semigroup of operators generated by a diagonalizable operator and exponential

Question

Let $A \colon D(A) \subset H \to H$ be the generator of a $C^0$ semigroup. Suppose that $e_k$ is an orthonormal basis and $A$ is diagonalizable $A e_k=\lambda_k e_k$ with eigenvalues $e_k$. Then is it true that $<e^{At}x,e_k>=e^{t \lambda_k} <x,e_k>$ ?

Answer 1

Hint: you might have a look at the Spectral Inclusion Theorem at the book of Engel & Nagel (p. 276), especially, $$e^{t\sigma_p(A)}\subset \sigma_p(e^{tA}).$$ In fact the spectral mapping theorem holds for the point spectrum, that is: $$e^{t\sigma_p(A)}= \sigma_p(e^{tA}) \setminus \{0\}.$$