Finding the kernel and image of a linear transformation

Question

Let $P_{3}(\mathbb{R})$ be the vector space of real polynomials of degree 2 or less, given by: $f=aX^2+bX+c$

Let the function L be a linear transformation given by

$L:P_3(\mathbb{R}) \mapsto P_3(\mathbb{R})$,

$f \mapsto X\cdot f'-f$

(1) Find the kernel and the image of L.

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Here is what I've tried:

The basis of $P_{3}(\mathbb{R})=\{1,x,x^2\}$.

The image of L is then given by $span(\{L(1),L(x),L(x^2)\})$

With $L(1)=-1, L(x)=1-x,L(x^2)=2x-x^2=x(2-x)$

So the image is spanned by: $span(\{-1,1-x,x(2-x)\})$

Assuming I've understood and calculated the correct image of L, how do I go about finding the kernel?

Answer 1

The image is spanned by $L(1),L(x)$ and $L(x^2)$ as you said. When you compute you get $L(1)=-1,L(x)=0$ and $L(x^2)=x^2$. Hence the image is span$\{1,x^2\}$.

To compute the kernel, you find the roots of the equation $x\cdot f'=f$. Note that for $f=ax^2+bx+c$ you have $f'=2ax+b$ so $x\cdot f' = 2ax^2+bx$. Hence $f$ is in the kernel iff $2a=a$ and $c=0$, i.e. $a=c=0$. Hence the kernel is the subspace spanned by $x$.