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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
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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. –  Pietro Majer Jun 28 '10 at 7:22
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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
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3 Answers

up vote 11 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,

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

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.

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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 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 $GL_n({\mathbb R})$.

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

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