Is $\exp(\log(A))=A$ for any matrix $A$ where $\log(A)$ is defined?

Question

It is possible to show that:

$$\exp(\log(A)) = A$$

When $A$ is diagonalizable (and the logarithm exists). Furthermore it is possible to show that when $\| A - I\| < 1$ then $\exp(\log(A))=A$. The proof for that relies of the fact that any matrix can be approximated by a series of diagonalizable matrices, and on the fact that when $\| A - I\| < 1$ the logarithm always exists. This let's our approximation $A_n \to A$ eventually (for large enough $n$) lie in the ball of radius $1$ around $I$ which allows us to invoke the continuity of $\exp, \log$ and finish the theorem.

My question is, is the equality $\exp(\log(A)) = A$ always true when $\log(A)$ exists? The same proof could be applied only if the $\log$ exists in some open neighborhood of $A$, but that's not always the case. For example taking $A = 2\cdot I$ then for any $\epsilon$ we have $\log((2 + \epsilon)\cdot I)$ doesn't exist. Despite that, $\exp(\log(2\cdot I)) = 2\cdot I$.

Is there a proof that $\exp(\log(A)) = A$ when $\log(A)$ exists? Is there a counterexample?

In this question the matrix logarithm is defined via the power series:

$$\log(A) = \sum_{n=1}^\infty (-1)^{n+1} \frac{(A-I)^n}{n}$$

Answer 1

This answer is based on the observation that a form of complex analysis can be done in the matrix algebra.

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Let $\mathfrak{A}$ denote the set of all $d\times d$ complex matrices. Then an $\mathfrak{A}$-valued function $F(z) = [f_{ij}(z)]_{i,j=1}^{d}$ on an open subset $U$ of $\mathbb{C}$ is holomorphic if each entry $f_{ij}(z)$ is holomorphic on $U$. Now it is easy to note:

Lemma 1. Let $F$ and $G$ be two $\mathfrak{A}$-valued functions on a domain $U$ of $\mathbb{C}$. If $F = G$ on some subset $S$ of $U$, where $S$ has an accumulation point in $U$, then $F = G$ on all of $U$.

Proof. Apply the identity theorem entry-wise. $\square$

Lemma 2. Let $F$ be an $\mathfrak{A}$-valued holomorphic function on an open set $U$ of $\mathbb{C}$. Then $\exp(F(z))$ is also holomorphic on $U$.

Proof. The sum $\exp(F(z)) = \sum_{n=0}^{\infty} \frac{1}{n!} F(z)^n$ converges locally uniformly on $U$, and so does each entry of the partial sum. $\square$

Now let $X \in \mathfrak{A}$ be such that $S_N = \sum_{n=1}^{N}\frac{(-1)^{n-1}}{n}X^n$ converges to some $S \in \mathfrak{A}$ as $N \to \infty$. Then

\begin{align*} \sum_{n=1}^{N} \frac{(-1)^{n-1}}{n}X^n z^n &= \sum_{n=1}^{N} (S_n - S_{n-1}) z^n \\ &= \sum_{n=1}^{N} S_n z^n - \sum_{n=1}^{N-1} S_n z^{n+1} \\ &= S_N z^N + (1 - z) \sum_{n=1}^{N-1} S_n z^n. \end{align*}

(Of course, this is nothing but a consequence of summation by parts.) Assuming $|z| < 1$ and letting $N \to \infty$, both sides converges and the following equality holds:

$$ \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}X^n z^n = (1 - z) \sum_{n=1}^{\infty} S_n z^n. \tag{1} $$

Let us denote the left-hand side of $\text{(1)}$ by $f(z)$. Since $(S_n)$ converges, each entry of $S_n$ is bounded. So, if we write $S_n = [S_{n;ij}]_{i,j=1}^{d}$, then each $\sum_{n=1}^{\infty} S_{n;ij} z^n$ has radius of convergence at least $1$. From this, we find that $f$ is holomorphic on $\mathbb{D} = \{z \in \mathbb{C} : |z| < 1\}$. Then by Lemma 2, $\exp(f(z))$ is also holomorphic on $\mathbb{D}$. Moreover, we already know that if $\|zX\| < 1$, then

$$ \exp(f(z)) = 1 + zX . $$

By Lemma 1, this equality extends to all of $\mathbb{D}$. Finally, by Abel's theorem, $f(r) \to S$ as $r \to 1^-$. (Although the linked article only covers the case of $\mathbb{C}$-valued coefficients, the case of $\mathfrak{A}$-valued coefficients can be proved mutatis mutandis.) So,

$$ \exp(S) = \lim_{r \to 1^-} \exp(f(r)) = \lim_{r \to 1^-} (1 + rX) = 1 + X. $$

This is enough to establish the desired claim.