To determine the eigenvalues of $L$ I need a representation matrix of:
$$L: \mathbb{R} [X]_p \to \mathbb{R} [X]_p,\quad f \mapsto [(1-X^2)f']' $$
where $\mathbb{R} [X]_p$ is the vector space of polynomials with real coefficients with degree at most $p$. How I can find a matrix representation?
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$\begingroup$Matrix of linear transformation depends on a choice of basis, so let $$\mathcal{A}=\left\{\alpha_1=1,\alpha_2=X,\alpha_3=X^2,...,\alpha_p+1=X^p\right\}$$ be a basis of $\mathbf{R}[X]_p$.
So,$$L(\alpha_n)=\left(\left(1-X^2\right)\left(X^{n-1}\right)'\right)'=\left(\left(1-X^2\right)\left(n-1\right)X^{n-2}\right)'=-n\left(n-1\right)X^{n-1}+\left(n-1\right)\left(n-2\right)X^{n-3}=\left(n-1\right)\left(n-2\right)\alpha_{n-2}-n\left(n-1\right)\alpha_n$$ for $n>2$, you can find the values of $L(\alpha_1),L(\alpha_2)$ easily.
Now, by definition of matrix of linear transformation $A= \left[L(\alpha_1)_\mathcal{A}\,L(\alpha_2)_\mathcal{A}\cdots L(\alpha_p)_\mathcal{A}\right]$ where columns $L(\alpha_i)_\mathcal{A}$ are coordinates of $L(\alpha_i)$, for example if $L(\alpha_5)=12\alpha_3-20\alpha_5$ then $$L(\alpha_5)_\mathcal{A}=\begin{pmatrix}0\\0\\12\\0\\-20\\0\\\vdots\end{pmatrix}$$ Therefore matrix of $L$ in basis $\mathcal{A}$ is $$A=\begin{pmatrix} 0&0&2&0&0&0&\cdots\\ 0&-2&0&6&0&0&\cdots\\ 0&0&-6&0&12&0&\cdots\\ 0&0&0&-12&0&20&\cdots\\ 0&0&0&0&-20&0&\cdots\\ \end{pmatrix}=[a_{ij}],$$ where $a_{ij}=\left\{ \begin{align} -i(i-1)&&1<i=j\leq p\\ (j-1)(j-2)&&0<j\leq p\,\textrm{and}\, j=i+2\\ 0&&\textrm{otherwise} \end{align}\right.$
$\endgroup$ $\begingroup$You need to take a basis for $\mathbb{R}[X]_p$, $f_1, \dots, f_n$.
Compute the coefficients of $L(f_1), \dots, L(f_n)$ in that basis.
Then the $A_{ij}$ entry of the matrix is the $f_i$ coefficient of $L(f_j)$.
See 15.10 of for a more general formulation of this:
See for some examples:
$\endgroup$ $\begingroup$To define a matrix that represent $L$ choose a basis of $\mathbb{R} [X]_p$ that I will name $v_0,v_1,\ldots,v_p$. Now the coefficients of a matrix $A$ (noted as $A_{j,k}$) are defined by
$$Lv_k=\sum_{j=0}^p A_{j,k}v_j,\quad\forall j,k\in\{0,1,\ldots,p\}$$
If you can find a basis such that
$$Lv_k\in\operatorname{span}(v_0,\ldots,v_j),\quad\forall k\le j$$
then the representation matrix would be upper-triangular (then the eigenvalues of $L$ lie in the diagonal of this matrix). However the existence of upper-triangular matrices are not ensured for operators on real vector spaces.
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