# Find $\operatorname{Ker}(L)$ and $\operatorname{Im}(L)$

Let $P_3 (\mathbb{R})$ denote the vector space of real polynomials: $f=ax^2+bx+c$ of degree $\le 2$.

Where:

\begin{alignat}{2} L:P_3&(\mathbb{R}) &\longrightarrow &\, P_3(\mathbb{R}) \\ &f &\longmapsto &\, x\cdot f'-f \qquad \text{($f'$ is the derivative of $f$)} \end{alignat}

Find the kernel and image of $L$.

This is what I have so far:

I know that the elements $v \in P_3(\mathbb{R})$ where $x\cdot f'-f = 0$, which is equivalent with $ax^2-c=0$. This means that if $a = 0$ and $c=0$ or $ax^2=c$ then we get the zero polynomial and that would be the zero polynomial, which is want we want to find as the $\operatorname{Ker}(L)=0$

The image is identical to the column space, which means that the image is the polynomials of the form $ax^2-c$ for $a,c \in \mathbb{R}$ ?

Would this be the kernel and the image of $L$ and is there a more rigid/formal method to use?

• For the image: all good, although the $-$ sign is superfluous. For the kernel, you lost track. $ax^2-c=0$ for all $x$ means that $a=c=0$, which implies that $f(x)=bx$. Commented Feb 25, 2018 at 12:47
• So the kernel would be that for any value b, bx maps to 0? However, why is it for all x that $a = c = 0$ would it not be the case that $ax^2-c\neq0$ for some x? Commented Feb 25, 2018 at 15:21
• Let me put it this way. Let $f\in Ker L$. Then $xf'=f$. Therefore $f(0)=0\cdot f'(0)=0$. So $c=0$. Same for $a$. Commented Feb 25, 2018 at 15:30
• I see, it makes perfect sense then that f(x) = bx, which is the Ker(L), right? Commented Feb 25, 2018 at 15:47
• Ker L is the set of all polynomials of that form, right. Commented Feb 25, 2018 at 15:51

## 1 Answer

$\{1,x,x^2\}$ is a base of your space. Now let's call $\phi:P_3(\mathbb{R})\to P_3(\mathbb{R})$ with $\phi(f)=xf'-f$

Then $\phi(1)=x 0-1=-1=-1+0(x)+0(x^2)$

$\phi(x)=x1-x=0+0(x)+0(x^2)$

$\phi(x^2)=x(2x)-x^2=x^2=0+0(x)+1(x^2)$

So the matrix of $\phi$ in the above base is $\begin{pmatrix} -1 & 0 & 0 \\ 0 & 0 & 0\\ 0 & 0 & 1 \end{pmatrix}=A$

Now for the kernel you can solve $AX=0$ for $X\in \mathbb{R^{3\times 1}}$

If I'm right this gives $x=z=0\Rightarrow ker(A)=\langle(0 ,1, 0)^{t}\rangle \Rightarrow ker(\phi)=\langle0(1)+1(x)+0(x^2)\rangle=\langle x \rangle$

Can you do the same for $im(\phi)$?