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Let $M_n(\mathbb{F})$ be the set of all $n\times n$ with entries in $\mathbb{F}$ and let $\exp:M_n(\mathbb{C})\to M_n(\mathbb{C})$ be defined by $$ \exp(A)=\sum_{k=0}^{\infty}\frac{A^k}{k!},$$ for all $A\in M_n(\mathbb{C}).$

I want to prove that $\exp$ is a surjective map from $M_n(\mathbb{C})$ to $GL(n,\mathbb{C})=\left\{A\in M_n(\mathbb{C})\,\middle| \det(A)\neq0\right\}$, how do I go about that?

I mean saying that $\exp:M_n(\mathbb{R})\to GL(n,\mathbb{R})$ is an analogous to saying $\exp:\mathbb{R}\to \mathbb{R}_{>0}$ and this is also pretty intuitive, since, in analogy with the case of numbers, $A^0=I\;\forall A$, so $\exp(0)=I+0+\frac{0^2}{2!}+\dots=I$, so even for $A=0$ we get $\det\left(\exp(A)\right)\neq0$ and so because of the first term we can never get a zero determinant. But I have no idea how to prove the subjectiveness. Thanks in advance.

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  • $\begingroup$ You can probably use Jordan canonical form, and reduce the problem to showing that any Jordan block can be written as the exponential of some matrix, which is not too difficult. It's just an idea, I don't know for sure if it'll work for large Jordan blocks... $\endgroup$
    – shalop
    Commented Apr 20, 2015 at 6:03
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    $\begingroup$ NB $\exp$ is not surjective as a map $M_n({\Bbb R}) \to GL(n, {\Bbb R})$, or even as a map $M_n({\Bbb R}) \to GL_+(n, {\Bbb R})$. $\endgroup$ Commented Apr 20, 2015 at 6:09
  • $\begingroup$ Okay but it is as a map $M_n(\mathbb{C})\to GL(n,\mathbb{C})$, though right? $\endgroup$ Commented Apr 20, 2015 at 6:18
  • $\begingroup$ Yes, in that case the map is surjective. $\endgroup$ Commented Apr 20, 2015 at 6:42
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    $\begingroup$ This was asked on Math Overflow at mathoverflow.net/questions/29749/… , though the answer given there is basically the content of Shalop's hint (which, to be precise, should be restricted to Jordan blocks of nonzero eigenvalue). Of course, that hint must work, as such blocks are in $GL(m, {\Bbb C})$ for appropriate $m$. $\endgroup$ Commented Apr 20, 2015 at 6:47

2 Answers 2

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First, this result is false if $\Bbb C$ is replaced by $\Bbb R$ (even after making reasonable additional hypotheses to avoid the most obvious counterexamples) as Travis remarks, so one should use something specific for the complex numbers.

Given a linear operator $\phi$ that we wish to realise as in the image of $\exp$, one may decompose the vector space as direct sum of its generalised eigenspaces $E_\lambda$, and since $\phi$ is assumed invertible, one always has $\lambda\neq0$. If the restriction of$~\phi$ to each$~E_\lambda$ lies in the image of $\exp$ applied to the space of linear operators of$~E_\lambda$, then one can combine a choice of pre-images of $\phi|_{E_\lambda}$ to obtain a a pre-image of$~\phi$. In other words it suffices to consider the case where there is just one generalised eigenspace: so $\def\id{\mathrm{id}}\phi-\lambda\id$ is nilpotent for some $\lambda\neq0$. Since $\exp(A+B)=\exp(A)\exp(B)$ for commuting $A,B$, and multiples of $\id$ commute with anything, one may factor $\phi=\lambda\id\circ (\id-N)$ where $N=\lambda^{-1}(\lambda\id-\phi)$ is nilpotent, which reduces us to the cases of a nonzero scalar operator $\lambda\id$ on one hand, and of that of a unipotent operator $\id-N$ on the other. For the former case it suffices to choose a logarithm of the complex number$~\lambda$. For the latter case one has an algebraic formula for the logarithm: $$ \log(\id-N)=-\sum_{k=1}^{n-1}\frac{N^k}k \qquad\text{when $N^n=0$;} $$ this completes the proof.

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    $\begingroup$ I do not know if I am missing something, but I think that to end the proof you must prove that $\exp$ and $\log$ are really mutually inverse, i.e. $\exp(\log(\id-N))=\id-N$. Is there any elegant way or must you go through all the mess with subindices and summations? $\endgroup$
    – Miguel
    Commented Apr 14, 2016 at 18:20
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    $\begingroup$ @MiguelAtencia: In the ring $\Bbb Q[[X]]$ of formal power series in one indeterminate, one has the identity $\exp(\log(1-X))=1-X$ (there is some interesting combinatorial content to that fact, but it obviously must be true). Substituting for $X$ any matrix $A$ for which this makes sense, i.e., for which $\log(\id-A)$ converges, will give a valid identity of matrices, and $A=N$ is certainly such a case. $\endgroup$ Commented Apr 15, 2016 at 4:19
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    $\begingroup$ "Interesting combinatorial content" is an euphemism for "all the mess with subindices and summations", isn't it? :) $\endgroup$
    – Miguel
    Commented Apr 15, 2016 at 10:16
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    $\begingroup$ No, that was not my intention. I meant that if you interpret this obvious identity in the light of what composition of formal power series means, it gives you combinatorial facts that must be true. For sure, you can prove those things combinatorially, and it can be fun to do so, but if you hadn't started with an identity from analysis, maybe you wouldn't have come up with those facts in the first place, or you might find them unintuitive. This is one of those cases where changing point of view can be very fruitful. $\endgroup$ Commented Apr 15, 2016 at 11:48
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    $\begingroup$ @Phibetakappa Well, I talked about nilpotency after saying "it suffices to consider the case where there is just one generalised eigenspace", so that assumption makes it unnecessary to do any restriction. But the reduction to that case was indeed based on the consideration that facts about restrictions can be combined to a globally valid conclusion. $\endgroup$ Commented Dec 11, 2020 at 17:16
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I am thinking about a simple solution (perhaps simple-minded). For a diagonalizable matrix, it can be obtained by exponentiation, because for each eigenvalue $w\neq 0$, $e^z=w$ has a solution in $\mathbb{C}$. Diagonalizable matrices are dense in $GL(n,\mathbb{C})$, so for any $M \in GL(n,\mathbb{C})$, there exists a sequence of diagonalizable matrix $M_k$ that converges to $M$, and each can be written as $M_k = e^{A_k}$. Convergence imposes bound on the norm of $M_k$ hence on $A_k$, so we have a subsequence of $A_k$ converging to $A$. By continuity of the exponential map we should have $e^A=M$.

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    $\begingroup$ $A_k$ might not be bounded though; for example in $C,$ $M = 1$ and $A_k = k2\pi i $ is not bounded $\endgroup$
    – The One
    Commented Nov 5, 2021 at 18:18
  • $\begingroup$ I think you are right. I have to restrict the imaginary part of $z$ to be in $[0,2\pi)$ for the solution of $e^z=w$. $\endgroup$ Commented Nov 7, 2021 at 2:11

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