One-parameter subgroups and exponentials

[Fredrik]

I'm trying to understand this proof of the claim that if A is a one-parameter subgroup of $M_n(\mathbb C)$, there exists a unique matrix X such that $A(t)=e^{tX}$. (Theorem 2.13, page 37. Proof on page 38).

To prove uniqueness, just note that A'(0)=X.

To prove existence, the author defines

$$X=\frac{1}{t_0} \log A(t_0)$$

and then shows that $A(t)=e^{tX}$ for all t in a dense subset of the real numbers. Continuity then implies that this identity holds for all real t.

My problem with this is that he doesn't clearly state what t₀ is, so did he really define what X is? I mean, if this t₀ does the job, it looks like we could have used t₀/2 instead, or any other positive real number that's smaller than t₀. I'm probably missing something obvious, and I'm hoping someone can tell me what that is.

 

[tiny-tim]

My problem with this is that he doesn't clearly state what t₀ is, so did he really define what X is? I mean, if this t₀ does the job, it looks like we could have used t₀/2 instead, or any other positive real number that's smaller than t₀.

Hi Fredrik!

Yes, any value for t₀ will do …

just choose whichever value of t is most convenient!

 

[Fredrik]

That contradicts the uniqueness unless we can prove that $\frac{1}{t_0}\log A(t_0)$ is independent of t₀. I don't see how to do that.

 

[Fredrik]

I think I got it. The proof shows that regardless of what t₀ is (assuming that it's positive and small enough), we have $$A(t)=e^{t(\frac{1}{t_0}\log A(t_0))}$$ for all t. That, together with the result A'(0)=X, which guarantees uniqueness, implies that $\frac{1}{t_0}\log A(t_0)$ is independent of t₀. This seemed circular to me at first, but I don't think it is.

 

[tiny-tim]

… This seemed circular to me at first, but I don't think it is.

Hi Fredrik!

Yes, it's ok …

it's just one of those things everyone has to work out for themselves!