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Problem:

Define $C=\left(\begin{matrix} a & -b \\ b & a \end{matrix}\right)$ and $\Delta=\left(\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right)$.

Define $L=\left(\begin{matrix} C & \Delta & \\ &C & \Delta & \\ & & \ldots \\ & & & C & \Delta \\ & & & & C \end{matrix}\right) \in \mathscr{M}(n \times n,\mathbb{R})$ where $\mathscr{M}(n \times n,\mathbb{R})$ denote the space of real matrix $n \times n$.

Remembering that we define $e^B=\sum_{k \geq 0}\frac{B^k}{k!}$ for $B \in \mathscr{M}(n \times n,\mathbb{R})$, how can we conlclude that there exists $M \in \mathscr{M}(n \times n,\mathbb{R})$ such that $e^M=L^2$?

I had this problem reading this answer and I would like to know if we can conclude the statement above without using such advanced (for me) arguments of algebra (i.e. Lie groups, Lie algebra and so on).

Remark: I am not sure if this post really needs the tag "lie-algebras".

Filippo Giovagnini
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Define $D$ to be the block-diagonal $D = \operatorname{diag}(C,\dots,C)$, and let $N = L - D$.

First, we find an $Q$ such that $e^{M_1} = D^{-1}L = I + D^{-1}N$. We see that $D^{-1}N$ is nilpotent of order $n/2$, just like $N$. Define $$ Q = \ln(I + D^{-1}N) = D^{-1}N - \frac 12 [D^{-1}N]^2 +\cdots + (-1)^{n/2 - 1}[D^{-1}N]^{n/2 - 1}. $$

Let $K$ be such that $e^K = C$, and let $P = \operatorname{diag}(K,\dots,K)$. It follows that $D = e^P$.

Because $P$ and $Q$ commute, we have $$ e^{P + Q} = e^P e^{Q} = D(D^{-1}L) = L. $$ So, taking $M = P + Q$ gives us $e^M = L$.

For the case of $L^2$, we simply have $L^2 = (e^M)^2 = e^{2M}$.

Ben Grossmann
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  • You thus have proven that $e^{\mathscr{M}(n \times n,\mathbb{R})}=\mathscr{GL}(n,\mathbb{R})$ or I am wrong? Is it true? – Filippo Giovagnini Jun 24 '20 at 14:23
  • @FilippoGiovagnini That is not true. For example, there is no matrix $M$ for which $$ e^M = \pmatrix{1&0\0&-1}. $$ – Ben Grossmann Jun 24 '20 at 14:27
  • I understand your proof, it is very clear. I did not understand only why can you say that there exists $K$ (where $K$ is taken from?) such that $e^K = C$. I would need that $M = P + Q \in \mathscr{M}(n \times n,\mathbb{R})$. – Filippo Giovagnini Jun 24 '20 at 15:00
  • @FilippoGiovagnini Use what you know about complex numbers. Define $$ I = \pmatrix{1&0\0&1}, \quad J = \pmatrix{0&-1\1&0}. $$ if $K = pI + qJ$, then we will have $e^{K} = C$ whenever $e^{p + qi} = a + bi$. That is, it suffices to take $p + qi$ to be a complex logarithm of $a + bi$; note that $p,q$ are real. – Ben Grossmann Jun 24 '20 at 15:10