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


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.

  • $\begingroup$ 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? $\endgroup$ – Zhoe Dec 16 '13 at 5:31
  • 1
    $\begingroup$ 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. $\endgroup$ – JohnD Dec 16 '13 at 6:18
  • $\begingroup$ No problem, just checking that I had the correct train of thought. Thanks again! $\endgroup$ – Zhoe Dec 16 '13 at 6:23
  • $\begingroup$ 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. $\endgroup$ – Zhoe Dec 16 '13 at 17:19

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.