Basis of eigenvectors of a linear transformation Let $\mathbb R_n[x]$ the vector space of polynomials with degree less or equal $n$ and we consider the linear transformation $f$ defined by
$$\forall P\in \mathbb R_n[x]\quad  f(P)=(x^2-1)P''+2xP'$$
I proved that $f$ has the spectrum
$$\mathrm{sp}(f)=\{k(k+1),\ k=0,\ldots,n\}$$
I'm stuck in this question: Prove that there's a unique basis $(P_0,\ldots,P_n)$ of $\mathbb R_n[x]$ such that:
$$\forall k=0,\ldots,n\quad P_k \ \text{is a monic polynomial with degree }\ k\ \text{which's an eigenvector of }\ f $$
Any help would be appreciated.
 A: Use induction.  Note that the operator is independent of $n$, so you can recycle your previous polynomials.
A: You can prove the assertion without uniqueness by completing the following sketch.
Note that for $n=0$, we need a constant $P_0 \equiv c \neq 0$ for which $f(c) = (x^2-1)c'' - 2xc' = \lambda c$. This is true for $\lambda = 0$ and $c = 1$ (or any $c$, for that matter). Hence we can set $P_0 = 1$, the unique monic polynomial of degree $0$ and the basis for $\Bbb R_0[x]$ is of course just $\{P_0\}$.
Assume that there is a unique basis $\{P_0,\ldots,P_n\}$ for $\Bbb R_n[x]$ for which $P_k$ is a monic eigenvector of $f$ of degree $k$ for $k=0,\ldots,n$. If we can extend this to a basis of $\Bbb R_{n+1}[x]$, then the result follows by induction. So we want $P_{n+1}(x) = \sum_{k=0}^{n+1}a_kx^k$ such that $f(P_{n+1}) = \lambda P_{n+1}$. This last equation says that
$$
(x^2-1)\sum_{k=2}^{n+1}k(k-1)a_kx^{k-2} + 2x\sum_{k=1}^{n+1}ka_kx^{k-1} = \lambda\sum_{k=0}^{n+1} a_k x^k.
$$
Find $P_{n+1}$ by rewriting the left hand side of this equation in the form $\sum_{k=0}^{n+1} c_kx^k$, concluding that $c_k = \lambda a_k$ for each $k$ and solving these equations for $a_k$. Requiring that $a_k = 1$ gives you only one option for the remaining coefficients.
(Edited to take into account that the basis is supposed to consist of monic polynomials.)
