TITLE: Confusion about convergence for the logarithm of a matrix
QUESTION [1 upvotes]: How can I prove that for a matrix $A$,
$$
\text{log}A = \sum_{m=1}^\infty (-1)^{m+1} \frac{(A-I)^m}{m}
$$
is absolutely convergent if $||A - I|| < 1$? (I'm using the Hilbert-Schmidt norm.)
In Hall's textbook Lie Groups, Lie Algebras, and Representations, the following points are made (around section 2.3).

For $z\in\mathbb{C}$, the function
$$
\text{log}z = \sum_{m=1}^\infty (-1)^{m+1} \frac{(z-1)^m}{m}
$$
is defined and holomorphic in a circle of radius 1 about $z=1$.
$||(A-I)^m||\le ||A-I||^m$ for $m\ge 1$.

I see that $$\sum_{m=1}^\infty (-1)^{m+1} \frac{||A-I||^m}{m}$$ will converge if $||A-I||<1$ due to point 1 but I can't figure out how to bound $||\text{log}A||$ by that series. I tried the triangle inequality:
$$||\text{log}A||\le \sum_{m=1}^\infty ||(-1)^{m+1} \frac{(A-I)^m}{m}||=\sum_{m=1}^\infty \frac{||(A-I)^m||}{m} \le \sum_{m=1}^\infty \frac{||A-I||^m}{m}$$
But now I've lost the $\text{log}(z)$ form. What simple idea am I missing? Thanks!

REPLY [2 votes]: Gary supplied the missing link I needed; thank you, Gary!
Pick up at the sequence of inequalities in my question:
$$||\text{log}A||\le \sum_{m=1}^\infty ||(-1)^{m+1} \frac{(A-I)^m}{m}||=\sum_{m=1}^\infty \frac{||(A-I)^m||}{m} \le \sum_{m=1}^\infty \frac{||A-I||^m}{m}.$$
Note that
$$
-\hbox{log}(1-||A-I||) = -\sum_{m=1}^\infty (-1)^{m+1} \frac{(1- ||A-I|| - 1)^m}{m} = \sum_{m=1}^\infty \frac{||A-I||^m}{m}.
$$
Now $z = 1-||A-I||$ is inside a circle of radius 1 centered at 1 so log converges at this point. Therefore, $||\hbox{log}A|| < \infty$ is bounded above and $\hbox{log}A$ is absolutely convergent.
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