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Given two similar, diagonizable square matrices $A$ and $B$ that do not commute, can the Baker-Campbell-Hausdorff formula be simplified exploiting the similarity to obtain a nice expression for $\ln(AB)$? (It's probably not simply $\ln A + \ln B$ due to the lack of commutivity)

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Do you know if commutators of similar matrices have any simple form? If commutators of similar matrices have no simpler form due to being similar, I don't see a way that the BCH simplifies for similar matrices. –  robjohn Aug 16 '11 at 15:45
@robjohn: not that I know of. Maybe I was hoping for some non-existent free lunch here... –  Tobias Kienzler Aug 16 '11 at 15:51
I'd be somewhat surprised if there were any significant simplification. What helps is common eigenvectors; I don't see how common eigenvalues would. –  joriki Aug 17 '11 at 5:52
@joriki common Eigenvectors would imply simultaneous diagonalizability so the "usual" logarithm formula $\ln(AB)=\ln A + \ln B$ could be applied, right? But yes, I was hoping similarity would still provide ... something at least –  Tobias Kienzler Aug 17 '11 at 6:04
Yes, that's what I meant; I just wanted to emphasize the common context of those two terms, "similarity" and "simultaneous diagonalizability", one implying the same eigenvalues and the other the same eigenvectors. –  joriki Aug 17 '11 at 6:54

2 Answers 2

I would say you can simply use Baker-Campbell-Hausdorf in this way:

$$ \log AB = \log e^{\tilde A} e^{\tilde B} = \log e^{\tilde A + \tilde B + \frac{1}{2}[\tilde A,\tilde B] + \dots} = \tilde A + \tilde B + \frac{1}{2}[\tilde A,\tilde B] + \dots $$

where $\tilde A = \log A$ and $\tilde B = \log B$. I dont think you can simplify much more.

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@ Enrico , When $AB=BA$, $\tilde{A}\tilde{B}=\tilde{B}\tilde{A}$ and your formula becomes $\log(AB)=\log(A)+\log(B)$, that is false. –  loup blanc Oct 3 at 14:44
Yet, your formula is valid when $||A-I||<1$ and $||B-I||<1$. –  loup blanc Oct 3 at 15:54

I read many mistakes. Assume that $\log(.)$ is the principal logarithm, that is $\log(re^{i\theta})=\log(r)+i\theta$ where $\theta\in (-\pi,\pi)$.

  1. When we write $\log(AB)=\log(A)+\log(B)$, we implicitly assume that $A,B,AB$ have no negative eigenvalues.

  2. The previous equality is not true when $AB=BA$; in fact, it is false even when $n=1$ over $\mathbb{C}$ : $\log(e^{2i\pi/3}e^{2i\pi/3})=-2i\pi/3\not= 2i\pi/3+2i\pi/3$ and when $n=2$ over $\mathbb{R}$: cf. $\log(Rot(2\pi/3)Rot(2\pi/3))$.

  3. About the Enrico's post, $\log(e^X)\not= X$ because $e^{0_2}=exp(\begin{pmatrix}0&2\pi\\-2\pi&0\end{pmatrix})$.

EDIT. @ Tobias Kienzler.

Let $A,B\in M_n(\mathbb{C})$ s.t. $AB=BA$ and $A,B,AB$ have no eigenvalues in $(-\infty,0]$; let $spectrum(A)=(\lambda_i)_i$. Note that $A,B$ are not necessarily diagonalizable.

Proposition: There is a permutation $(\mu_i)_i$ of the eigenvalues of $B$ s.t. $\log(AB)-\log(A)-\log(B)$ is similar to $diag((\log(\lambda_i\mu_i)-\log(\lambda_i)-\log(\mu_i))_i)$.

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2's a good point - for $\mathbb C$ the identity hold $(\mod 2\pi i)$ though, maybe there is a generalization of congruency for linear operators? –  Tobias Kienzler Oct 1 at 11:31

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