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I'm trying to compute

$$I(t) = \int_0^t e^{ A \tau} e^{A^T \tau} \ d \tau $$

where $A$ is a real matrix, and $A^T$ its transpose.

I know that if $A$ was symmetric and $B = A + A^T$ nonsingular, I could use the rule for the Integral of matrix exponential and the result would be

$$ I(t) = B^{-1} \left( e^{t B } - I \right) $$

What if $A$ is not symmetric, though?

Rodrigo de Azevedo
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Jessica
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  • Since $A+A^T$ is symmetric it's diagonalizable (irrespective of whether it's singular or not), and then you should be able to use the fact that for a diagonlizable matrix $B$, $e^{B}=e^{UDU^T}=Ue^DU^T$. – Set Nov 27 '17 at 19:49
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    Yes, but if $A$ is not symmetric it might be that $A$ and $A^T$ do not commute. Then I cannot write the integrand as $\exp\left(A + A^T\right)$ in the first place. – Jessica Nov 27 '17 at 21:22

1 Answers1

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You must do a numerical calculation. Let $Z(t)=\int_0^t e^{\tau A}e^{\tau A^T}d\tau$.

For example, $Z(t)$ is the solution of the ODE: $AZ(t)+Z(t)A^T=Z'(t)-I_n$, s.t. $Z(0)=0$.