Multiple roots and Complex roots

In the previous section, we consider the case where the eigen equation has the distinct real roots. In this section, we consider the case of multiple roots and complex roots.

Let ${\bf X}^{\prime} = A{\bf X}$ , where the matrix $A$ is real matrix. Suppose that $\lambda = a+bi$ is the eigenvalue and ${\bf C} = {\bf C}_{1} + i{\bf C}_{2} \ ({\bf C}_{1},{\bf C}_{2} \mbox{real vector})$ is the eigenvector for $\lambda$ . Then by the eigenvalue equation

$\displaystyle (A - \lambda I){\bf C} = {\bf0}$

The conjugate of the eigenvalue equation satisfies

$\displaystyle \overline{(A - \lambda I){\bf C}} = (\bar{A} - \bar{\lambda}\bar{I})\bar{\bf C} = {\bf0}$

Thus, $\bar{\lambda} = a - bi$ is also an eigenvalue and corresponding eigenvector is $\bar{\bf C}$ . Thus, ${\bf X}_{1} = {\bf C}e^{\lambda t}$ and ${\bf X}_{2} = \bar{\bf C}e^{\bar{\lambda} t}$ are solutions of ${\bf X}^{\prime} = A{\bf X}$ . Then the linear combination of ${\bf X}_{1}$ and ${\bf X}_{2}$ is also a solution. Therefore,

$\displaystyle \Re{\bf C}e^{\lambda t} = \Re{\bf X}_{1} = \frac{{\bf X}_{1} + {\bf X}_{2}}{2} = {\bf C}_{1}e^{at}\cos{bt} - {\bf C}_{2}e^{at}\sin{bt}$

and

$\displaystyle \Im{\bf C}e^{\lambda t} = \Im{\bf X}_{1} = \frac{{\bf X}_{1} - {\bf X}_{2}}{2i} = {\bf C}_{1}e^{at}\sin{bt} - {\bf C}_{2}e^{at}\cos{bt}$

are linearly independent solutions.

Example 3..3 Solve the following differential equation

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

SOLUTION

$\displaystyle \det(A - \lambda I) = \det\left(\begin{array}{rrr} 1-\lambda&0&0\... ...ambda&-2\\ 2&2&-\lambda \end{array}\right) = (1-\lambda)(\lambda^{2} + 4) = 0$

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

For $\lambda = 1$ , we find the eigenvector.

$\displaystyle A - I$	$\displaystyle =$	$\displaystyle \left(\begin{array}{rrr} 0&0&0\\ 1&-1&-2\\ 2&2&-1 \end{array}\r... ...ghtarrow} \left(\begin{array}{rrr} 1&-1&-2\\ 0&0&0\\ 0&4&3 \end{array}\right)$
		$\displaystyle \stackrel{\begin{array}{c} {}_{\frac{1}{4}R_{2}}\\ {}_{\frac{1}{... ...arrow} \left(\begin{array}{rrr} 1&0&-5/4\\ 0&1&3/4\\ 0&0&0 \end{array}\right)$

Note that we $c_3$

can be chosen arbitrary. Thus we let $c_3 = 4$

. Then the eigenvector is ${\bf C} = \left(\begin{array}{r} 5\\ -3\\ 4 \end{array}\right)$ .

For $2i$ , we find the corresponding eigenvector

$\displaystyle A - 2i I$	$\displaystyle =$	$\displaystyle \left(\begin{array}{rrr} 1+2i&0&0\\ 1&2i&-2\\ 2&2&-2i \end{array}\right)$
	$\displaystyle \stackrel{\begin{array}{c} {}_{-2 R_{2} + R_{3}}\\ {}_{R_{1} \leftrightarrow R_{2}} \end{array}}{\longrightarrow}$	$\displaystyle \left(\begin{array}{rrr} 1&2i&-2\\ 1+2i&0&0\\ 0&2-4i&4+2i \end{array}\right)$
	$\displaystyle \stackrel{\begin{array}{c} {}_{-(1+2i)R_{1} + R_{2}}\\ {}_{\frac... ...}R_{2} + R_3}\\ {}_{\frac{-2i}{4-2i}R_{2} + R_1} \end{array}}{\longrightarrow}$	$\displaystyle \left(\begin{array}{rrr} 1&0&0\\ 0&1&i\\ 0&0&0 \end{array}\right)$

Now

is free to choose. Thus the eigenvector is ${\bf C} = \left(\begin{array}{r} 0\\ -i\\ 1 \end{array}\right)$ . Now we find the real part and imaginary part of ${\bf C}e^{\lambda t}$ Then

$\displaystyle {\bf C}e^{\lambda t}$	$\displaystyle =$	$\displaystyle \left(\begin{array}{c} 0\\ -i\\ 1 \end{array}\right)e^{-2it} = \left(\begin{array}{c} 0\\ -i\\ 1 \end{array}\right)(\cos{2t} - i\sin{2t})$
	$\displaystyle =$	$\displaystyle \left(\begin{array}{c} 0\\ \sin{2t} - i\cos{2t}\\ \cos{2t} - i\... ...ray}{c} 0\\ -\cos{2t}\\ -\sin{2t} \end{array}\right)}_{\mbox{imaginary part}}$

Thus the general solution is

$\displaystyle {\bf X}(t) = c_{1}\left(\begin{array}{c} 5\\ -3\\ -e^{2t} \end{... ...c} 0\\ -\cos{2t}\\ -\sin{2t} \end{array}\right). \ensuremath{\ \blacksquare}$

Let ${\bf X}^{\prime} = A{\bf X}$ , where the matrix $A$ is real matrix. Suppose that $\lambda$ is multiple eigenvalues and $A$ is not diagonalizable. Then cosider ${\bf X} = {\bf C}e^{At}$ . Since

$\displaystyle e^{at} = 1 + at + \frac{(at)^{2}}{2!} + \frac{(at)^{3}}{3!} + \cdots,$

we have the exponential matrix $e^{At}$ such that

$\displaystyle e^{At} = I + tA + \frac{t^{2}}{2!}A^{2} + \frac{t^{3}}{3!}A^{3} + \cdots$

Thus

$\displaystyle \frac{d}{dt}e^{At} = A + tA^{2} + \frac{t^{2}}{2!}A^{3} + \cdots = Ae^{At}$

and ${\bf X} = {\bf C}e^{At}$ is a solution of ${\bf X}^{\prime} = A{\bf X}$ .

$\displaystyle {\bf C}e^{At}$	$\displaystyle =$	$\displaystyle {\bf C}e^{(A - \lambda I)t}e^{\lambda t} = e^{\lambda t}({\bf C} + t(A - \lambda I){\bf C} + \frac{t^{2}}{2!}(A - \lambda I)^{2}{\bf C}$
	$\displaystyle +$	$\displaystyle \frac{t^{3}}{3!}(A - \lambda I)^{3}{\bf C} + \cdots + \frac{t^{k}}{k!}(A - \lambda I)^{k}{\bf C} + \cdots)$

Now note that if $k$

is the least number satisfing $(A - \lambda I)^{k}{\bf C} = 0$ , then

$\displaystyle {\bf C}e^{At} = e^{\lambda t}({\bf C} + t(A - \lambda I){\bf C} +... ... I)^{2}{\bf C} + \cdots + \frac{t^{k-1}}{(k-1)!}(A - \lambda I)^{(k-1)}{\bf C})$

Example 3..4 Solve the following differential equation

$\displaystyle \left\{\begin{array}{rl} x_{1}^{\prime} =& 2x_{1} - x_{2} + 2x_{3... ...2}^{\prime} =& 2x_{2} + 2x_{3}\\ x_{3}^{\prime} =& 2x_{3} \end{array}\right .$

SOLUTION Since

$\displaystyle \det(A - \lambda I) = \det\left(\begin{array}{rrr} 2-\lambda&-1&2\\ 0&2-\lambda&2\\ 0&0&2-\lambda \end{array}\right) = (2 - \lambda)^{3}$

then eigenvalues are $\lambda = 2$ . Now we find the eigenvector C for $\lambda = 2$ .

$\displaystyle A - 2I = \left(\begin{array}{rrr} 0&-1&2\\ 0&0&2\\ 0&0&0 \end{a... ...ghtarrow} \left(\begin{array}{rrr} 0&1&0\\ 0&0&1\\ 0&0&0 \end{array}\right)$

Thus ${\bf C} = \left(\begin{array}{c} 1\\ 0\\ 0 \end{array}\right)$ and the eigenvector is

$\displaystyle \left(\begin{array}{c} 1\\ 0\\ 0 \end{array}\right)e^{2t}$

Since the degree of the matrix $A$ is 3, we have to find three linearly independent solutions. Thus we need to find C such that

$\displaystyle (A - 2I)^{2}{\bf C} = {\bf0},\ (A - 2I){\bf C} \neq {\bf0}$

Since

$\displaystyle (A - 2I)^{2} = \left(\begin{array}{rrr} 0&0&-2\\ 0&0&0\\ 0&0&0 \end{array}\right),$

$c_{1},c_{2}$ can be chosen arbitrary. Thus we let ${\bf C} = \left(\begin{array}{c} \alpha\\ \beta\\ 0 \end{array}\right)$ . Then $(A - 2I)^{2}{\bf C} = {\bf0}$ . Now we choose $\alpha ,\beta$ such that $(A - 2I){\bf C} \neq {\bf0}$ .

$\displaystyle (A - 2I){\bf C} = \left(\begin{array}{rrr} 0&-1&2\\ 0&0&2\\ 0&0... ...nd{array}\right) = \left(\begin{array}{c} -\beta\\ 0\\ 0 \end{array}\right).$

Then we choose $\beta = 1$ . ${\bf C} = \left(\begin{array}{c} 0\\ 1\\ 0 \end{array}\right)$ and the second solution is

$\displaystyle e^{At}{\bf C}$	$\displaystyle =$	$\displaystyle e^{2t}e^{(A - 2I)t}{\bf C} = e^{2t}[{\bf C} + t(A - 2I){\bf C}] =... ...0&0&0 \end{array}\right)\left(\begin{array}{c} 0\\ 1\\ 0 \end{array}\right)\}$
	$\displaystyle =$	$\displaystyle e^{2t}\{\left(\begin{array}{c} 1\\ 1\\ 0 \end{array}\right) + \... ...array}\right)\} = e^{2t}\left(\begin{array}{c} 1-t\\ 1\\ 0 \end{array}\right)$

For the third solution, we need to find ${\bf C}$ satisfying

$\displaystyle (A - 2I)^{3}{\bf C} = {\bf0},\ (A - 2I)^{2}{\bf C} \neq {\bf0}$

Note that $(A - 2I)^{3} = O$ . Then ${\bf C} = \left(\begin{array}{c} \alpha\\ \beta\\ \gamma \end{array}\right)$ satisfies $(A - 4I)^{3}{\bf C} = {\bf0}$ . Now

$\displaystyle (A - 2I)^{2}{\bf C} = \left(\begin{array}{rrr} 0&0&-2\\ 0&0&0\\ ... ...{array}\right) = \left(\begin{array}{c} -2\gamma\\ 0\\ 0 \end{array}\right).$

Thus we choose $\gamma = 1$ . Then ${\bf C} = \left(\begin{array}{c} 0\\ 0\\ 1 \end{array}\right)$ Now calculate the third solution

$\displaystyle e^{At}{\bf C}$	$\displaystyle =$	$\displaystyle e^{2t}e^{(A - 2I)t}{\bf C}$
	$\displaystyle =$	$\displaystyle e^{2t}[{\bf C} + t(A - 2I){\bf C} + \frac{t^{2}}{2!}(A - 2I)^{2}{\bf C}]$
	$\displaystyle =$	$\displaystyle e^{2t}[\left(\begin{array}{c} 0\\ 0\\ 1 \end{array}\right) + t\... ... 0&0&0 \end{array}\right)\left(\begin{array}{c} 0\\ 0\\ 1 \end{array}\right)]$
	$\displaystyle =$	$\displaystyle e^{2t}\left(\begin{array}{c} 2t - t^{2}\\ 2t\\ 1 \end{array}\right)$

Note that $\left(\begin{array}{c} 1\\ 0\\ 0 \end{array}\right),\ \left(\begin{array}{c} ... ...array}\right),\ \left(\begin{array}{c} 2t - t^{2}\\ 2t\\ 1 \end{array}\right)$ are linearly independent. Thus

$\displaystyle {\bf X} = e^{2t}[c_{1}\left(\begin{array}{c} 1\\ 0\\ 0 \end{arr... ...{array}{c} 2t - t^{2}\\ 2t\\ 1 \end{array}\right)]\ensuremath{\ \blacksquare}$

Subsections

Exercise