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Vector Spaces of Polynomials

Two examples are presented here to illustrate the computation of transformation matrices in the context of vector spaces of polynomials.

It is important to remember that a transformation matrix can be used to convert the coordinate vectors of objects in one vector space to the coordinate vectors of objects in another vector space. (Note: The two vector spaces can be the same vector space.)

It is also important to remember that an object in a vector space can be computed as a linear combination of the elements of a basis of that vector space from the object's coordinate vector relative to that basis.

Let $P_2$ be the vector space of polynomials of degree 2, with the ordered basis $\beta$:

$$$\beta =\left\{\begin{array}{ccc}{1}^{\phantom{0}}& x^{\phantom{1}}& x^2\end{array}\right\}$

Let $P_3$ be the vector space of polynomials of degree 3, with the ordered basis $\gamma$:

$$$\gamma =\left\{\begin{array}{cccc}{1}^{\phantom{0}}& x^{\phantom{1}}& x^2& x^3\end{array}\right\}$

Example 1

Let $T$ be a linear transformation from $P_3$ into $P_2$:

           $T:P_3\to P_2$   where $T\left(f\right)$$=$$f^{\prime }$

This means that the transformation $T$ maps a polynomial, by taking its derivative, in $P_3$ to a polynomial in $P_2$. We can compute the corresponding transformation matrix ${\left[T\right]}_{\gamma }^{\beta }$ as follows. First we compute the derivative of each basis vector in the basis $\gamma$ of $P_3$ as a linear combination of the elements of the basis $\beta$ of $P_2$:

$T\left(1^{\phantom{0}}\right)=0\ast 1^{\phantom{0}} + 0\ast x^{\phantom{1}} + 0\ast x^2$
$T\left(x^{\phantom{1}}\right)=1\ast 1^{\phantom{0}} + 0\ast x^{\phantom{1}} + 0\ast x^2$
$T\left(x^2\right)=0\ast 1^{\phantom{0}} + 2\ast x^{\phantom{1}} + 0\ast x^2$
$T\left(x^3\right)=0\ast 1^{\phantom{0}} + 0\ast x^{\phantom{1}} + 3\ast x^2$

Then we take the coefficients in each linear combination above and put them into the columns of the transformation matrix ${\left[T\right]}_{\gamma }^{\beta }$: we put the three coefficients in the first equation into the first column of the matrix; next, put the three coefficients in the second equation into the second column; then, put the three coefficinets in the third equation into the third column; finally, put the three coefficients in the last equation into the fourth column of the matrix ${\left[T\right]}_{\gamma }^{\beta }$:

${\left[T\right]}_{\gamma }^{\beta }=\left(\begin{array}{cccc}0& 1& 0& 0\\ 0& 0& 2& 0\\ 0& 0& 0& 3\end{array}\right)$

This transformation matrix converts the coordinate vector relative to $\gamma$ of an object (a poynomial) in $P_3$ to the coordinate vector relative to $\beta$ of an object (a polynomial) in $P_2$. Using this fact, we now compute the derivative $h^{\prime }\left(x\right)$ of a polynomial $h\left(x\right)$:

           $h\left(x\right)=5+7x^{\phantom{1}}+3x^2+x^3$

We first write down the coordinate vector of the polynomial $h\left(x\right)$:

${\left[h\left(x\right)\right]}_{\gamma }=\left(\begin{array}{c}5\\ 7\\ 3\\ 1\end{array}\right)$

Then we multiply it with the transformation matrix ${\left[T\right]}_{\gamma }^{\beta }$:

           ${\left[T\right]}_{\gamma }^{\beta }{\left[h\left(x\right)\right]}_{\gamma }=\left(\begin{array}{cccc}0& 1& 0& 0\\ 0& 0& 2& 0\\ 0& 0& 0& 3\end{array}\right)\left(\begin{array}{c}5\\ 7\\ 3\\ 1\end{array}\right)=\left(\begin{array}{c}7\\ 6\\ 3\end{array}\right)$

to get the coordinate vector of $h^{\prime }\left(x\right)$:

           ${\left[h^{\prime }\left(x\right)\right]}_{\beta }=\left(\begin{array}{c}7\\ 6\\ 3\end{array}\right)$

From the coordinate vector ${\left[h^{\prime }\left(x\right)\right]}_{\beta }$, we compute $h^{\prime }\left(x\right)$ itself as a linear combination of the elements of the basis $\beta$:

           $T\left(h\right)\left(x\right)=$$h^{\prime }\left(x\right)=7+6x^{\phantom{1}}+3x^2$

by using the coordinate vector ${\left[h^{\prime }\left(x\right)\right]}_{\beta }$ which contains the coefficients of the linear combination required to construct $T\left(h\right)\left(x\right)$ out of the elements of the basis $\beta$ of $P_2$.

Example 2

Now let $U$ be a linear transformation from $P_2$ into $P_3$:

           $U:P_2\to P_3$   where $U\left(f\right)\left(x\right)$$=$${\int }_{-\infty }^{x}fdt$

The transformation $U$ computes the integral of a polynomial in $P_2$. To find the transformation matrix ${\left[U\right]}_{\beta }^{\gamma }$, we start by transforming each basis vector in the basis $\beta$ of $P_2$ to the corresponding object (a polynomial) in $P_3$ by taking the linear combination of the elements of the basis $\gamma$ of $P_3$:

$U\left(1^{\phantom{0}}\right)=0\ast 1^{\phantom{0}} + 1\ast x^{\phantom{1}} + 0\ast x^2 + 0\ast x^3$
$U\left(x^{\phantom{1}}\right)=0\ast 1^{\phantom{0}} + 0\ast x^{\phantom{1}} + \frac{1}{2}\ast x^2 + 0\ast x^3$
$U\left(x^2\right)=0\ast 1^{\phantom{0}} + 0\ast x^{\phantom{1}} + 0\ast x^2 + \frac{1}{3}\ast x^3$

Then, we fill in the columns of the transformation matrix ${\left[U\right]}_{\beta }^{\gamma }$ by using the same method as we have used for the transformation matrix in Example 1:

${\left[U\right]}_{\beta }^{\gamma }=\left(\begin{array}{ccc}0& 0& 0\\ 1& 0& 0\\ 0& \frac{1}{2}& 0\\ 0& 0& \frac{1}{3}\end{array}\right)$

The transformation matrix ${\left[U\right]}_{\beta }^{\gamma }$ converts the coordinate vectors relative to $\beta$ of objects (polynomials) in $P_2$ to the coordinate vectors relative to $\gamma$ of objects (polynomials) in $P_3$. That is, this transformation matrix enables us to compute the coordinate vector of an object in $P_3$ from the coordinate vector of an object in $P_2$. Armed with this knowledge, we now compute the integral $U\left(g\right)\left(x\right)$ of a polynomial $g\left(x\right)$ in $P_2$:

           $g\left(x\right)=5+2x^{\phantom{1}}+3x^2$

The coordinate vector of $g\left(x\right)$ is ${\left[g\left(x\right)\right]}_{\beta }$:

           ${\left[g\left(x\right)\right]}_{\beta }=\left(\begin{array}{c}5\\ 2\\ 3\end{array}\right)$

Multiplying the coordinate vector ${\left[g\left(x\right)\right]}_{\beta }$ by the transformation matrix ${\left[U\right]}_{\beta }^{\gamma }$ produces the coordinate vector of the result, ${\left[U\left(g\right)\left(x\right)\right]}_{\gamma }$:

           ${\left[U\left(g\right)\left(x\right)\right]}_{\gamma }={\left[U\right]}_{\beta }^{\gamma }{\left[g\left(x\right)\right]}_{\beta }$

           ${\left[U\left(g\right)\left(x\right)\right]}_{\gamma }=\left(\begin{array}{ccc}0& 0& 0\\ 1& 0& 0\\ 0& \frac{1}{2}& 0\\ 0& 0& \frac{1}{3}\end{array}\right)\left(\begin{array}{c}5\\ 2\\ 3\end{array}\right)=\left(\begin{array}{c}0\\ 5\\ 1\\ 1\end{array}\right)$

Now we can compute $U\left(g\right)\left(x\right)$ as a linear combination of the elements of the basis $\gamma$:

           $U\left(g\right)\left(x\right)$$=$${\int }_{-\infty }^{x}gdt$$=$$01^{\phantom{0}}+5x^{\phantom{1}}+1x^2+1x^3$$=$$5x^{\phantom{1}}+x^2+x^3$

by using the coordinate vector ${\left[U\left(g\right)\left(x\right)\right]}_{\gamma }$ which contains the coefficients of the linear combination required to construct $U\left(g\right)\left(x\right)$ out of the elements of the basis $\gamma$ of $P_3$.

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