I have the following linear transformation: $D : V(\mathbb{R})\rightarrow V(\mathbb{R}) $ defined by:
$$ D:p(x)=ax^3 +bx^2 + cx + d \mapsto Dp(x) = 3ax^2 + 2bx + c$$
And I want to find the kernel and the image of the transformation. Regarding the kernel, I have done the following
$$\operatorname{ker}(D) = \{ Dp(x) = 0 \mid x \in V\} $$
$$ p(x)=ax^3 +bx^2 + cx + d = Dp(x) = 3ax^2 + 2bx + c = 0 $$
Which can be solved through the quadratic formula. Is the answer then:
$$\operatorname{ker}(D) = \left\{\frac{-b\pm\sqrt{b^2-3ac}}{3a} \right\}$$
And what about the image? I know that $\operatorname{Im}(D) = \{ Dp(x) \mid x \in V \}$, but how can I show it?
Best Answer
Your answer for the kernel is not correct. The kernel is going to be the polynomials $p$ sent to $0$ under $D$, i.e. those with $Dp=0$. You naively solved for the variable $x$, but that is about assigning a particular value to $x$. You need to find the polynomials with $Dp=0$, not solve a quadratic equation; you will arrive a set of polynomials, not a set of $x$ values that are the root of some equation.
To figure out the image, you need to figure out what $D$ is really sending polynomials to. Based on the previous hint, you should have a guess of what happens to a polynomial $p$ and in particular its degree; so you should be able to guess "the image is all polynomials of degree $N$" (for some number $N$ you should determine). Then you just need to show that each of those polynomials has a preimage under $D$, i.e. for each $q$ in the image construct a $p$ such that $Dp=q$.