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There is an isomorphism between

$P_n(x) = \{p(x) : p(x) = a_0 + a_1x + a_2x^2 +\ldots+ a_nx^n,\ \forall a_i \in \Bbb R\}$

and $\Bbb R^{n+1}$, in the sense that $a_0 + a_1x + a_2x^2 + ... + a_nx^n \in P_n(x)$

may be viewed as $(a_0, a_1, a_2, \ldots , a_n) \in \Bbb R^{n+1}$.

A linear transformation $D : P_n(x) \to P_{n-1}(x)$ such that

$$D[p(x)] = \dfrac{d}{dx}(p(x))$$

a. What is the kernel of $D$;

b. What is the image space of $D$;

c. Define $D$ as a matrix product equation (In the form $A \mathbf v = \mathbf w$). Be sure to specify the domain and the codomain.

I do not particularly understand these concepts, but from what I've read, I've gathered:

The image consists of all the values the function takes in its codomain.

image $D =$ {$f(x) : x \in P_{n-1}(x)$}

The kernel is the set of all zeros of the transformation ie solutions of the equations $A \mathbf v = \mathbf w$ where $\mathbf w = \mathbf 0$

Not sure where to go from here?

Attempted some more.

ker($D$) $=$ {$p(x) \in P_n(x) | D(p(x)) = 0$}

so $p(x)$ is such that

$\dfrac{d}{dx}(p(x)) = 0$

$\dfrac{d}{dx}[a_0 + a_1x + a_2x^2 +...+ a_nx^n] = 0$

$0 + a_1 + 2a_2x +...+ na_nx^{n-1} = 0$

so ker($D$) $=$ {$a_0 + a_1x + a_2x^2 +...+ a_nx^n | 0 + a_1 + 2a_2x +...+ na_nx^{n-1} = 0$}

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Zhoe
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2 Answers2

2

I'll illustrate these concepts for $n=2$ and let you generalize. Going back to the concept of finding the matrix for a linear transformation, let's first write the matrix of your operator $D$ (with respect to the standard basis on $P_n$), which we will denote by $[D]$: $$ D(a+bt+ct^2)={d\over dt}(a+bt+ct^2)=b+2ct, $$ so $$ \begin{bmatrix} a\\b\\c\end{bmatrix}\overset{D}{\longmapsto}\begin{bmatrix} b\\2c\\0\end{bmatrix}, \tag{$*$} $$ and therefore $$ [D]=\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 2\\ 0 & 0 & 0\end{bmatrix} $$ since $$ D\left(\begin{bmatrix} a\\b\\c\end{bmatrix}\right)=\begin{bmatrix} 0 & 1 & 0\\ 0 & 0 & 2\\ 0 & 0 & 0\end{bmatrix}\begin{bmatrix} a\\b\\c\end{bmatrix}=\begin{bmatrix} b\\2c\\0\end{bmatrix}. $$

So, now to your questions.

(1) Since $\text{ker}(D)=\{p\in P_2:D(p)=0\}$, we only need to consider $[D]\mathbf{v}=\mathbf{0}$, which says $b=0$, $c=0$, and $a$ is free. Hence, $$ \text{ker}(D)=\left\{\begin{bmatrix} a\\ 0\\ 0\end{bmatrix}:a\in\mathbb{R}\right\}=\{a+0\cdot t+0\cdot t^2:a\in\mathbb{R}\}. $$

(2) The image space (range space) of $D$ is the set of all possible "outputs" of $D$, and revealed by $(*)$ above: $$ \text{range}(D)=\left\{\begin{bmatrix} b\\ 2c\\ 0\end{bmatrix}:b,c\in\mathbb{R}\right\}=\{b+2c\cdot t+0\cdot t^2:b,c\in\mathbb{R}\}. $$

(3) This was answered above by finding $[D]$. Note that $[D]$ is $3\times 3$ since $D$ was a mapping from $P_2$ to $P_2$ which is a three dimensional space.

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JohnD
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  • A few questions, not sure if they'll make any sense but..The question itself didn't specify that it was a polynomial of degree two, but you only went up the squared term, is that acceptable? Will they yield the same results? And $a$ is free because in $[D]$ it doesn't have a leading one? – Zhoe Dec 16 '13 at 05:31
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    Oops for some reason I read it as $n=2$, maybe because of your other question that I linked. See if you can generalize this to $P_n$, where $[D]$ will be $(n+1)\times (n+1)$. Yes on $a$ being free. – JohnD Dec 16 '13 at 06:18
  • No problem, just checking that I had the correct train of thought. Thanks again! – Zhoe Dec 16 '13 at 06:23
  • For $n=2$, this was acceptable. Ker$(D) =\lbrace p(x) :p(x) = a_0, $ where $a_0 \in R \rbrace$, $\space$ Img$(D)=P_{n-1}(x)$, $\space$ and $D: R^{n+1}\to R^n$ were the solutions my teacher gave. – Zhoe Dec 16 '13 at 17:19
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a. Which polynomials in $P_n$ have derivative equal to zero? (Take the derivative of a general polynomial and see what you need the coefficients to be)

b. What polynomials in $P_{n-1}$ can you get after taking a derivative of a polynomial in $P_n$? (Take the derivative of a general polynomial and see what coefficients you could end up with)

c. Compute $d/dx[a_0+a_1x+...+a_nx^n]$. This is an element of $P_{n-1}$. Which element of $R^n$ does it correspond to? Find the matrix that takes the element of $R^{n+1}$ you started with before taking d/dx, and returns the element of $R^n$ that you found above.

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