Let $f:\Re_{3}[x] \rightarrow \Re_{4}[x]$
such as
$f(p(x))=xp(x)$ for any $x \in \Re_{3}[x]$
Determine if the next polynomials belong to the rank of $f$ and if they belong to the nullity of $f$.
$a) x^3$
I'm confused about this.
We have that $f(x)=xp(x)$ so
$f(x^2)=xx^2=x^3$? I don't quite get how to do it. From the solutions it says
$x^3 \in Im f$ and $x^3 \notin Nuc f$
But isn't $x^3=0 (=) x=0?$ How come its not in $Nuc f?$
(Note: I'm trying to be better on the site, so if I should edit anything or am not being clear, I accept criticism.)
Let $F: R_n[x] \rightarrow R_{n+1}[x]$
By definition a vector [polynomial] P belongs to the kernel if $F(P)=0$
Here we have $F(x^3)=x^4$ which is a nonzero polynomial, hence $x^3$ doesn't lie in the kernel.
If we want to check wether a vector [polynomial] $P$ belongs to the image we have to check if there is such a vector [polynomial] $W$ such that $F(W)=P$
Here: $F(W)=xW=x^3 \rightarrow W=x^2$ ; By definition $x^3$ lies in the image.