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 Post subject: Lyapunov norm
PostPosted: Sat, 16 Feb 2008 13:07:50 UTC 
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I want to find the Lyapunov norm for the linear system \dot x=Ax,x\in\mathbb{R}^2 with A=
\left(
\begin{array}{cc}
-\frac{1}{5}&1\\
0&-\frac{1}{5}
\end{array}
\right)
I just learned what the Lyapunov form is, but I thought I could diagonalize the matrix but it's not diagonalizable, so how do I do this?


Lyapunov norm
||\cdot||_2

||e^{tA}x||_2\leq e^{-\alpha t}||x||_2, \ \ \  t>0

for some \alpha>0


Last edited by Sarah on Sat, 16 Feb 2008 21:07:10 UTC, edited 1 time in total.

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PostPosted: Sat, 16 Feb 2008 20:29:23 UTC 
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Take a look at (5A)^2 = I.

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PostPosted: Sat, 16 Feb 2008 21:09:59 UTC 
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I am really really sorry, I made a typo!! It's twice -\frac{1}{5}

But the thing is, I actually don't really get the idea.... :(


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PostPosted: Sat, 16 Feb 2008 21:27:58 UTC 
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You can now re-write A = D + N, where D is a diagonal matrix, and N is nilpotent (N^2 = 0) and commutes with D. Then e^A = e^De^N.

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