For a matrix $A\in M_n (\mathbb R),$ we consider the exponential $e^{tA}, t\in \mathbb R$. For $x\in \mathbb R^n\setminus\{0\},$ let $f : t\longmapsto e^{tA}x.$ My question concerns the surjectivity of the mapping $f$ from $\mathbb R$ to $\mathbb R^n$: is there a class of matrices $A$ for which $f$ is surjective for all $x\in \mathbb R^n\setminus\{0\}?$
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It seems that $t \mapsto e^{tA}x$ is continuous, hence cannot be surjective if $n>1$. – lisyarus May 19 '16 at 11:14
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1@lisyarus there are continuous surjective $\mathbb{R}\to\mathbb{R}^n$ functions: http://math.stackexchange.com/questions/357052/does-there-exists-a-continuous-surjection-from-mathbbr-to-mathbbr2 . – matb May 19 '16 at 11:21
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@matb Oh. My intuition said something like in Martin's answer, but I was completely wrong with surjectivity. Thanks. – lisyarus May 19 '16 at 11:23
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Note that $A $ will always have a nonzero eigenvector, say $Ax=\lambda x $. In that case, $f (t)=e^{\lambda t}\,x ,$ so the range of $f $ is contained in a one-dimensional subspace.
Martin Argerami
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