Let $S:P_n \rightarrow P_n$ (where $P_n$ is a space of polinomials where $\deg(p)\leq n$) be defined as $$(Sp)(t)=\left(tp(t-2)\right)''\\$$ Determine the $\dim(\operatorname{Im} S)$, $ \dim(\ker S)$, a basis for the image and a basis for the kernel.

What I've tried so far: Obviously, all polynomials with $\deg=0$ are from the kernel (because of the double derivative). Also, all polynomials such that $p(t-2)=0$, but the basis for such a vector space would be $\{(t+2),(t+2)^2,...,(t+2)^n\}$, which leads to the conclusion that $\dim(\ker S)=n+1=\dim(P_n)$. Did I make any mistakes?

Thanks in advance.

  • $\begingroup$ Looks good. You could already ask yourself for what polynomials $p$ does $p(t-2)=0$ hold? $\endgroup$
    – flawr
    Apr 23, 2017 at 19:13

1 Answer 1


The polynomials such that $p(t-2)=0$ are indeed in the kernel, but $t$ is the variable. Therefore, such polynomials are identically $0$.

A start could be the following. First we determine the polynomials $P$ such that $\left(tP\left(t-2\right)\right)''=0$. Since the second derivative of a polynomial $Q$ is $0$ (identically) if and only if $Q(t)=at +b$ for some constants $a$ and $b$, this means that $tP\left(t-2\right)=at+b$. Evaluating at $t=0$, we get $b=0$ and finally that $P(t-2)=a$, so that $P$ is constant. Conversely, all constant polynomials are in the kernel.

You do not need to compute the image if you know the relationship of its dimension with that of the kernel.

Otherwise, look at $S\left(P_j\right)$ where $P_j=\left(t+2\right)^j$ for $1\leqslant j\leqslant n-1$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .