I am to prove the following proposition using S+N decomposition.
Let $A\in M_n(\mathbb{F})$ - that is, an $n\times n$ matrix over some field $\mathbb{F}$, and $\mu_1,...,\mu_l$ its eigenvalues with algebraic multiplicities $n_1,...,n_l$. Write $\mu_q=\alpha_q+i\beta_q$, with $\alpha_q,\beta_q\in\mathbb{R}$ Then \begin{align*} &\mathbb{F}=\mathbb{C}\\ &\{e^{tA}\}_{ij}\in \text{Span}_\mathbb{C}\{t^p e^{\mu_q t} | 1\leq q\leq l, 0\leq p \leq n_q\}\\ &\mathbb{F}=\mathbb{R}\\ &\{e^{tA}\}_{ij}\in \text{Span}_\mathbb{R}\{t^p e^{\mu_q t}\cos(\beta_q t),t^p e^{\mu_q t}\sin(\beta_q t) | 1\leq q\leq l, 0\leq p \leq n_q\} \end{align*}
But I'm unsure how to get started, even. For S+N decomposition, I know that a square matrix A can be 'decomposed' into a diagonalizable matrix S and a nilpotent matrix N which commute, such that the exponential $e^{tA}=e^{t(S+N)}=e^{tS}e^{tN}$. Furthermore, the exponential of diagonal or nilpotent matrices can be computed with relative ease.
For a nilpotent $n\times n$ matrix, the exponential series terminates for $i=n$, e.g. \begin{equation*} e^{tN} = \mathbf{Id} +tN+...\frac{t^{n-1}}{(n-1)!}N^{n-1} \end{equation*}
And for a diagonal matrix $D=\text{Diag}\{\lambda_1,...\lambda_n\}$, you get a diagonal matrix where every entry is just the exponentials, e.g. $e^{tD}=\text{Diag}\{e^{t\lambda_1},...,e^{t\lambda_n}\}$.
However, as I understand it the decomposition matrix $S$ is diagonalizable, but not necessarily diagonal. So how would the product $e^{tN}e^{tS}$ shape up? I'm only certain of the $e^{tN}$-part.
Secondly, I don't see the connection with the spans given in the proposition, and I'm not sure how to verify that all the matrix entries belong to those spans.