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I want to know whether following are true or false:

for any given natural number $n$, $T>0$ a rational, suppose that $Q_1, \cdots, Q_n$ are $m\times m$ matrices with rational entries, $t_1, \cdots, t_n$ are positive real numbers with $t_1+\cdots+t_n=T$,

the number $\alpha \cdot \Pi_{i=1}^n e^{Q_it_i} \cdot \beta$ is

(1) NOT a rational number; (2) is NOT a rational number almost surely? Here, $\alpha$ is arow vector and $\beta$ is a column vector, both with rational entries.

I know that for n=1, (1) (and hence (2)) is true, which can be proved by Lindemann-Weierstrass theorem (note that T is a rational). However, I do not know if $n=2$.


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Got something from the answer below? – Did Aug 11 '12 at 10:08
What about if $Q$ has complex eigenvalues? – noether Jul 14 '13 at 8:28

This number may be rational, for example if $t_iQ_i=s_iQ$ for every $i$, for some given matrix $Q$, with $s_1+s_2+\cdots+s_n=0$. But in general it is not, if only for connexity reasons. Consider for example the case $n=2$ and define $x(t)=\alpha\cdot\mathrm e^{tQ_1}\cdot\mathrm e^{(T-t)Q_2}\cdot\beta$. Then $x:[0,T]\to\mathbb R$ is continuous hence if $x(0)\ne x(T)$, that is, if $\alpha\cdot\mathrm e^{TQ_2}\cdot\beta\ne\alpha\cdot\mathrm e^{TQ_1}\cdot\beta$, then $x(t)$ cannot be rational for every $t$ in $[0,T]$.

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The result you state for $n=1$ seems to be false without some additional assumptions (consider the case $Q=0$). – Did Jul 8 '12 at 17:24
thanks. Yes, I exclude those "trivial" cases, e.g., $Q=0$. – maomao Jul 8 '12 at 19:22
Could you elaborate the argument that if $x(0)\neq x(T)$, then $x(t)$ cannot be rational for every $t$. Why? just because that $x(t)$ is continuous? – maomao Jul 8 '12 at 19:27
Because the interval $(x(0),x(T))$ contains some irrational numbers. – Did Jul 8 '12 at 19:31

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