I have come across this statement. If $\mathbf{u}$ is a generalized eigenvector, so that $(\mathbf{A}-\lambda \mathbf{I})^m \mathbf{u} = 0,$ then $$e^{t\mathbf{A}}\mathbf{u} = e^{t\lambda} e^{t(\mathbf{A}-\lambda \mathbf{I})} \mathbf{u} = e^{t\lambda} \sum_{j=0}^{m-1} \frac{(\mathbf{A}-\lambda \mathbf{I})^j t^j}{j!}\mathbf{u}. $$
I have two questions.
For the first equality ($e^{t\mathbf{A}}\mathbf{u} = e^{t\lambda} e^{t(\mathbf{A}-\lambda \mathbf{I})} \mathbf{u}$), the right-hand side can be rewritten as $e^{t\lambda} e^{t(\mathbf{A}-\lambda \mathbf{I})} \mathbf{u} = e^{t(\lambda + \mathbf{A} - \lambda \mathbf{I})}\mathbf{u}.$ That seems to suggest that $e^{\lambda - \lambda \mathbf{I}} = e^0 = 1$, but $\lambda$ is a scalar and $\lambda \mathbf{I}$ is not, so how does that work?
For the second equality (involving the sum), why is the sum from $j=0$ to $j=m-1$, when normally we have that $e^{tx} = \sum_{j=0}^{\infty} \frac{(tx)^j}{j!}$?
UPDATE
Regarding the second question, I think I have figured out the answer. Since $(\mathbf{A}-\lambda \mathbf{I})^m \mathbf{u} = 0$, any term involving the power of $m$ or higher will be equal to $0$. The first question still stands, though.