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Let $A$ and $B$ be two $n \times n$ real matrices. (In my application, $A$ and $B$ are $6\times 6$ traceless singular real matrices) I am interested in finding the smallest $T$ such that the integral $\int_0^T \exp(tA) \exp(tB)dt$ becomes a singular matrix.

If in the integrand there is only one exponential (such is the case when $A$ and $B$ commute) then there is a result by Kalmann-Ho-Narendra which states that the matrix $\int_0^T \exp(tA)dt$ is invertible if $T(\mu−\lambda) \ne 2k\pi i$ for any nonzero integer $k$, where $\mu$ and $\lambda$ are any pair of eigenvalues of $A$.

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  • $\begingroup$ You mean that the value of this integral is a singular matrix? Usually "singular integral" means something very different. $\endgroup$
    – Fedor Petrov
    Commented Aug 5, 2016 at 10:50
  • $\begingroup$ I mean a singular matrix, thanks Fedor! $\endgroup$
    – nadia
    Commented Aug 5, 2016 at 10:58
  • $\begingroup$ I would just like to remark that using the Baker-Campbell-Hausdorff formula (e.g. en.wikipedia.org/wiki/…) you can convert the integrand into the form $\exp(tf(A,B))$. This could be helpful at least in special cases, e.g. if $[A,B]$ is proportional to the unit matrix so that the Baker-Campbell-Hausdorff formula consists of only three terms. $\endgroup$
    – Andreas Rüdinger
    Commented Aug 7, 2016 at 21:12
  • $\begingroup$ Yes, I am aware of that but the commutator of $A$ and $B$ is not multiple of the identity matrix. $\endgroup$
    – nadia
    Commented Aug 8, 2016 at 7:25

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