Eigenspace and eigenvalues of a complex polynomial $P_3$ is a complex vector space of the complex polynomials of the degree $\leq 3$. An endomorphism $\varphi$ of $P_3$ is defined as:
$$\varphi(p)= p''+p'+p$$
I have to determine the eigenvalues of $\varphi$ and its eigenspace.
Here is what I found out until now:
$$\varphi p= \lambda p \Leftrightarrow (\varphi-\lambda \cdot id)p = 0$$
$$p^2+p+1=\lambda p$$
$$p^2+1+(1-\lambda)p = 0$$
$$\lambda = \frac{p^2+1}{p}+1$$
Is this how you should calculate the eigenvalues? Or do you have a better idea?
Or do you have to do it like this:
$$p^2+p+1=0$$
$$p_1 = -\frac{1}{2} + i \frac{\sqrt{3}}{2}$$
$$p_2 = -\frac{1}{2} - i \frac{\sqrt{3}}{2}$$
These are the complex solutions for the differential equation. Are they needed to compute the eigenvalues?
Thank you for your time.
 A: Note how $\varphi$ acts on a polynomial $p(t) = at^3+bt^2+ct + d \in P_3$:
\begin{align}
\varphi(at^3+bt^2+ct + d) &= (6at + 2b) + (3at^2+2bt+c) + (at^3+bt^2+ct + d) \\
&= at^3+(3a+b)t^2+(6a+2b+c)t + (2b+c+d)
\end{align}
Assume $\varphi(p) = \lambda p$. Comparing coefficients of $t^3$ gives $a = \lambda a$ so $\lambda = 1$ or $a = 0$.
If $a = 0$, then comparing coefficients of $t^2$ gives $b = \lambda b$ so $\lambda = 1$ or $b = 0$.
If $a = b = 0$, then comparing coefficients of $t$ gives $c = \lambda c$ so $\lambda = 1$ or $c = 0$.
If $a = b = c = 0$, then comparing coefficients of $1$ gives $d = \lambda d$ so $\lambda = 1$ or $d = 0$.
Hence, the only eigenvalue is $\lambda = 1$.
To find the eigenspace of $\lambda = 1$ assume $p'' + p' + p = \varphi(p) = p$. It follows $p'' = -p'$ so integrating gives $p = -p'+C$ for some constant $C \in \mathbb{C}$.
This means
$$at^3+bt^2+ct + d = -3at^2-2bt-c + C$$
so $a = b = c = 0$ and $d = C$ is arbitrary. Hence $\ker (\varphi - I) = \operatorname{span}\{1\}$, which is the space of constant polynomials.

Note:

Perhaps more general way is to write the matrix of $\varphi$ w.r.t. the basis $\{t^3, t^2, t, 1\}$:
$$\begin{bmatrix}
1 & 0 & 0 & 0\\
3 & 1 & 0 & 0\\
6 & 2 & 1 & 0\\
0 & 2 & 1 & 1
\end{bmatrix}$$
We can now immediately see that the only eigenvalue is $1$ and easily calculate the eigenspace.
