Question 7.6.1: Find a general solution of the system x′ = [9 4 0 -6 -1 0 6 ...

The characteristic equation of the coefficient matrix A in Eq. (5) is

|A – λI| = \left | \begin{matrix} 9-\lambda & 4 & 0 \\ -6 & -1-\lambda & 0 \\ 6 & 4 & 3-\lambda \end{matrix} \right |

= (3 – λ)[(9 -λ)(-1-λ)+24]

= (3 – λ)(15 – 8λ + λ²)

= (5 – λ)(3 – λ)² = 0.

Thus A has the distinct eigenvalue λ_{1} = 5 and the repeated eigenvalue λ_{2} = 3 of multiplicity k = 2.

CASE 1: λ_{1} = 5. The eigenvector equation (A – λI)v = 0, where v = [a    b    c]^{T} , is

(A – 5I)v = \left [ \begin{matrix} 4 & 4 & 0 \\ -6 & -6 & 0 \\ 6 & 4 & -2 \end{matrix} \right] \left [ \begin{matrix} a \\ b \\ c \end{matrix} \right ] = \left [ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right ] .

Each of the first two equations, 4a + 4b = 0 and -6a – 6b = 0, yields b = -a. Then the third equation reduces to 2a – 2c = 0, so that c = a. The choice a = 1 then yields the eigenvector

\textbf{v}_{1} =[1     -1       1]^{T}

associated with the eigenvalue λ_{1} = 5.

CASE 2: λ_{2} = 3. Now the eigenvector equation is

(A – 3I)v =\left [ \begin{matrix} 6 & 4 & 0 \\ -6 & -4 & 0 \\ 6 & 4 & 0 \end{matrix} \right] \left [ \begin{matrix} a \\ b \\ c \end{matrix} \right ] = \left [ \begin{matrix} 0 \\ 0 \\ 0 \end{matrix} \right ] .

so the nonzero vector \textbf{v}=[a     b    c]^{T} is an eigenvector if and only if

6a + 4b = 0;                           (6)

that is, b = -\frac{3}{2}a. The fact that Eq. (6) does not involve c means that c is arbitrary, subject to the condition v ≠ 0. If c = 1, then we may choose a = b = 0; this gives the eigenvector

\textbf{v}_{2} = [0     0     1]^{T}

associated with λ_{2} = 3. If c = 0, then we must choose a to be nonzero. For instance, if a = 2 (to avoid fractions), then b = -3, so

\textbf{v}_{3} =[2     -3       0]^{T}

is a second linearly independent eigenvector associated with the multiplicity 2 eigenvalue λ_{2} = 3. Thus we have found a complete set \textbf{v}_{1}, \textbf{v}_{2}, \textbf{v}_{3} of three eigenvectors associated with the eigenvalues 5, 3, 3. The corresponding general solution of Eq. (5) is

\textbf{x}(t) = c_{1} \textbf{v}_{1} e^{5t} + c_{2}\textbf{v}_{2} e^{3t} + c_{3} \textbf{v}_{3} e^{3t}          (7)

= c_{1} \left [ \begin{matrix} 1 \\ -1 \\ 1 \end{matrix} \right ] e^{5t} +c_{2} \left [ \begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right ] e^{3t} + c_{3} \left [ \begin{matrix} 2 \\ -3 \\ 0 \end{matrix} \right ] e^{3t}.

with scalar component functions given by

x_{1}(t) = c_{1}  e^{5t} +           2c_{3}  e^{3t} ,

x_{2}(t) = -c_{1} e^{5t}           –  3c_{3} e^{3t},

x_{3}(t) = c_{1} e^{5t}  +   c_{2} e^{3t}.