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.

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?

  • $\begingroup$ $L(x)=x\cdot 1-x=0$ and $L(x^2)=x^2$. When you calculate you forgot to multiply $f'$ by $x$. $\endgroup$ – Levent Mar 7 '18 at 14:03

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 $x$.

  • $\begingroup$ I asked an instructor about it and was told that it is not $\{x\}$, so I thought I had misunderstood how to find the kernel. Maybe I've misunderstood how to find the image instead? In this example the kernel equals the subspace spanned by x, is it then correct to say that the kernel $=\{0\}$ $\endgroup$ – Sirmimer Mar 7 '18 at 14:48
  • $\begingroup$ Kernel is not $\{x\}$, it is the subspace spanned by $x$, maybe that's what your instructor meant. Of course it is NOT correct to say that the kernel is $\{0\}$ if you have already shown that it is spanned by $\{x\}$. $\endgroup$ – Levent Mar 7 '18 at 14:50
  • $\begingroup$ So the kernel is spanned by $\{x\}.$. But if I wish to write it in terms of elements, how would I go about that? Is it correctly understood that the kernel in this example is given by the all the $f$ such that $L(f)=0$ ? $\endgroup$ – Sirmimer Mar 7 '18 at 15:21
  • $\begingroup$ The definition of the kernel of a map $L$ is the set of all elements $f$ such that $L(f)=0$. If you want to write the elements of kernel, they are all of the form $c\cdot x$ for some $c\in\mathbb{R}$. $\endgroup$ – Levent Mar 7 '18 at 15:50

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