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Example Find the fundamental matrix of the following linear system.．

$\displaystyle {\bf X}^{\prime} = \left(\begin{array}{rrr} 2&-1&-1\\ 2&1&-1\\ 0&-1&1 \end{array}\right){\bf X}$

Answer

$\displaystyle \det(A - \lambda I) = \det\left(\begin{array}{rrr} 2-\lambda&-1&-... ...-1&1-\lambda \end{array}\right) = (2-\lambda)(\lambda^{2} - 2\lambda + 2) = 0. $

Then we have $\lambda = 2, 1 \pm i$.

We find the eigen vector corresponds to $2$.

$\displaystyle A - 2I$ $\displaystyle =$ $\displaystyle \left(\begin{array}{rrr} 0&-1&-1\\ 2&-1&-1\\ 0&-1&-1 \end{array... ...rrow} \left(\begin{array}{rrr} 1&-1/2&-1/2\\ 0&1&1\\ 0&0&0 \end{array}\right)$  
  $\displaystyle \stackrel{\frac{R_{2}}{2}+ R_{1}\to R_{1}} {\longrightarrow}$ $\displaystyle \left(\begin{array}{rrr} 1&0&0\\ 0&1&1\\ 0&0&0 \end{array}\right)$  

Thus, the eigen vector is ${\bf C} = \left(\begin{array}{r} 0\\ -1\\ 1 \end{array}\right)$.

We next find the eigen vector corresponds to the complex eigen values $1+i$.

$\displaystyle A - (1+i) I$ $\displaystyle =$ $\displaystyle \left(\begin{array}{rrr} 1-i&-1&-1\\ 2&-i&-1\\ 0&-1&-i \end{array}\right)$  
  $\displaystyle \stackrel{\begin{array}{c} {}_{\frac{1}{1-i} \times R_{1}\to R_{1}}\\ {}_{-(1+i)R_{1} + R_{2}\to R_{2}} \end{array}}{\longrightarrow}$ $\displaystyle \left(\begin{array}{rrr} 1&-1/1-i&-1/1-i\\ 0&1&i\\ 0&-1&-i \end{array}\right)$  
  $\displaystyle \stackrel{\begin{array}{c} {}_{\frac{R_{2}}{1-i} + R_{1}\to R_{1}}\\ {}_{R_{2} + R_{3}\to R_{3}} \end{array}}{\longrightarrow}$ $\displaystyle \left(\begin{array}{rrr} 1&0&-1\\ 0&1&i\\ 0&0&0 \end{array}\right)$  

Thus, the eigen vector is ${\bf C} = \left(\begin{array}{r} 1\\ -i\\ 1 \end{array}\right)$. Now we find the real part and the imaginary part of ${\bf C}e^{\lambda t}$.
$\displaystyle {\bf C}e^{\lambda t}$ $\displaystyle =$ $\displaystyle \left(\begin{array}{c} 1\\ -i\\ 1 \end{array}\right)e^{(1+i)t} = \left(\begin{array}{c} 1\\ -i\\ 1 \end{array}\right)e^{t}(\cos{t} + i\sin{t})$  
  $\displaystyle =$ $\displaystyle \left(\begin{array}{c} e^{t}\cos{t} + ie^{t}\sin{t}\\ e^{t}\sin{... ...}\\ - e^{t}\cos{t}\\ e^{t}\sin{t} \end{array}\right)}_{\mbox{imaginary part}}$  

Thus, the fundamental matrix is given.

$\displaystyle \Phi = \begin{pmatrix}0& e^{t}\cos{t} & e^{t}\sin{t}\\ e^{2t} & ... ...}\sin{t} & -e^{t}\cos{t}\\ -e^{2t} & e^{t}\cos{t} & e^{t}\sin{t} \end{pmatrix}$


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