# Does there exist an $n\times n$ real matrix $A$ exist such that $e^{e^{A}} - I_n$ is singular?

I have one doubt whether an $n\times n$ real matrix $A$ exist such that $e^{e^{A}} - I_n$ is singular?

I think I have to show that 1 is the eigenvalue of $e^{e^{A}}$ in case answer is yes. But I am not finding a way to proceed with this idea. I need help with this.

Hint: What complex numbers $\lambda$ solve $e^\lambda=1$? And next: What complex numbers $\mu$ solve $e^\mu=\lambda$ for one of those values of $\lambda$? Can you arrange for $A$ to have such $\mu$ among their eigenvalues?
Clearly no such $A$ exists when $n=1$. For $n=2$ and $A=\pmatrix{\log(2\pi)&-\pi/2\\ \pi/2&\log(2\pi)}$, we have $e^{e^A}=I$. Therefore a desired $A$ exists for every $n>1$.
• are you sure? I computed your $e^{e^A}$ using Matlab and got a (numerical) answer that was nowhere near $I$. It had very large elements and was not even symmetric. – Stefan Smith Apr 23 '13 at 1:03
• @StefanSmith I am 100% sure. You have probably called the wrong function. In Matlab, matrix exponential is performed by $\mathtt{expm}$, while $\mathtt{exp}$ is the entrywise exponential function. – user1551 Apr 23 '13 at 5:46
• @StefanSmith Solve $e^B=I$ and then solve $e^A=B$. In the scalar case, $B=2\pi i$ and $A=\log(2\pi)+i\pi/2$ are two solutions, then $A$ is turned into a 2x2 real Jordan block. – user1551 Apr 26 '13 at 8:06