3

$\textbf{Problem}$

Trying to that $\exp: M_n(\mathbb{C}) \rightarrow GL(n,\mathbb{C})$ is surjective.

$\textbf{Attempt}$

I started by considering a complex matrix A $\in M_n \mathbb{C}$, to this matrix I applied the Jordan normal form so $ A = \lambda I + N $ where $I$ is the identity matrix and $N$ is an upper triangular matrix.

Thus, using the exponential map I get:

$$ M_n \mathbb{C} \rightarrow GL(n \mathbb{C}) \\ A = (\lambda I + N) \mapsto exp(\lambda I + N) $$

Expanding the exponential we get and using the following Jordan Formula:

$\textbf{Jordan Formula:}$

$$\mathbf J=\begin{pmatrix}\lambda&1&&\\&\lambda&\ddots&\\&&\ddots&1\\&&&\lambda\end{pmatrix}$$

$$f(\mathbf A)=\begin{pmatrix}f(\lambda)&f^\prime(\lambda)&\cdots&\frac{f^{(n-1)}(\lambda)}{(n-1)!}\\&f(\lambda)&\ddots&\vdots\\&&\ddots&f^\prime(\lambda)\\&&&f(\lambda)\end{pmatrix}$$

$$exp(\mathbf A)=\begin{pmatrix}exp(\lambda)&exp(\lambda)&\cdots&\frac{exp(\lambda)}{(n-1)!}\\&exp(\lambda)&\ddots&\vdots\\&&\ddots&exp(\lambda)\\&&&exp(\lambda)\end{pmatrix}$$

and I can identify this last matrix as an element of $GL(N, \mathbb C)$? If yes, is this enough to prove the surjectiveness of the map? What am I missing?

Thanks in advance!

Edit:

I used @BenGrossmann and @user1551 hint, so I defined:

$ M = log(\lambda I ) + N $ and $J = \lambda I + N $ then

$\exp(M) = \exp(\log ( \lambda I) + N) = \lambda I \exp(N) $ as N is nilpotent of degree 3 thus $exp(N) \approx I + N + \frac{1}{2} N^2$, I plug this in the above equation and I get:

$$ \exp(M) = \lambda I (\approx I + N + \frac{1}{2} N^2) $$

How can I prove that $\exp(M) = J$ from this?

RKerr
  • 332
  • 1
  • 9
  • What you need to prove is not that $\exp(A)\in GL_n(\mathbb C)$, but that every matrix in $GL_n(\mathbb C)$ is a matrix exponential. – user1551 Mar 05 '21 at 17:08
  • @RFeynman One way to proceed from what you've done is to note that $\exp(A)$ is similar to the Jordan block of the same size associated with $\exp(\lambda)$. – Ben Grossmann Mar 05 '21 at 17:15
  • @user1551 How can I do that? Can you give me a hint? – RKerr Mar 05 '21 at 17:18
  • @BenGrossmann Maybe I must be tired because I can't seem to grasp your idea, can you recommend some notes or can you give me a "taste" on how to do it? Thanks! – RKerr Mar 05 '21 at 17:20
  • As Ben says, let $B\in GL_n(\mathbb C)$ and let $J$ be its Jordan form. Let $M$ be a Jordan form that has the same block structure as $J$, but with all eigenvalues of $J$ replaced by their logarithms. Then show that the Jordan form of $e^M$ is $J$. Hence $e^M=PJP^{-1}$ for some $P$. Now $e^A=J$ where $A=P^{-1}MP$. – user1551 Mar 05 '21 at 17:24
  • @user1551 but how does that last two condition prove the surjectiveness? – RKerr Mar 05 '21 at 18:06
  • @Ruy I didn't understand that answer that's why I did this post. – RKerr Mar 05 '21 at 18:07
  • @user1551 I edited my question, can you help me please? – RKerr Mar 06 '21 at 17:14

1 Answers1

1

Let $A$ be any unital, commutative algebra over any field with characteristic zero. Denote by $$ N = \{x\in A: x^n=0, \text{ for some } n>0\} $$ and $$ U = \{y\in A: (y-1)^n=0, \text{ for some } n>0\}. $$ Since $A$ commutative, it easy easy to see that $N$ is a subalgebra (the fact that it is closed under addition follows from Newton's Binomial Theorem). Observe also that $U=1+N$.

Next consider the functions $$ \ln:U\to N, \quad\text{and}\quad \exp:N\to U, $$ given by $$ \ln(y) = -\sum_{n=1}^\infty \frac{(1-y)^n}n, \quad\text{and}\quad \exp(x) = \sum_{n=0}^\infty \frac{x^n}{n!}. $$

Despite the occurence of "$\infty $" as the upper limits of the above sums, only finitely many terms are nonzero because $1-y$ and $x$ are nilpotent elements. Notice that the fact that $N$ is an ideal guarantees that $\ln$ is indeed $N$-valued, while $\exp$ is $U$-valued.

Although it is a bit of a pain to prove it, these maps are each others inverse.

Given an invertible complex matrix $y$ whose Jordan form consists of a single Jordan block, we may write $$ y=\lambda I+x $$ where $\lambda \neq 0$ and $x$ is nilpotent. Rewritting $y$ as $$ y=\lambda (I+x/\lambda ), $$ writing $\lambda =\exp(z)$, for some complex number $z$, and working within the unital commutative algebra of matrices generated by $y$, we then have that $$ y=\lambda (I+x/\lambda ) = $$$$ = \exp(z) \exp\big (\ln(I+x/\lambda )\big ) = $$$$ = \exp\big (z+ \ln(I+x/\lambda )\big ). $$

For a general invertible matrix $y$, one may repeat the above procedure for each of its Jordan blocks.

Ruy
  • 19,160
  • 1
  • 15
  • 37
  • I don't get it, how this proves the surjectiveness of the map? – RKerr Mar 06 '21 at 16:22
  • Given an invetible matrix $U$, decompose it in Jordan blocks $U=\bigoplus_i U_i$. Then each $U_i$ is the exponential of some matrix $A_i$, so $$ U=\bigoplus_i U_i=\bigoplus_i \exp(A_i)= \exp \left(\bigoplus_i A_i\right). $$ – Ruy Mar 06 '21 at 18:46