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

Question

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

Answer 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.

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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.

Answer 2

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 http://www.umpa.ens-lyon.fr/~serre/DPF/exobis.pdf . 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 $GL_n({\mathbb R})$.

Answer 3

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