Finding all subspaces invariant under F.

Question

The question and its answer are given below:

But I do not know why is this the answer, could anyone explain this for me please?

Answer 1

If the characteristic polynomial of a linear operator $\alpha$ on $\Bbb C^n$ has no multiple roots, then it has $n$ distinct eigenvalues, and since eigenvectors of distinct eigenvalues are linearly independent, we actually get a basis $v_1,\dots, v_n$ of eigenvectors, and coordinating in this basis, $\alpha$ becomes diagonal.
Now, coordinated in this basis, $e^{t\alpha}$ is still diagonal, so it has the same eigenvectors ($v_1,\dots, v_n$).