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Let $A_1,A_2\in GL(n,\mathbb C)$ be two commuting invertible matrix, we want to show $\log(A_1)$ commutes with $\log(A_2)$ (for some pair).

To be more precise, let $A\in GL(n,\mathbb C)$, we define $\log(A)$ to be the preimage of $A$ under exponential map, i.e., the set of matrix $B\in M_{n\times n}(\mathbb C)$ such that $\exp(B):=\sum_{k\ge 0}B^k/k!=A$. We want to show there exists $B_i\in \log(A_i)$, $i=1,2$, such that $B_1B_2=B_2B_1$.

Note that when $A_1, A_2$ are unipotent, the proof is easy: The canonical logarithm is given by the Taylor series $$B_i=\sum_{k\ge 1}(-1)^{k+1}\frac{(Id-A_i)^k}{k}$$ which is actually a finite sum, and $B_1$ commutes with $B_2$ follows from $(Id-A_1)^k$ commutes with $(Id-A_2)^k$.

In the general situation, the difficulty is that the representation of matrix logarithm is complicated.

To find a logarithm for $A\in GL(n,\mathbb C)$, first by the identity $S^{-1}\exp(B)S=\exp(S^{-1}BS)$, it reduce to assume that $A=\mathrm {diag}(J_1,...,J_p)$ is in the standard Jordan form, where $$J_i=\begin{bmatrix} \lambda_i & 1 & & \\ & \lambda_i & \ddots & \\ & & \ddots & 1\\ & & & \lambda_i \end{bmatrix}$$ is a standard Jordan block with eigenvalue $\lambda_i$ $(1\le i\le p)$. Note that $\lambda_i\neq 0$, $J_i/\lambda_i$ is unipotent, so its logarithmic is (multivalued) $ \log(\lambda_i)\log(J_i/\lambda_i)$.

So the possible logarithm of $A\in GL(n,\mathbb C)$ is

$$B=S^{-1}\begin{bmatrix} \log(\lambda_1)\log(J_1/\lambda_1) & & & \\ & \log(\lambda_2)\log(J_2/\lambda_2) & \ddots & \\ & & \ddots & \\ & & & \log(\lambda_p)\log(J_i/\lambda_p) \end{bmatrix}S$$

I stopped here... Hope someone could help.

Arash
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AG learner
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  • Note: if $A_1$ and $A_2$ commute, then there is $P\in GL_n$ with $P^{-1}A_1P$ and $P^{-1}A_2P$ both in Jordan canonical form. – David Hill Aug 15 '18 at 02:48
  • @David Hill The answers on this post provide counterexamples: https://math.stackexchange.com/questions/49378/simultaneous-jordanization – AG learner Aug 15 '18 at 04:26
  • @Yilong Zhang , did you understand or at least read my answer ? –  Sep 11 '18 at 22:16
  • Dear @loupblanc, sorry for the late action. I was involved in some other problems after I asked this question... Honestly, I haven't carefully read your solution yet, but I'll do so in the following few days when I get time. Thanks for your solution. – AG learner Sep 12 '18 at 00:45

1 Answers1

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Let $\theta\in\mathbb{R}$ s.t. for every $\lambda_1\in spectrum(A_1),\lambda_2\in spectrum(A_2)$, $arg(\lambda_1)-\theta\notin 2\pi \mathbb{Z},arg(\lambda_2)-\theta\notin 2\pi\mathbb{Z}$.

Note that $\log_{\theta}: re^{i\alpha}\in\{re^{i\alpha}\in\mathbb{C};r>0,\alpha\in(\theta,\theta+2\pi)\}\rightarrow \log(r)+i\alpha$ is a holomorphic function; then $\log_{\theta}(A_i)$ can be defined using the Hermite's interpolation (cf. Higham, functions of matrices); therefore $e^{\log_{\theta}(A_i)}=A_i$. Then $\log_{\theta}(A_i)$ is a polynomial in $A_i$ and, finally, $\log_{\theta}(A_1)$ and $\log_{\theta}(A_2)$ commute.

Remark. That does not imply that $\log_{\theta}(A_1)+\log_{\theta}(A_2)$ can be written in the form $\log_{\beta}(A_1A_2)$. For example, let $A_1=A_2=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$. Then $\log_{-\pi}(A_i)=\begin{pmatrix}0&\pi/2\\-\pi/2&0\end{pmatrix}$ and $\log_{-\pi}(A_1)+\log_{-\pi}(A_2)=\begin{pmatrix}0&\pi\\-\pi&0\end{pmatrix}$; on the other hand, $A_1A_2=-I_2$ and any $\log_{\beta}(-I)$ ($\beta\notin -\pi+2\pi\mathbb{z}$) is a scalar matrix.