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

In summary, the conversation discusses applying the Gram-Schmidt orthogonalization procedure to find an orthonormal basis for the space P4([0, 1]) with respect to a specific inner product. The basis is used to find the adjoint of a linear map. The process involves finding the map of each basis vector with respect to the given map and using the results to compute the 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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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.

## 1. What is a Gram-Schmidt orthonormal basis?

A Gram-Schmidt orthonormal basis is a set of vectors that are both orthogonal (perpendicular) and normalized (unit length). This means that each vector in the set is perpendicular to all other vectors in the set, and each vector has a length of 1.

## 2. How is a Gram-Schmidt orthonormal basis calculated?

A Gram-Schmidt orthonormal basis is calculated by taking a set of linearly independent vectors and applying the Gram-Schmidt process, which involves finding the orthogonal projection of each vector onto the subspace spanned by the previous vectors in the set. The resulting vectors are then normalized to have a length of 1.

## 3. What is the significance of a Gram-Schmidt orthonormal basis?

A Gram-Schmidt orthonormal basis is significant because it allows us to express any vector in a given vector space as a unique linear combination of the basis vectors. This makes it easier to perform calculations and solve problems involving linear transformations.

## 4. What is the adjoint for a linear map in P4([0,1])?

The adjoint for a linear map in P4([0,1]) is a linear transformation from the space of polynomials of degree 4 or less on the interval [0,1] to its dual space. It is defined as the transpose of the matrix representation of the linear map with respect to a given basis.

## 5. How is the adjoint for a linear map in P4([0,1]) calculated?

The adjoint for a linear map in P4([0,1]) can be calculated by finding the matrix representation of the linear map with respect to a given basis, taking the transpose of this matrix, and then converting it back to a linear transformation. Alternatively, the adjoint can also be calculated by finding the inner product of the linear map with respect to a given basis and then taking the conjugate transpose of this inner product matrix.