Let $L : V \to V$ be an operator on an $n$-dimensional complex vector space. Show if $p \in \mathbb{C}[t]$, the operator $p(L)$ is nilpotent if and only if the eigenvalues of $L$ are roots of $p$. What goes wrong in the real case when $p(t) = t^2 + 1$ and the dimension of $V$ is odd?
I understand the first part, since an operator is nilpotent if and only if all its eigenvalues are $0$, and the complex eigenvalues of $p(L)$ are given by $p(\lambda)$ where $\lambda$ is a complex eigenvalue of $L$.
I don't see why this doesn't also apply in the real case though, since we could just view the operator as a operator on $\mathbb{C}^n$ and conclude that the complex eigenvalues must be roots of $p$. In fact the claim seems necessarily false, since if $V$ is an odd-dimensional real vector space $L$ must have a real eigenvalue $\lambda$. But then $\lambda^2 + 1 \neq 0$ is an eigenvalue of $L^2 + I$, thus $L^2 + I$ cannot be nilpotent.