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Hi, I've been looking for a clear reference which shows that the matrix exponential is surjective from $M_{n}(C)$ to $Gl_{n}(C)$. Wikipedia claims this is true, but I haven't seen it proven... Also, can someone suggest how to create a power series for a function log(x) defined for a given $A\in Gl_{n}(C)$ thats outside our standard set B(I,1)??? Specifically, what if $A=e^{B}$ with $\det(B)=0$? Thanks in advance? Tom Petrillo

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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. – Robin Chapman Jun 28 '10 at 6:14
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. – Pietro Majer Jun 28 '10 at 7:08
Define $\log(A)$ using the operator calculus, starting with any branch of the logarithm. Check e.g. For details, you should perhaps try e.g.…. I've got the impression this is not the proper site to address your query; please check the FAQ. – Pietro Majer Jun 28 '10 at 7:22
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. – Victor Protsak Jun 28 '10 at 9:46
sorry, you are right, I was not concentrate – Pietro Majer Dec 13 '11 at 13:17
up vote 16 down vote accepted

My recollection is that Rossmann's book on Lie groups has a detailed discussion of the exponential map and surjectivity issue. Matrix exponential map is equivariant under conjugation,


and, as Robin has already remarked, one can easily check that a matrix in Jordan normal form is in the image of $\exp: M_n(\mathbb{C})\to GL_n(\mathbb{C}),$ establishing surjectivity for $G=GL_n(\mathbb{C}).$

As for your second question, you can always (a) rescale (b) shift by scalar matrices and re-center the $\log$ series:

$$(a)\ \exp(nB)=\exp(B)^n\quad (b)\ \exp(B+\lambda I_n)=e^{\lambda}\exp(B).$$

For topological reasons, there isn't a canonical formula for $\log$ that works locally everywhere.

Warning:. Exponential map is not always surjective. The following family of matrices is not in the image of the exponential map from the Lie algebra $\mathfrak{g}=\mathfrak{sl}_2(\mathbb{C})$ (traceless $2\times 2$ matrices) to the Lie group $G=SL_2(\mathbb{C})$ (unit determinant $2\times 2$ matrices):

$$h_a=\begin{bmatrix} -1 & a\\ 0 & -1 \end{bmatrix},\ a\ne 0.$$

No preimage in $M_2(\mathbb{C})$ of $h_a$ under $\exp$ can have trace $0$. Indeed, if $\exp(X_a)=h_a$ then the eigenvalues of $X_a$ must be $\pi i+2\pi n, -\pi i - 2\pi m (n, m\in \mathbb{Z})$ and the trace condition implies that $n=m,$ so $X_a$ has distinct eigenvalues, hence it is diagonalizable, but $h_a$ is not — contradiction.

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Contrary to Pietro's claim, $\exp$ is not a group homomorphism on $M_n({\mathbb C})$. However, it is one, when restricting to a commutative sub-algebra ${\mathcal C}[M]$, $M$ a given matrix. This, plus some easy topological arguments, is used to prove that the exponential is surjective onto $GL_n({\mathbb C})$. See Exercise 66 in my web-site . This exercise contains in addition the result that the image of $\exp$ over $M_n({\mathbb R})$ coincides with the image of $M\mapsto M^2$ over $GL_n({\mathbb R})$.

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A proof is sketched as an exercise in Warner's book

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