Let $f: V_3 \rightarrow V_3$ be the function such that $p(X) \mapsto p''(X)$, calculate the eigenvalues of f

Question

Let $V_3$ be the vector space of all polynomials of degree less than or equal to 3. The linear map $f: V_3 \rightarrow V_3$ is given by $p(X) \mapsto p''(X)$. Calculate the eigenvalues of f.

First of all I'm not entirely sure how to derive the matrix of f. If I try to give the matrix with regards to the basis $B=\{1,X,X^2,X^3\}$ I get: $$f(b_1)=0$$$$f(b_2)=0$$$$f(b_3)=2$$$$f(b_4)=6X$$, which yields the following matrix:

$$ A = \begin{pmatrix} 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 6 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$

Then I get for $det(A-\lambda I) = (-\lambda)^4 = \lambda$, therefore $\lambda = 0$ is the only eigenvalue.

I kind of have my doubts that this is the correct solution? How do I get the eigenvalues of f?

Answer 1

Without much thought you should be able to see that repeating this operation on any polynomial (even if one would allow ones of degree larger than$~3$) ultimately makes it zero. Then if there were a nonzero polynomial for which the operation amounts to multiplication by a constant, this constant can only be zero (otherwise repeating the operation would never reach$~0$). So it should not be surprising that $0$ is the only eigenvalue of$~f$.

Answer 2

You only need to delete the part $ "\; = \lambda\; "$ from $det(A-\lambda I) = (-\lambda)^4 = \lambda$ and the remaining is correct. You can also notice that $A^2=0$ which means that $A$ is a nilpotent matrix so the only possible eigenvalue is $0$.

Answer 3

We want $$f''(x)=\lambda f(x) \\ 6ax+2b=\lambda(ax^3+bx^2+cx+d)$$ So either $$\lambda=0, a=b=0$$ or $$a=b=0\implies 0\equiv\lambda(cx+d)\implies c=d=0\implies f(x)\equiv0$$ You are correct, the only eigenvalue is $0$.