While teaching a course on ODE, I needed to introduce the notion of matrix logarithms. I intend to define it as follows.
Definition (Matrix Logarithm)
Let $A\in GL_n(\mathbb{C})$. We define
A) Unipotent case: When $A$ is unipotent, i.e. $A=I+N$, where $N\in M_n(\mathbb{C})$ is nilpotent, we define $\ln A\in M_n(\mathbb{C})$ as $$ \ln A=\ln(I+N):=\sum_{k=1}^{n}\frac{(-1)^{k+1}}{k}N^k. $$ B) Diagonalizable case: When $A$ is diagonalizable, i.e. $A= PDP^{-1}$, where $P\in GL_n(\mathbb{C})$ and $D:=\operatorname*{diag}(\lambda_1,\ldots,\lambda_n)$ with $\lambda_1,\ldots,\lambda_n\in \mathbb{C}\setminus\{0\}$, we define $\ln A\in M_n(\mathbb{C})$ as $$ \ln A:= P\operatorname*{diag}(\ln\lambda_1,\ldots,\ln\lambda_n)P^{-1}, $$ where $\ln \lambda_i = \ln |\lambda_i|+i\arg \lambda_i$, for all $i=1,\ldots,n$.
C) Invertible case: When $A\in GL_n(\mathbb{C})$, we define $$ \ln A:=\ln D + \ln (I+D^{-1}N), $$ where $D, N\in M_n(\mathbb{C})$ are diagonalizable and nilpotent respectively, $A=D+N$, and $DN=ND$. The existence of $D,N$ follows from the Jordan-Chevalley Decomposition Theorem.
The problem arises when I want to prove that $$ e^{\ln A}=A,\text{ for all }A\in GL_n(\mathbb{C}).\ \ \ \ (1) $$
While it is easy to verify equation (1) when $A$ is unipotent or diagonalizable, I don't see any quick way to prove (1) when $A$ is a general invertible matrix. This will be easy to prove if one can show that
$$ \ln D\ln (I+D^{-1}N)=\ln (I+D^{-1}N)\ln D.\ \ \ \ \ (2) $$ Then, the matrix exponential laws and Equation (1) in unipotent or diagonalizable cases will settle (2) in the general case.
QUESTION: Is there any easy way to prove (2), i.e. $$ \ln D\ln (I+D^{-1}N)=\ln (I+D^{-1}N)\ln D, $$ when $D$ is diagonalizable, $N$ is nilpotent and $DN=ND$?
Thank you.
The question reduces to prove that $\ln(D)(D^{-1}N)^k = (D^{-1}N)^k\ln(D)$ for every $k$, so proving that $\ln(D)$, $D^{-1}$ and $N$ all commute is sufficient.
If $D=P\Sigma P^{-1}$ where $\Sigma$ is diagonal, then $$\ln(D)D^{-1}=P\ln(\Sigma)P^{-1}P\Sigma^{-1}P^{-1} \\= P\ln(\Sigma)\Sigma^{-1}P^{-1} \\=P\Sigma^{-1}\ln(\Sigma)P^{-1} \\=P\Sigma^{-1}P^{-1}P\ln(\Sigma)P^{-1} =D^{-1}\ln(D).$$
Moreover, $DN=ND$ by hypothesis, so $$\Sigma P^{-1}NP = P^{-1}NP \Sigma$$ meaning that $P^{-1}NP$ is block diagonal with blocks corresponding to the eigenspaces of $\Sigma$, that are the same as those of $\ln(\Sigma)$, so $$\ln(\Sigma) P^{-1}NP = P^{-1}NP \ln(\Sigma)\implies \ln(D)N = N\ln(D).$$