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I am studying the book "Matrix Groups: An Introduction to Lie Group Theory" by Andrew Baker. There is a proposition in the book about the exponential and logarithm functions for matrices. Let me first give the definitions: $$ \mbox{For any} A\in M_n(K), \mbox{ where }K=\mathbb{R} \mbox{ or }K=\mathbb{C}, exp(A)=\sum_{n\geq0}\frac{1}{n!}A^n $$ and $$ \mbox{For any} A\in M_n(K) \mbox{ such that } ||A-I||<1, \mbox{ where }K=\mathbb{R} \mbox{ or }K=\mathbb{C}, log(A)=\sum_{n\geq0}\frac{(-1)^{n-1}}{n}(A-I)^n $$ The proposition basically says that they are inverses of each other. I will just write one of them: $$ \mbox{If } ||A-I||<1, \mbox{then } exp(log(A))=A $$ In the proof, book states that the following equality follows from comparing coefficients. $$ exp(log(A))=\sum_{m\geq0}\frac{1}{m!}(\sum_{n\geq0}\frac{(-1)^{n-1}}{n}(A-I)^n)^m=A $$ What I did: I tried to calculate the coefficients of the terms seperately. Unfortunately, all I got was divergent series. I will show you what I did for the coefficient of the constant term of the RHS of the last equation above.
I observed that the coefficients of the constant term will come only from the coefficients of $(-I)^n$. Hence, the coefficient of the constant term is the following: $$ \sum_{m\geq0}\frac{1}{m!}(\sum_{n\geq0}\frac{(-1)^{n-1}}{n}(-1)^n)^m $$ But, the series $\sum_{n\geq0}\frac{(-1)^{n-1}}{n}(-1)^n$ inside is divergent. Because it is equal to the series $\sum_{n\geq0}\frac{(-1)^{2n-1}}{n}=-\sum_{n\geq0}\frac{1}{n}$. Similar to this one, I always find divergent series when I try to find the coefficients. Obviously, I am making an horrible mistake but I could not figure it out. Can you help me? Thank you for any help.

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You can't prove this identity with a series expansion in $A$ because $\log(A)$ diverges as $A \to 0$. The convergence condition is that $\| A - I \| < 1$ and so you should be doing a series expansion in terms of $X = A - I$ instead. This means you want to check

$$\exp(\log(I + X)) = \sum_{m \ge 0} \frac{1}{m!} \left( \sum_{n \ge 1} \frac{(-1)^{n-1}}{n} X^n \right)^m = I + X$$

and this time, while tedious, it's actually doable: for example the constant term is now $1$.

A much cleaner way to do this is to appeal to the uniqueness theorem for solutions to ODEs, as follows. Set

$$f(t) = \exp(\log(I + tX)).$$

Check that (for sufficiently small $t$)

$$\frac{df}{dt} = \exp(\log(I + tX)) (I + tX)^{-1} X$$

and hence that $f$ satisfies the ODE

$$\frac{df}{dt} = \frac{X}{I + tX} f(t)$$

with initial condition $f(0) = 1$. Now it's easy to check that $1 + tX$ also satisfies this ODE with the same initial condition, so by the uniqueness theorem $f(t) = 1 + tX$.

Similarly to check inverses in the other direction, set

$$g(t) = \log (\exp (tX)).$$

Check that, again for sufficiently small $t$,

$$\frac{dg}{dt} = \frac{X \exp(tX)}{\exp(tX)} = X$$

(here it's a little easier to see what's going on), hence that $g$ solves the ODE

$$\frac{dg}{dt} = X$$

with initial condition $g(0) = 0$. Now it's again easy to check that $tX$ satisfies the same ODE with the same initial condition, so $g(t) = tX$.

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