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I spent too many time trying to solve this problem...and finals are coming. Please help me!

I just can't see a method to do this demonstration:

"For an $A_{n \times n}$ matrix, demonstrate that a function f:$R^n$\{0} $\rightarrow$ $R$ defined by

\begin{eqnarray} f(x)=lim_{t\rightarrow \infty} \frac{1}{t}log\|e^{At}x\| \end{eqnarray}

can take at most, $n$ distinct values."

It seems to me that it's related to eigenvalues.

I tried to use $e^{At}=\sum_{k=0}^\infty \frac{t^k}{k!}A^k$, and $e^{At}=Se^JS^{-1}$, but I didn't get it out.

What would you do?

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  • $\begingroup$ shouldn't that give you the component of x in the direction of the eigenvector with the largest real part? $\endgroup$ – abel Jan 22 '15 at 14:55
  • $\begingroup$ How did you get there? $\endgroup$ – Élio Pereira Jan 22 '15 at 14:59
  • $\begingroup$ suppose $A$ has full set of eigenvctors $u_1, u_2, \cdots.$ and if $x(0) = a_1u_1 + a_2 u_2 + \cdots,$ then $x(t) = a_1e^{\lambda_1 t} + a_2e^{\lambda_2 t} + \cdots$ now compute the quotient and take the limit. $\endgroup$ – abel Jan 22 '15 at 15:03
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it is eigenvalue with the largest real part assuming that the matrix $A$ is diagonalizable. here is the reason. let $\{u_1, u_2, \cdots \}$ is a basis made up of eigenvectors of $A$ arranged such that $re(\lambda_1) > re(\lambda_2) \ge \cdots$.

suppose $x(0) = a_1u_1 + a_2u_2 + \cdots.$ then $x(t) = a_1e^{\lambda_1t} + a_2e^{\lambda_2 t} + \cdots.$

computing $|x|^2 = |a_1|^2 e^{2re(\lambda_1)} + \cdots$ and now taking the log gives you

$$\lim_{t \to \infty}\dfrac{|x(t)|}{t} = re(\lambda_1) $$

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If the matrix is diagonalizable, then this reduces to a simple matter of going into the eigenspace and proving that in the limit, the following is true:

$$\lim_{t\to \infty}\frac{1}{t}\log (\sum_i a_i e^{\lambda_i t})=\max(\lambda_i)$$ where $\max$ selects from those $i$ that have $a_i> 0$.

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  • $\begingroup$ I can see it! Thank you! $\endgroup$ – Élio Pereira Jan 22 '15 at 15:11

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