x^2-2*x^3


Could somebody please guide me as to how I would

USE THIS BASIS TO FIND THE ADJOINT OF THE LINEAR MAP A(ax^4 + bx^3 + cx^2 + dx + e) = cx^2 (=cx^2 ... no bx^2... latex acts funny?)

I honestly have no idea where to start?

thought a good place to start would be to find the map of e_i with respect to A,

A(e1) = 0
A(e2) = 0
A(e3) = x^2
A(e4) = 3/2*x^2
A(e5) = 9/7*x^2
 
Last edited:
Physics news on Phys.org
The adjoint A* of a map A is the linear transformation satisfying <Ax,y> = <x,A*y> for all x, y in your vector space. Since you have supposedly correctly found a basis which is orthnormal w.r.t. the given inner product, find A* should be straightforward.

Compute <A(ei), ej> for 1 < i, j < 5. This will give you <ei, A*(ej)> for each i, j. Since the basis is orthonormal, you get:

A*(ej) = <A(e1),ej>e1 + ... + <A(e5),ej>e5

You should be able to go from here.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

Similar threads

Replies
58
Views
4K
Replies
16
Views
3K
Replies
2
Views
1K
Replies
3
Views
2K
Replies
1
Views
2K
Replies
2
Views
1K
Replies
5
Views
2K
Replies
14
Views
4K
Back
Top