[tex] A = \[ \left( \begin{array}{ccc}(adsbygoogle = window.adsbygoogle || []).push({});

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]?

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# Matrix exponentials

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