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?)
Best Answer
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 $f\notin D$.