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It is easy to see that within the disk algebra $A(D)$ $$\Delta(z):= \begin{pmatrix} 1&0\\z&1 \end{pmatrix}\; \begin{pmatrix} 1&1\\0&1 \end{pmatrix}= \begin{pmatrix} 1&1\\z&1+z \end{pmatrix} $$ is a product of two exponential matrices. Is $\Delta(z)$ itself an exponential matrix of a holomorphic matrix $M(z)$ ? I don't think so, but could not come up with a proof. Is there a general method to deal with these questions?

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  • $\begingroup$ See again my answer, the upper-left entries lacked the correct sign. $\endgroup$ – Denis Serre May 2 '16 at 12:38
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$\Delta(z)$ does not have a nonpositive real eigenvalue for $|z| < 4$, so the principal branch of the logarithm is defined and analytic on a neighbourhood of its spectrum, and thus the holomorphic functional calculus produces the desired analytic logarithm $M(z)$ for $|z| < 4$.

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It seems that $\Delta(z)$ is the exponential of an holomorphic $M(z)$. Using the eigenvalues and eigenvectors of $\Delta$, I find $$M(z)=\frac\mu{\sqrt{z(z+4)}}\begin{pmatrix} -z & 2 \\ 2z & z \end{pmatrix},$$ where $\mu$ is the solution of $$\sinh\mu=\frac12\sqrt{z(z+4)}.$$ Because $\sinh^{-1}$ is an odd holomorphic function, the quotient $\frac\mu{\sqrt{z(z+4)}}$ can be chosen holomorphic, at least in a domain where $$\left|\frac14z(z+4)\right|<1,$$ because the convergence radius of the power series $$\sinh^{-1} x=x-\frac1{2\cdot3}x^3+\frac{1\cdot3}{2\cdot4\cdot5}x^5+\frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot7}x^7+\cdots$$ equals $1$. To have a positive answer to the question, we need to prove that $\sinh^{-1}$ extends holomorphically to a slightly larger domain, containing the image of $D$ under the multi-valued map $$z\mapsto\frac12\sqrt{z(z+4)}\,.$$ A good news is that this image does not reach the singularities $\pm i$ of the series above.

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To address the part of the question about a general method - seems like the Baker-Campbell-Hausdorff formula can give something tractable in this case. Denoting $e=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ and $zf=\left(\begin{smallmatrix}0&0\\z&0\end{smallmatrix}\right)$ (so that $\Delta=\operatorname{Exp}(e).\operatorname{Exp}(zf)$), one has $$ zh:=[e,zf]=z(e.f-f.e)=\left(\begin{smallmatrix}z&0\\0&-z\end{smallmatrix}\right) $$ so that the whole Lie algebra generated by $e$ and $zf$ comes out the vector space spanned by $e_n=(2z)^{n-1}e$, $f_n=2^{n-1}z^nf$, $h_n=2^{n-1}z^nh$, $n=1,2,...$, with brackets $$ [e_m,e_n]=[f_m,f_n]=[h_m,h_n]=0, $$ $$ [h_m,e_n]=2e_{m+n}, $$ $$ [h_m,f_n]=-2f_{m+n} $$ and $$ [e_m,f_n]=h_{m+n-1}. $$ In other words the Lie algebra is (almost all of the) $\mathfrak{sl}_2[z]$, its exponential map must be well studied by physicists.

In fact if I am not mistaken the BCH formula gives in this case \begin{multline*} \Delta=\operatorname{Exp}\left(e_1+f_1-\frac12h_1-\frac1{12}(e_2+f_2-\frac12h_2)+\frac1{120}(e_3+f_3-\frac12h_3)+...\right.\\\left....+(-1)^{n-1}\frac{(n-1)!}{4^{n-1}(2n-1)!!}(e_n+f_n-\frac12h_n)+...\right), \end{multline*} and since $$ e_n+f_n-\frac12h_n=\frac{(2z)^{n-1}}2\begin{pmatrix}-z&2\\2z&z\end{pmatrix}, $$ we get $$ M(z)=\frac12\sum_{n=1}^\infty(-1)^{n-1}\frac{(n-1)!}{(2 n-1)\text{!!}}\left(\frac z2\right)^{n-1}\begin{pmatrix}-z&2\\2z&z\end{pmatrix} $$ which (I think) boils down to the same $M(z)$ as in another answer.

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