It is easy to see that within the disk algebra $A(D)$ $$\Delta:= \begin{pmatrix} 1&0\\z&1 \end{pmatrix}\; \begin{pmatrix} 1&1\\0&1 \end{pmatrix}= \begin{pmatrix} 1&1\\z&1+z \end{pmatrix} $$$$\Delta(z):= \begin{pmatrix} 1&0\\z&1 \end{pmatrix}\; \begin{pmatrix} 1&1\\0&1 \end{pmatrix}= \begin{pmatrix} 1&1\\z&1+z \end{pmatrix} $$ is a product of two exponential matrices. Is $\Delta$$\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?
