I'm looking for a nice textbook that proves this elementary result:
Theorem. Suppose $V$ is a finite-dimensional real or complex vector space, and suppose that $\{U(t)\}_{t \in \mathbb{R}}$ is a continuous 1-parameter group of linear operators on $V$, meaning
- $U(t) : V \to V$ is a linear operator for each $t \in \mathbb{R}$;
- $U(t)$ depends continuously on $t$ (in the usual topology on the space of linear operators from $V$ to itself);
- $U(0) = 1_V$;
- $U(s+t) = U(s) U(t)$ for all $s, t \in \mathbb{R}$.
Then there exists a unique linear operator $H : V \to V$ such that $U(t) = \exp(tH)$. Conversely, for any linear operator $H : V \to V$, if $U(t) = \exp(tH)$ then $\{U(t)\}_{t \in \mathbb{R}}$ is a continuous 1-parameter group of linear operators on $V$.
I'd really like a reference to a proof that only does the finite-dimensional case—and please give me the theorem number, since I need this for a paper I'm writing. The referee is demanding a reference!
I find it sad that no reference to a proof of this result is given in the Wikipedia articles "Matrix exponential" and "One-parameter group".
I would accept a proof of the more general result where $V$ is a Banach space and $\{U(t)\}_{t \in \mathbb{R}}$ is a 1-parameter group of bounded linear operators that's continuous in the norm topology. The proof is exactly the same! But I'd rather not bring in these extra concepts.
I really don't want a theorem that considers the case where $V$ is infinite-dimensional and $\{U(t)\}_{t \in \mathbb{R}}$ is only continuous in the strong topology. This is distractingly technical for what I need!
Take your favorite book on Lie groups, a morphism $f:G\rightarrow H$ between Lie groups induces a morphism $Lie(f):Lie(G)\rightarrow Lie(H)$ between their Lie algebra which commutes with the exponential. Here take $G=\mathbb{R}$ and $H=Gl(n,\mathbb{R})$.