# Linear operator on n-th degree polinomial vector space

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?

• Looks good. You could already ask yourself for what polynomials $p$ does $p(t-2)=0$ hold? Apr 23, 2017 at 19:13
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.
Otherwise, look at $S\left(P_j\right)$ where $P_j=\left(t+2\right)^j$ for $1\leqslant j\leqslant n-1$.