# Basis and Adjoint.

## 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?)

## 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

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## Answers and Replies

AKG
Science Advisor
Homework Helper
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.