# Invertible $T \in L(R^n, R^n)$ such that there is no $S \in L(R^n,R^n)$ with $e^S = T.$

I am having a hard time coming up with an example such that an invertible $$T \in L(R^n, R^n)$$ such that there is no $$S \in L(R^n,R^n)$$ with $$e^S = T.$$

Any matrix $$T$$ with negative determinant will do, since you have $$\det (e^S) = e^{\text{tr}(S)}>0$$ This however is only because you require $$S$$ to be a real metrix. In the space of complex matrices you can always calculate $$S=\ln T$$ for invertible $$T$$.
• what's an example of a matrix of $(R^n, R^n)$ with negative determinant? – pop Mar 5 at 6:56
• @pop Take any matrix with a positive determinant. Multiply one row by $-1$. Done. For example $diag(-1,1,1,\dots,1)$. – Adam Latosiński Mar 5 at 7:05