Matrix exponentials

  • #1
[tex] A = \[ \left( \begin{array}{ccc}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \end{array} \right)\] [/tex]

eigenvalues are [tex] \lambda_{1} = -1, \ \lambda_{2} = \lambda_{3} = 1 [/tex]

[tex](A-\lambda_{1}I)u^{(1)} = 0 \ => \ u^{(1)} = \[ \left( \begin{array}{c}
-1 \\
0 \\
1 \end{array} \right)\] [/tex]

[tex] (A-\lambda_{2}I)u^{(2)} = 0 \ => u^{(1)} = \[ \left( \begin{array}{c}
1 \\
0 \\
1 \end{array} \right)\] [/tex]

[tex] (A -\lambda_{3}I)u^{(3)} = u^{(2)} \ => \
\left(
\begin{array}{ccc|c}
1&0&-1&-1\\
0&0 &0&1\\
0&0&0&0
\end{array}
\right)
[/tex]

since we cannot have 0 = 1, we can say that there is only one eigenvector for

[tex]\lambda = 1 [/tex]

which means that the Jordan form will be

[tex] \[ \left( \begin{array}{ccc}
1 & 1 & 0 \\
0 & 1 & 0 \\
0 & 0 & -1 \end{array} \right)\] [/tex]

Am I correct?

Now I need to find exp(Jt) and I'm not sure how.

If I only have 2 eigenvectors, how can I find the fund. matrix T such that

[tex]e^{At} = Te^{Jt}T^{-1} [/tex]?
 
Last edited:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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In general, to find such a function of other than a "number", use the Taylor's series

eA= I+ A+ (1/2)A2+ (1/3!)A3+ ...

If A happens to be diagonal, that's easy: An is just the diagonal matrix with exponentials of the diagonal elements of A on its diagonal.

For Jordan form, its a little more complicated. Try calculating J2, J3, ... for this particular J yourself and see if you spot a pattern.
 
  • #3
I noticed that A2k = I and A(2k+1)'s = A, but I don't know how to get eA from that.
 
  • #4
Hurkyl
Staff Emeritus
Science Advisor
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I noticed that A2k = I and A(2k+1)'s = A, but I don't know how to get eA from that.
How exactly are you having trouble summing the series?
 
  • #5
I'm just an idiot. I didn't realize that the Taylor expansions' limits came out to be cosh and sinh. I think I finished this problem. Thanks for the help so far.
 
  • #6
Hurkyl
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For the record, there's another method that can be useful. Note that you can separate that Jordan block into

[tex]
\left[ \begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array} \right]
=
\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]
+
\left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]
[/tex]

....
 
  • #7
Ah, I see, and the second matrix is nilpotent.

But still, if I got that, how would I compute

[tex]
e^{At} = Te^{Jt}T^{-1}
[/tex]

Not sure what T would be in this case.
 
  • #8
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
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Right. And you'll find the results of doing arithmetic look a lot like derivatives -- a nilpotent matrix behaves a little bit like an infinitessimal.

(Again, look to the series to get started)
 

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