Basic properties of the matrix logarithm

Question

Let $|| \cdot ||$ be any submultiplicative norm on $\operatorname{Mat}_n(\mathbb C)$. For complex matrices $A$ with $||I_n - A|| < 1$, we can define the matrix logarithm

$$\log(A) = \sum\limits_{k=1}^{\infty} (-1)^{k+1} \frac{1}{k} (I_n - A)^k$$

which satisfies $\exp \log A = A$. I have three questions. Assuming that $A, B, AB$ are all in the domain of $\log$, do we have

$$\log(AB) = \log(A) + \log(B)$$

? If $A$ is invertible with $||I_n - A|| < 1$, do we have $||I_n - A^{-1}|| < 1$? Finally, do we have

$$\log(A^{-1}) = - \log A$$

for all invertible $A$ such that $||I_n - A||$ and $||I_n - A^{-1}||$ are $< 1$? I believe this should follow from the first two questions.

Answer 1

In this answer, I will assume that $$\log(A) = -\sum_{k=1}^\infty \frac{1}{k} (I- A)^k$$ defined on $\Omega = \{A : \left\|I_n - A\right\| < 1\}$ which is an convex open subset of $GL_n(\mathbb{C})$.

First I will prove that $\exp \log A = A$. Indeed, for $A \in \Omega$, let $$f(t) = \exp (\log ((1-t) I_n + tA)$$ and $$g(t) = ((1-t)I_n + tA)^{-1}$$ $$g'(t) = - (I_n - A) \left(((1-t)I_n + tA)^{-1}\right)^2 = -(I_n -A) (g(t))^2$$ As if $HX = XH$, $$d\log(X)(H) = X^{-1}H$$ and $$d\exp(X)(H) = H\exp(X)$$ then $$f'(t) = -((1-t)I_n + tA)^{-1}(I_n - A) \exp ((1-t)I_n + tA) = g'(t) (g(t))^{-1} f(t) \implies (fg)' (t) = f(t)g'(t) + f'(t)g(t) = 0$$ As $f(0)g(0) = I_n$ so $f(t)g(t) = I_n$, $$\exp\log (A) A^{-1} = f(1)g(1) = I_n$$

Now we will prove that if $A$, $B$ are in $\Omega$ such that $BA = AB \in \Omega$ then $$\log(AB) = \log (A) + \log (B)$$ In fact let $$f(t) = \log(A((1-t) I_n + tB)) - \log((1-t)I_n + tB)$$ then $$f'(t) = -(A(I_n - B)) (A((1-t)I_n + tB)))^{-1} - \left( -(I_n -B) ((1-t)I_n + tB)^{-1})\right) = 0$$ So $$\log(A) = f(0) = f(1) = \log(AB) - \log(B)$$

Answer 2

I think the first one is false in general even if we assume that $AB = BA$. For instance let us assume that for every $A$ and $B$ such that $A$, $B$ and $AB$ are in the domain of $\log$ and $AB = BA$, $$\log(AB) = \log(A) + \log(B)$$

Now let $A$ be a fixed matrix in the domain of $\log$. For every $t\in [0,1)$, $$\left\|I_n - tA\right\| = \left\|I_n - A + (1-t)A\right\| \le \left\|I_n - A\right\| + (1-t) \left\|A\right\| < 1$$ if $t > \epsilon$.

This implies that $t \mapsto \log(tA)$ is well defined on $(\epsilon, 1)$ and it is also differentiable with :

$$\frac{\partial}{\partial t} \log(tA) = \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k} \frac{\partial}{\partial t} (I_n - tA)^k = \sum_{k=1}^\infty (-1)^{k+1}(-A) (I_n-tA)^{k-1} = -A \sum_{k=0}^\infty (-1)^k(I_n -tA)^k = -A (I_n -(tA - I_n))^{-1} = -A (2I_n - tA)^{-1}$$

Or for every $t\in (\epsilon, 1)$, $(tI_n)A = A(tI_n) = tA$. With our assumption : $$\log(tA) = \log (tI_n) + \log(A) = \log(2-t) I_n + \log(A)$$ After differetiating the two side of the equation we have $$-A (2I_n - tA)^{-1} = -\frac{1}{2-t} I_n$$ wich implies that $$(2-t)A = 2I_n - tA \implies A = I_n$$

For the second question, I think is also false in general. We have $\left\|I_n\right\| \ge 1$ and let $$A = \left(1-t\right)I_n$$ with $t = \frac{1}{2\left\|I_n\right\|}\left(1 + \frac{\left\|I_n\right\|}{1 + \left\|I_n\right\|}\right) < 1$. Then $$\left\|I_n - A\right\| = t\left\|I_n\right\| = \frac{1}{2}\left(1 + \frac{\left\|I_n\right\|}{1 + \left\|I_n\right\|}\right) < 1$$ and $$\left\|I_n - A^{-1}\right\| = \frac{t}{1-t}\left\|I_n\right\| = \frac{\frac{1}{2}\left(1 + \frac{\left\|I_n\right\|}{1 + \left\|I_n\right\|}\right)}{1-\frac{1}{2\left\|I_n\right\|}\left(1 + \frac{\left\|I_n\right\|}{1 + \left\|I_n\right\|}\right)} = \frac{\left\|I_n\right\|\left(1 + \frac{\left\|I_n\right\|}{1 + \left\|I_n\right\|}\right)}{2\left\|I_n\right\|-\left(1 + \frac{\left\|I_n\right\|}{1 + \left\|I_n\right\|}\right)} = \frac{2\left\|I_n\right\|^2 + \left\|I_n\right\|}{2\left\|I_n\right\|^2 - 1} > 1$$

For the last question is also false in general try to construct a counter example with $A = tI_n$ (it shouldn't be difficult).

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With you definition you don't have $\exp \log(A) = A$, indeed $\log (tI_n) = \log(2-t) I_n$ and $\exp \log (tI_n) = (2-t) I_n \neq tI_n $. So try to change your definition to :

$$\log (A) = -\sum_{k=1}^{\infty} \frac{1}{k} (I_n -A)^k$$ with this last definition I think may the first question and the last one be correct.