Find eigenvalues anf eigenvectors of $f(p(x)):=p'(x)$

Question

Find eigenvalues anf eigenvectors of $f \in \operatorname{End}(P_{3}^{R}[x])$, where $f(p(x)):=p'(x)$

I have calculated the linear map matrix in standard basis and then I've applied the Cayley-Hamilton theorem, which gave me this determinant:

$$ \begin{bmatrix} -\lambda&1&0&0\\ 0&-\lambda&2&0\\ 0&0&-\lambda&3\\ 0&0&0&-\lambda \end{bmatrix} = 0$$

which only eigenvalue is $\lambda = 0$. Hence, the only eigenvector would be $(1,0,0,0)$.

Is this process correct?

Thanks in advance.

Answer 1

The computations are correct, but the answer is not, since $(1,0,0,0)\notin P_3^{\mathbb R}[x]$. The eigenvectors corresponding to the null eigenvalue are the constant polynomials.

Answer 2

$f$ is a nilpotent endomorphism since $f^4\equiv0$. Indeed, $\displaystyle \frac{d^4}{dx^4}f(x)=0$ for every polynomial of degree at most $3$. Hence $0$ is the unique eigenvalue and it has algebraic multiplicity $1$. The eigenspace corrisponding to $0$ is the kernel of $f$, namely the subspace of those polynomiasl whose first derivative is zero, namely constant polynomials.

Now: $P_3^{\Bbb R}$ naturally identifies with $\Bbb R^4$ by the map $a+bx+cx^2+dx^3\longmapsto (a, b, c, d)$. By using this identification, the subspace corresponding to constant polynomials is the subspace generated by $(1,0,0,0)$. I think you have used this identification in your mind. However, what you have written is not properly correct unless you specify that you are using this identification and working with $\Bbb R^4$ instead of $P_3^{\Bbb R}$.

Answer 3

An alternative method without matrices. If $p(x)$ is an eigenvector then $\exists \lambda$ such that

$$ p'(x) = \lambda p(x).$$

If $p(x)$ is a non constant polynomial then we now that the degree of $p'(x)$ is less than the degree of $p(x)$ and we cannot have $p'(x) = \lambda p(x)$.

Therefore $p(x)$ must be a constant polynomial and we have the equation $$ p'(x) = 0 = \lambda p(x)$$ and we must have $\lambda = 0$ and $p(x)$ constant.