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Let A be an n × n real matrix. Find necessary and sufficient conditions on A so that $e^{At}$ is bounded for all $t>0$.

I am trying to figure this out. Can anyone give me a hint? Does it have to do with the eigenvalues of A?

lulu
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MathIsHard
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1 Answers1

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Yes. It has to do with the eigenvalues of $A$. Start by answering the question when $A$ is diagonalizable over $\Bbb C$. In the general case, Jordan canonical form will help.

Ted Shifrin
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  • Thank you. I think that it is that the eigenvalues need to be negative so that $e^{A}=I+\frac{1}{1!}A + \frac{1}{2!}A^2+$... will alternate in signs and remain bounded? Is that right? – MathIsHard Sep 12 '17 at 16:56
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    Think about $e^{t\Delta}$ more explicitly, for $\Delta$ a diagonal matrix. Also remember that you may have complex eigenvalues. – Ted Shifrin Sep 12 '17 at 17:04
  • Ok darn :/ I am having a hard time on this problem... – MathIsHard Sep 12 '17 at 17:20
  • One idea I had is can't I say that $|e^A| = | \sum_{k=0}^\infty \frac{A^k}{A!}| \leq \sum_{k=0}^\infty \frac{|A^k|}{A!}=e^{|A|}$. So if I can bound the norm of A then I can bound the exponential of A? – MathIsHard Sep 12 '17 at 17:33
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    There are typos, but you can say it; it won't help, however. Start by working out the exponential of a diagonal matrix, then a diagonalizable matrix, as I already said. – Ted Shifrin Sep 12 '17 at 17:36
  • Ok. Thank you. I will try that. – MathIsHard Sep 12 '17 at 17:37