Timeline for How to show the matrix exponential is onto? And, how to create a powerseries for log that works outside B(I,1)

Current License: CC BY-SA 2.5

11 events

when toggle format	what		by	license	comment
Dec 13, 2011 at 13:17	comment	added	Pietro Majer		sorry, you are right, I was not concentrate
Apr 12, 2011 at 21:44	answer	added	Enrique Macias		timeline score: 1
Aug 26, 2010 at 13:57	answer	added	Denis Serre		timeline score: 10
Jun 28, 2010 at 19:31	vote	accept	user7133
Jun 28, 2010 at 10:33	history	edited	Victor Protsak		tags
Jun 28, 2010 at 10:27	answer	added	Victor Protsak		timeline score: 19
Jun 28, 2010 at 9:46	comment	added	Victor Protsak		I think it is a legitimate question. Pietro: 1 In the matrix case, it is not generally true that $\exp(A)\exp(B)=\exp(A+B),$ so the image of the exponential map need not be a subgroup. Indeed, the exponential map is not surjective already for $SL_2(\mathbb{C}).$ 2 Instead, one has Baker-Campbell-Hausdorff formula for $\log(\exp(A)\exp(B))$, which converges only on a certain set. Hence the question whether $\log$ can be extended beyond that set.
Jun 28, 2010 at 7:22	comment	added	Pietro Majer		Define $\log(A)$ using the operator calculus, starting with any branch of the logarithm. Check e.g. en.wikipedia.org/wiki/Holomorphic_functional_calculus. For details, you should perhaps try e.g. en.wikipedia.org/wiki/Wikipedia:Reference_desk/…. I've got the impression this is not the proper site to address your query; please check the FAQ.
Jun 28, 2010 at 7:08	comment	added	Pietro Majer		Also, $\exp:M_n(\mathbb{C}),+\to GL_n(\mathbb{C}),\cdot$ is a group homomorphism whose image is an open subgroup, that is the whole $GL_n(\mathbb{C}),$ which is connected.
Jun 28, 2010 at 6:14	comment	added	Robin Chapman		For the surjectivity of the matrix exponential, it suffices to prove that each Jordan block (with nonzero diagonal) is the exponential of a complex matrix.
Jun 28, 2010 at 6:08	history	asked	user7133	CC BY-SA 2.5