Proving that the matrix exponential map is surjective onto the general linear group

Question

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.

Answer 1

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.

Answer 2

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 $e^A=M$.