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



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


$$\exp(gXg^{-1})=g\exp(X)g^{-1},$$ 


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.


