Does $D$ have real eigenvalues?

Question

Let $V$ be the subspace of the real vector space of real valued functions on $\mathbb{R}$, spanned by $\cos t$ and $\sin t$. Let $D : V \to V$ be the linear map sending $f(t) \in V $ to $\dfrac{ df(t)}{dt}$.

1. Prove that $D$ has a real eigenvalue.

2. What happens if we consider instead the complex vector space of complex valued functions? Does it have an eigenvalue?

My attempt: (I claim 1 is false)

Suppose $D$ has an eigenvalue. Let $v$ be an eigenvector $v=a \sin t + b \cos t$ then by using definition I get

$$a(1-\lambda) \sin t = (1+ \lambda) b \cos t$$

This holds for all $t$, $t\in\mathbb{R}$. Put $t=0$, $b=0$ or $\lambda=-1$ and both don't work. So it has no eigenvalue.

But I think my answer to 1 is concrete, I want to see a intuitive and rigorous proof.

And for 2, I know every operator over complex field has eigenvalue; is it real in this case?

Answer 1

Let $\mathscr{B}=\{\cos t,\sin t\}$ be the basis of $V$; then the matrix of $T$ with respect to this basis is \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} Thus the characteristic polynomial is $X^2+1$, which has roots $i$ and $-i$.

Answer 2

Your answer to 1) is fine. You can also say that if $f\colon\mathbb{R}\longrightarrow\mathbb R$ is such that $f'=af$ (for a real number $a$), then $f(x)=ke^{ax}$, for some $k\in\mathbb R$. SUch a function doesn't belong to your space.

For 2), use the fact that, if $f(t)=\cos(t)+\sin(t)i$, then$$f'(t)=-\sin(t)+\cos(t)i=if(t).$$