I have the following here:
Let $f:\mathcal{P_3 \rightarrow P_3}$ be the linear transformation defined by $f(p(x))=xp'(x)$. Find its kernel and image. What are the dimensions of the kernel and image?
So $p(x)=ax^3+bx^2+cx+d$. So that means:
$x(p'(x))=0$
$x(3ax^2+2bx+c)=0$
But that means $x=0$ or $3ax^2+2bx+c=0$.
I'm not sure how to get the basis however.
For these vector to be LI, that would mean that $a,b,c$ all have to be $0$, so does that mean the kernel would just be $p(x)=d$ and the dimension would just be $1$?
I'm not sure about the image.
If I go ahead and transform the standard basis vectors, I get:
$f((1))=x(0)=0, f((x))=x(1)=x, f((x^2))=x(2x)=2x^2, f(x^3)=x(3x^2)=3x^3.$
However $0$ isn't LI so does that mean the image is $\{x,2x^2,3x^3\}$ and the dimension is 3?
Best Answer
If $p(x)\in\ker f$, $xp'(x)$ is the null polynomial. Therefore, $p'(x)$ is the null polynomial, which means that $p(x)$ is a constant polynomial. And it is clear that if $p(x)$ is a constant polynomial, then $f\bigl(p(x)\bigr)=0$. So, $\ker f$ is the set of all constant polynomials, and therefore it is $1$-dimensional; it has a basis which consists of the constant polynomial $1$. So, by the rank-nullity theorem, $\dim\operatorname{Im}(f)=3$.
On the other hand, as you wrote, $\{1,x,x^2,x^3\}$ is a basis of $\mathcal P_3$, and therefore $\{f(1),f(x),f(x^2),f(x^3)\}$ spans $\operatorname{Im}(f)$. But $f(1)=0$. This, together with the fact that $\dim\operatorname{Im}(f)=3$, tells you that $\{f(x),f(x^2),f(x^3)\}$ is a basis of $\operatorname{Im}(f)$. But $f(x)=x$, $f(x^2)=2x^2$ and $f(x^3)=3x^3$. So, $\operatorname{Im}(f)$ is the space spanned by $\{x,2x^2,3x^3\}$ which is the space of those $p(x)\in\mathcal P_3$ whose constant term is $0$.