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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 $n=2$.

Thanks!

  • 0
    What about if $Q$ has complex eigenvalues?2013-07-14

1 Answers 1

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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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    Because the interval $(x(0),x(T))$ contains some irrational numbers.2012-07-08