Eigenvalue of a real vector space

Question

Let V be the subspace of the real vector space of real valued functions on R, spanned by cos t and sin t. Let D : V → V be the linear map sending f(t) ∈ V to$\cfrac{ df(t)}{dt}$. Then D has a real eigenvalue.

The answer was given to be false.

But let $f(t)= \sin t - \cos t$ .Then this is a Eigen vector with eigenvalue -1 .isn't it real?

Please explain me where i'm wrong and corrrct approach to the problem

(Sorry! I'm wrong while taking derivative ...but please explain why it has no real eigenvalue?)

Answer 1

Your function $f$ is not an eigenvector, because $Df$ is not $\lambda f$ for some real number $f$.

And the answer is negative, because if $f'=\lambda f$ for some real number $\lambda$, then$$(\forall t\in\mathbb{R}):f(t)=ke^{\lambda t},$$for some constant $K$. And therefore $f\notin D$.