0
$\begingroup$

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 $e^S = T.$

$\endgroup$
1
$\begingroup$

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 $T$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ what's an example of a matrix of $(R^n, R^n)$ with negative determinant? $\endgroup$ – pop Mar 5 at 6:56
  • 1
    $\begingroup$ @pop Take any matrix with a positive determinant. Multiply one row by $-1$. Done. For example $diag(-1,1,1,\dots,1)$. $\endgroup$ – Adam Latosiński Mar 5 at 7:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.