PlanetMath (more info)
 Math for the people, by the people. Sponsor PlanetMath
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
Owner confidence rating: High Entry average rating: No information on entry rating
[parent] dual homomorphism of the derivative (Example)

Let $\cP{n}$ denote the vector space of real polynomials of degree $n$ or less, and let $\D{n}:\cP{n}\rightarrow \cP{n-1}$ denote the ordinary derivative. Linear forms on $\cP{n}$ can be given in terms of evaluations, and so we introduce the following notation. For every scalar $k\in\reals$ , let $\ev{n}_k\in (\cP{n})\dual$ denote the evaluation functional $$\ev{n}_k:p\mapsto p(k),\quad p\in \cP{n}.$$ Note: the degree superscript matters! For example: $$\ev1_2 = 2\, \ev1_1- \ev1_0,$$ whereas $\ev2_0, \ev2_1, \ev2_2$ are linearly independent. Let us consider the dual homomorphism $\D2\dual$ , i.e. the adjoint of $\D2$ . We have the following relations:

\begin{displaymath} \begin{array}{rrrr} \mathrm{D}_{2}^*\left(\mathrm{Ev}^{(1)}_... ...m{Ev}^{(2)}_0 &&+\frac{1}{2}\, \mathrm{Ev}^{(2)}_2. \end{array}\end{displaymath}
In other words, taking $\ev1_0, \ev1_1$ as the basis of $(\cP1)\dual$ and $\ev2_0, \ev2_1, \ev2_2$ as the basis of $(\cP2)\dual$ , the matrix that represents $\D2\dual$ is just

\begin{displaymath} \left( \begin{array}{rr} -\frac{3}{2} & - \frac{1}{2} \ 2 & 0 \ - \frac{1}{2} & \frac{1}{2} \end{array}\right) \end{displaymath}
Note the contravariant relationship between $\D2$ and $\D2\dual$ . The former turns second degree polynomials into first degree polynomials, where as the latter turns first degree evaluations into second degree evaluations. The matrix of $\D2\dual$ has 2 columns and 3 rows precisely because $\D2\dual$ is a homomorphism from a 2-dimensional vector space to a 3-dimensional vector space.

By contrast, $\D2$ will be represented by a $2\times 3$ matrix. The dual basis of $\cP1$ is $$-x+1,\quad x$$ and the dual basis of $\cP2$ is $$\frac{1}{2} (x-1)(x-2),\quad x(2-x),\quad \frac{1}{2} x(x-1).$$ Relative to these bases, $\D2$ is represented by the transpose of the matrix for $\D2\dual$ , namely $$ \begin{pmatrix} -\frac{3}{2} & 2 & - \frac{1}{2} \\ - \frac{1}{2} & 0 & \frac{1}{2} \end{pmatrix} $$ This corresponds to the following three relations:

\begin{displaymath} \begin{array}{lcrr} \mathrm{D}_{2}\left[\frac{1}{2} (x-1)(x-... ...right]&=& - \frac{1}{2}\,(-x+1) & + \frac{1}{2}\, x \end{array}\end{displaymath}




"dual homomorphism of the derivative" is owned by rmilson.
(view preamble | get metadata)

View style:


This object's parent.
Log in to rate this entry.
(view current ratings)

Cross-references: transpose, bases, dual basis, homomorphism, rows, columns, represents, matrix, basis, relations, dual homomorphism, linearly independent, superscript, functional, scalar, terms, linear forms, derivative, degree, polynomials, real, vector space

This is version 1 of dual homomorphism of the derivative, born on 2002-04-16.
Object id is 2842, canonical name is DualHomomorphismOfTheDerivative.
Accessed 2085 times total.

Classification:
AMS MSC15A04 (Linear and multilinear algebra; matrix theory :: Linear transformations, semilinear transformations)
 15A72 (Linear and multilinear algebra; matrix theory :: Vector and tensor algebra, theory of invariants)

Pending Errata and Addenda
None.
Discussion
Style: Expand: Order:
forum policy

No messages.

Interact
post | correct | update request | add example | add (any)