# Matrix Exponential Jordan Form Linear System

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 $$T=\begin{pmatrix} 0 & -1/s_0^2 \\ 1 & -1/s_0 \end{pmatrix}$$?

using the letter $$c,$$
$$\left( \begin{array}{rr} 1&0 \\ c&1 \end{array} \right) \left( \begin{array}{rr} 0&1 \\ -c^2&-2c \end{array} \right) \left( \begin{array}{rr} 1&0 \\ -c&1 \end{array} \right) = \left( \begin{array}{rr} -c&1 \\ 0&-c \end{array} \right)$$
• @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 $u = (G+cI) v$ – Will Jagy Jan 7 at 22:38