0
$\begingroup$

Let $f: \mathbb{R}^{3} \rightarrow \mathbb{R}^{2}, f(x)=Ax$ where $$A = \begin{pmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{pmatrix}$$

On the vector space $\mathbb{R}_{3}[x]$ of the real polynomials $p$ of grade $n_{p} \leq 3$ we have derivated function $f: \mathbb{R}_{3}[x] \rightarrow \mathbb{R}_{3}[x], f(p) = p'$ and the composition is $g = f \circ f$.

Calculate the dimensions and basis of the kernel $\text{Ker(g)}$ and the image $\text{Im(g)}$

I think to do this we need to derivate $f$ first? So we have

$$f(x) = Ax$$

$$f'(x) = A = \begin{pmatrix} 1 & 2 & 3\\ 4 & 5 & 6 \end{pmatrix}$$

But then I have troubles getting the kernel because there is no variable left because we derivated..?

Because I cannot write that matrix $A$ equals zero-vector. Or I don't have to derivate here at all? :s

  • 0
    In the question, you defined two different functions named $f$, one mapping $\mathbb R^3$ into $\mathbb R^2$, and another mapping the vector space of polynomials over $\mathbb R$ of deegree $\leq 3$ into itself, that takes polynomial $p \in \mathbb R_3[x]$ into it's derivative $p'$. When you say that $g=f \circ f$, do you have in mind the first or the second? – 2017-01-23
  • 0
    @Zoran Loncarevic Oh sorry, the second is meant. – 2017-01-23
  • 0
    In that case, note that first function is defined, but not used anywhere else in the question. – 2017-01-23
  • 0
    Despite the title you gave it, I don't think this exercise has anything to do with differentiating a linear function, more with the fact that differentiating **is** a linear function. – 2017-01-23

3 Answers 3

1

The first function $f: \mathbb R^3 \to \mathbb R^2$ has nothing to do with your question and we can disregard it.

You defined function $f: \mathbb R_3[x] \to \mathbb R_3[x]$ by $$f(p)=p', \quad p \in \mathbb R_3[x]$$ that is $f$ takes every polynomial $ax^3+bx^2+cx+d$ into its derivative $3ax^2+2bx+c$ (note that this is a linear function). Then $g=f \circ f$ is another linear function, taking every polynomial $ax^3+bx^2+cx+d$ into it's second derivative $6ax+2b$.

Since, by definition $$\begin{align} \ker(g) &=\{ p \in \mathbb R_3[x] \mid g(p)=0 \} \\ \operatorname{Img}(g) &=\{ g(p) \mid p \in \mathbb R_3[x] \} \end{align}$$ from the above, it is easy to see that $\ker(g)= \operatorname{Img}(g)=\mathbb R_1[x]$.

1

On the vector space $\mathbb{R}_{3}[x]$ of the real polynomials $p$ of grade $n_{p} \leq 3$ we have derivated function $f: \mathbb{R}_{3}[x] \rightarrow \mathbb{R}_{3}[x], f(p) = p'$ and the composition is $g = f \circ f$.

Calculate the dimensions and basis of the kernel $\text{Ker(g)}$ and the image $\text{Im(g)}$

We have $$g[p(x)]=(f\circ f)[p(x)]=f[f[p(x)]]=f[p'(x)]=p''(x)$$ Then, $$\ker g=\{p(x)\in\mathbb{R}_3[x]:p''(x)=0\}=\{a+bx:a,b\in\mathbb{R}\}.$$ So, $\ker g=\text{Span }\{1,x\} $ and $\{1,x\}$ linearly indepedent, that is $B_{\ker g}=\{1,x\}.$

Also, easily proved: $B_{\text{Im } g}=\ldots=\{x^2,x^3\}.$

0

I'm not really sure about why do you have two different definitions for $f$. If we take the last one an canonical base for $\mathbb{R}_3[x]$, the represent matrix of $f$ is $$A=\left[\begin{array}{ccc} 0 & 0 & 0 \\ 2 & 0 & 0 \\ 0 & 1 & 0\end{array}\right]$$ So the represent matriz of g is: $$A=\left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 2 & 0 & 0\end{array}\right]$$ Then $Im(g)=\langle1\rangle$, and $Ker(g)=\langle1,x\rangle$, so $dim(Im(g))=1, dim(Ker(g))=2$

  • 0
    I don't think $A$ represents a matrix of $f$ with respect to any basis of $\mathbb R_3[x]$; note that $\mathbb R_3[x]$ is 4 dimensional vector space. – 2017-01-23