# dual homomorphism of the derivative

Let $\mathcal{P}_{n}$ denote the vector space  of real polynomials of degree $n$ or less, and let $\mathrm{D}_{n}:\mathcal{P}_{n}\rightarrow\mathcal{P}_{n-1}$ denote the ordinary derivative. Linear forms  on $\mathcal{P}_{n}$ can be given in terms of evaluations, and so we introduce the following notation. For every scalar $k\in\mathbb{R}$, let $\mathrm{Ev}^{(n)}_{k}\in(\mathcal{P}_{n})^{*}$ denote the evaluation functional    $\mathrm{Ev}^{(n)}_{k}:p\mapsto p(k),\quad p\in\mathcal{P}_{n}.$

Note: the degree superscript matters! For example:

 $\mathrm{Ev}^{(1)}_{2}=2\,\mathrm{Ev}^{(1)}_{1}-\mathrm{Ev}^{(1)}_{0},$

whereas $\mathrm{Ev}^{(2)}_{0},\mathrm{Ev}^{(2)}_{1},\mathrm{Ev}^{(2)}_{2}$ are linearly independent  . Let us consider the dual homomorphism $\mathrm{D}_{2}^{*}$, i.e. the adjoint  of $\mathrm{D}_{2}$. We have the following relations   :

 $\begin{array}[]{rrrr}\mathrm{D}_{2}^{*}\left(\mathrm{Ev}^{(1)}_{0}\right)=&-% \frac{3}{2}\,\mathrm{Ev}^{(2)}_{0}&+2\,\mathrm{Ev}^{(2)}_{1}&-\frac{1}{2}\,% \mathrm{Ev}^{(2)}_{2},\\ \mathrm{D}_{2}^{*}\left(\mathrm{Ev}^{(1)}_{1}\right)=&-\frac{1}{2}\,\mathrm{Ev% }^{(2)}_{0}&&+\frac{1}{2}\,\mathrm{Ev}^{(2)}_{2}.\end{array}$

In other words, taking $\mathrm{Ev}^{(1)}_{0},\mathrm{Ev}^{(1)}_{1}$ as the basis of $(\mathcal{P}_{1})^{*}$ and $\mathrm{Ev}^{(2)}_{0},\mathrm{Ev}^{(2)}_{1},\mathrm{Ev}^{(2)}_{2}$ as the basis of $(\mathcal{P}_{2})^{*}$, the matrix that represents $\mathrm{D}_{2}^{*}$ is just

 $\left(\begin{array}[]{rr}-\frac{3}{2}&-\frac{1}{2}\\ 2&0\\ -\frac{1}{2}&\frac{1}{2}\end{array}\right)$

Note the contravariant relationship between $\mathrm{D}_{2}$ and $\mathrm{D}_{2}^{*}$. 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 $\mathrm{D}_{2}^{*}$ has 2 columns and 3 rows precisely because $\mathrm{D}_{2}^{*}$ is a homomorphism          from a 2-dimensional vector space to a 3-dimensional vector space.

By contrast, $\mathrm{D}_{2}$ will be represented by a $2\times 3$ matrix. The dual basis  of $\mathcal{P}_{1}$ is

 $-x+1,\quad x$

and the dual basis of $\mathcal{P}_{2}$ is

 $\frac{1}{2}(x-1)(x-2),\quad x(2-x),\quad\frac{1}{2}x(x-1).$

Relative to these bases, $\mathrm{D}_{2}$ is represented by the transpose  of the matrix for $\mathrm{D}_{2}^{*}$, 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:

 $\begin{array}[]{lcrr}\mathrm{D}_{2}\left[\frac{1}{2}(x-1)(x-2)\right]&=&-\frac% {3}{2}\,(-x+1)&-\frac{1}{2}\,x\\ \mathrm{D}_{2}\left[x(2-x)\right]&=&2\,(-x+1)&+0\,x\\ \mathrm{D}_{2}\left[\frac{1}{2}x(x-1)\right]&=&-\frac{1}{2}\,(-x+1)&+\frac{1}{% 2}\,x\end{array}$
Title dual homomorphism of the derivative DualHomomorphismOfTheDerivative 2013-03-22 12:35:28 2013-03-22 12:35:28 rmilson (146) rmilson (146) 4 rmilson (146) Example msc 15A04 msc 15A72