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Given the Linear System $\dot{x}(t)=A x(t)$ with $x_0=(x_{01},x_{02})$ as initial state and $A=\begin{pmatrix} 0 & 1 \\ -k/M & -h/M \end{pmatrix}$, when $h^2=4Mk$ the matrix A has a single eigenvalue $s_0=s_{1,2}=-\frac{h}{2M}$ with algebraic multiplicity 2. In this case A is not diagonalizable and its Jordan Form is $J=\begin{pmatrix} s_0 & 1 \\ 0 & s_0 \end{pmatrix}$. My objective is to find the exponential matrix so that $x(t)=e^{At}x_0$ using the Jordan normal form. I see that $e^{Jt}=e^{s_0t}\begin{pmatrix} 1 & t \\ 0 & 1 \end{pmatrix}$ So how can I find a matrix $T$ such that $J=TAT^{-1}$? Why the book suggests to use $T=\begin{pmatrix} 0 & -1/s_0^2 \\ 1 & -1/s_0 \end{pmatrix}$?

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using the letter $c,$

$$ \left( \begin{array}{rr} 1&0 \\ c&1 \end{array} \right) \left( \begin{array}{rr} 0&1 \\ -c^2&-2c \end{array} \right) \left( \begin{array}{rr} 1&0 \\ -c&1 \end{array} \right) = \left( \begin{array}{rr} -c&1 \\ 0&-c \end{array} \right) $$

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    $\begingroup$ c is for Cookie and it's good enough for me. $\endgroup$ – James S. Cook Jan 7 at 19:42
  • $\begingroup$ Sorry, I cannot understand, how did you find T? $\endgroup$ – Timothy Jan 7 at 21:08
  • $\begingroup$ @Timothy let me call my matrix $G,$ I am going to construct $P$ so that $P^{-1} GP = J$ is the Jordan form. Since $(G+cI)^2 = 0,$ I chose whatever I want for the right column $v$ of $P,$ I chose $v = \left( \begin{array}{rr} 0 \\ 1 \end{array} \right)$ Then the rule just says the left column of $P$ is $u = (G+cI) v$ $\endgroup$ – Will Jagy Jan 7 at 22:38

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