<s>[INST] $ L(P)= (1+X^2)P''(X)-2XP'(X)$ 
*

*$E=\mathbb{R}[X]$

*$L$ is an endomorphism of $E$ and $L(P)= (1+X^2)P''(X)-2XP'(X)$
What are the possible eigenvalues of $L$ and the dimension of the eigenspace ?

Let's $p(x)=\sum_{k=0}^{n} a_k X^k$ be an eigenvector associated to the eigenvalue $\lambda$.
$a_{n+1} = a_{n+2} = .. =0$
$
\begin{cases}
p(x) &= \sum_{k=0}^{+\infty} a_k x^k \\
p'(x) &= \sum_{k=1}^{+\infty} a_k x^{k-1} k \\
p''(x) &= \sum_{k=2}^{+\infty} a_k x^{k-2} k (k-1) \\
\end{cases}
$
$
\begin{cases}
-2x p'(x) &= \sum_{k=1}^{+\infty} -2 a_k x^k \\
x^2 p''(x) &= \sum_{k=2}  a_k  k (k-1)  x^k \\
p''(x) & =  \sum_{k=0}^{+\infty} a_{k+2} x^{k} (k+2) (k+1) \\ 
\end{cases}
$
$
\begin{align*}
(1+x^2)P''(x) -2xP'(x) - x P'(x) &= \lambda P(x) \\
\sum_{k=1}^{+\infty} -2 k a_k x^k +\sum_{k=2}  a_k  k (k-1)  x^k + \sum_{k=0}^{+\infty} a_{k+2} x^{k} (k+2) (k+1) &= \lambda \sum_{k=0}^{+\infty} a_k x^k \\
\end{align*}
$
$
\begin{cases}
-2 k a_k  + a_k  k (k-1)   +  a_{k+2}  (k+2) (k+1) &= \lambda a_k    ~~ \text{for} ~ k \geq 2  \\
a_2 \times 2  &= \lambda a_0 \\
-2 a_1 +a_3 \times 6  &= \lambda a_1 \\
\end{cases}
$
\begin{cases}
 (k^2-3k-\lambda)a_k &=-(k+2)(k+1)a_{k+2}\tag{1} \\
a_2 \times 2  &= \lambda a_0 \\
-2 a_1 +a_3 \times 6  &= \lambda a_1 \\
\end{cases}
$(1)$ is available only for $k \geq 2$.
$
\begin{align*}
\psi : \mathbb{N} &\to \mathbb{N}\\
      n & \mapsto n^2-3n \\
\end{align*}
$
$\psi(n)-\psi(m) = (n-m)(n+m-3)$
 [/INST] Let $\lambda$ be an eigenvalue of $L$ and $p(x)=\sum_{k=0}^n a_kx^k=\sum_{k=0}^\infty a_kx^k$ with $a_n=1$ and $0=a_{n+1}=a_{n+2}=\cdots$ be a corresponding eigenvector. From $(1+x^2)p''(x)-2xp'(x)=\lambda p(x)$, we obtain the recurrence relation
$$
\left[(k+2)(k+1)a_{k+2}+k(k-1)a_k\right]-2ka_k=\lambda a_k
$$
or equivalently
$$
(k^2-3k-\lambda)a_k=-(k+2)(k+1)a_{k+2}\tag{1}
$$
for each $k\ge0$. It follows that $\lambda=n^2-3n$.
Since $(n^2-3n)-(m^2-3m)=(n-m)(n+m-3)$, if $\lambda=n^2-3n$ for some $n>3$, then $n$ is uniquely determined and the LHS of $(1)$ is nonzero for every $0\le k<n$. Thus $a_{n-1},a_{n-2},\ldots,a_0$ can be uniquely and recursively determined from $(1)$ using the boundary conditions that $a_{n+1}=0$ and $a_n=1$. Hence the eigenspace for $\lambda$ exists and is one-dimensional.
When $\lambda=n(n-3)$ for some $0\le n\le3$, complications arise because $\lambda$ has two factorisations $n(n-3)$ and $m(m-3)$, where $m=3-n$.

*

*When $\lambda=0$, the equation $\lambda=n(n-3)=0$ has two nonnegative integer solutions $n=3$ and $n=0$. The null space of $L$ therefore consists of only cubic polynomials and constant polynomials. Suppose $p$ be a monic cubic polynomial in the null space of $L$. The recurrence relation $(1)$ then gives $a_2=0$ and $a_1=3$, but $a_0$ is undetermined because $(1)$ reduces to $0a_0=0$ when $k=0$. Hence the null space of $L$ exists and it is a two-dimensional subspace spanned by $x^3+3x$ and $1$.

*When $\lambda=-2$, the equation $\lambda=n(n-3)$ has two nonnegative integer solutions $n=2$ and $n=1$. The corresponding eigenspace thus consists of (apart from the zero polynomial) only polynomials of degrees $1$ and $2$. Let $p$ be a monic quadratic polynomial in the eigenspace. The recurrence relation $(1)$ gives $a_0=1$ but $a_1$ is undetermined because $(1)$ reduces to $0a_1=0$ when $k=1$. Therefore the eigenspace for this eigenvalue exists and it is a two-dimensional subspace spanned by $x^2-1$ and $x$.

In short, the eigenvalues of $L$ are $\lambda=n^2-3n$ for each $n\ge2$. The corresponding eigenspace is two-dimensional when $n=2,3$ and one-dimensional when $ n>3$.
</s>