What are the eigenvalues and eigeinvectors of $f(P) = (X - \alpha)(P'(X) + P'(\alpha)) - 2(P(X) - P(\alpha) )$.

Question

Let $\alpha \in \mathbb{R}$ and $f$ is an endomorphism of $\mathbb{R}_n [X]$ defined as:

$f(P) = (X - \alpha)(P'(X) + P'(\alpha)) - 2(P(X) - P(\alpha) )$

1. Write the matrix of $f$ in the basis $ \mathcal{B}= (e_0, e_1,..., e_n)$ and such that $e_k = (X - \alpha)^k$, $0 \leq k \leq n$.

2. What are the eigenvalues and eigeinvectors of $f$.

3. Is $f $ surjectif?

1. $ M(f, \mathcal{B}) = \begin{pmatrix} 2(P(X) - P(\alpha) & 0 & ... & 0 \\0 & (P'(X) + P'(\alpha)) & ... & 0\\ \vdots & \vdots & \vdots & \vdots \\ 0 & 0 & 0 & 0 \end{pmatrix} $

2. Computing $det(M - \lambda.I_n)$ gives the eigeinvalues:

$ \lambda_1 = (P'(X) + P'(\alpha)) $,

$\lambda_2 = - 2(P(X) - P(\alpha) ) $

And solving: $f(P) = \lambda_1$ and $f(P) = \lambda_2$ gives the eigenvectors:

$v_1 = \frac{(X - \alpha)((P'(X) + P'(\alpha))}{2(P'(X) - P'(\alpha)) + 2}$

$v_2 = \frac{(X - \alpha)(P'(X) + P'(\alpha)) + 2P(\alpha)}{ 2(P'(X) - P'(\alpha)) - 1}$

3. For the surjectivity, I need to prove that every polynomial $P \in \mathbb{R_n}[X]$, there exists a $ Q \in \mathbb{R_n}[X]$ such that: $f(Q)=P$

Which I don't know how.

Are my answers correct?

Answer 1

Your matrix is wrong, since $f(e_0)=0$ and $$f(e_k)=f\bigl((x-\alpha)^k\bigr)=(k-2)(x-\alpha)^k=(k-2)e_k,\quad\text{for $k>0$.}$$

Furthermore, $f$ can't be surjective, because $\alpha$ is a zero of $f(P)$. Hence a polynomial with P$(\alpha)\neq0$ isn’t in the image of $f$.

You may also notice that the rank of $M$ is less than $n+1$.