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I have a matrix-valued inhomogenous linear ODE

$X' = F(t)X + G(t)$, $X(0) = I_{n \times n}$,

$F(t),G(t) \in \mathbb{R}^{n \times n}$,

and the entries of $f$ and $g$ are continuous functions. What assumptions do I need on $F$ and $G$ to ensure that $X$ is invertible on a given interval $[0,T]$? If anyone could provide a reference on the topic, I'd appreciate it.

Edit: In one dimension, it's sufficient that $G$ is positive on $[0,T]$. Could something like this carry over to higher dimensions?

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If $G=0$, then $X(t)$ will be invertible, since it is the state transition matrix for the 'unforced' equation. Otherwise, you might try looking at the topic of zeros in the control literature. – copper.hat May 8 '12 at 1:25
up vote 1 down vote accepted

Consider the associated equation for $\det X$. Using that

$$ \frac{d}{dt} \det X = \det X \operatorname{tr} (X^{-1} \frac{d}{dt}X) $$

we get that

$$ \frac{d}{dt} \det X = \det X \left[ \operatorname{tr} (X^{-1} F X) + \operatorname{tr} (X^{-1}G) \right] $$

The first term is okay, as $\operatorname{tr} (X^{-1}FX) = \operatorname{tr}(F)$. For the second term, you need to be able to effectively control $\operatorname{tr}\left( \operatorname{adj}(X)G\right)$. The direct analog of the positivity condition of one dimensions is that $\operatorname{tr}\left( \operatorname{adj}(X)G\right) \geq 0$. (Notice that in 1 dimension the matrix $\operatorname{adj}(X) = 1$ always.) This condition however requires some a priori control on the range of $X$: that $X$ has positive determinant is not sufficient to guarantee a sign on $\operatorname{tr}\left( \operatorname{adj}(X)G\right)$ for non-trivial $G$.

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Very helpful, thank you. – Simon May 8 '12 at 13:46

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