Question 7.6.4: Find three linearly independent solutions of the system x′ =...

The characteristic equation of the coefficient matrix in Eq. (25) is

|A – λI| = \left [ \begin{matrix} -λ & 1 & 2 \\ -5 & -3-λ & -7 \\ 1 & 0 & -λ \end{matrix} \right ]

=1 · [-7 – 2 · (-3 – λ)] + (-λ)[(-λ)(-3 – λ)+ 5]

= -λ³ -3λ² -3λ – 1 = -(λ + 1)³ = 0,

and thus A has the eigenvalue λ = -1 of multiplicity 3. The eigenvector equation (A – λI)v =0 for an eigenvector v = [a     b     c]^{T}is

(A + I)v = \left [ \begin{matrix} 1 & 1 & 2 \\ -5 & -2 & -7 \\ 1 & 0 & 1 \end{matrix} \right ] \left [ \begin{matrix} a \\ b \\ c \end{matrix} \right ] =\left [ \begin{matrix} 0 \\ 0 \\0 \end{matrix} \right ].

The third row a + c = 0 gives c = -a; then the first row a + b + 2c = 0 gives b = a. Thus, to within a constant multiple, the eigenvalue λ = -1 has only the single associated eigenvector v =[a        a        -a]^{T} with a ≠ 0, and so the defect of λ = -1 is 2. To apply the method described here for triple eigenvalues, we first calculate

(A + I)² = \left [ \begin{matrix} 1 & 1 & 2 \\ -5 & -2 & -7 \\ 1 & 0 & 1 \end{matrix} \right ] \left [ \begin{matrix} 1 & 1 & 2 \\ -5 & -2 & -7 \\ 1 & 0 & 1 \end{matrix} \right ]= \left [ \begin{matrix} -2 & -1 & -3 \\ -2 & -1 & -3 \\ 2 & 1 & 3 \end{matrix} \right ]

and

(A + I)³ = \left [ \begin{matrix} 1 & 1 & 2 \\ -5 & -2 & -7 \\ 1 & 0 & 1 \end{matrix} \right ] \left [ \begin{matrix} -2 & -1 & -3 \\ -2 & -1 & -3 \\ 2 & 1 & 3 \end{matrix} \right ] = \left [ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right ].

Thus any nonzero vector v_{3} will be a solution of the equation (A +I)³ v_{3} = 0. Beginning with v_{3} = [1    0    0]^{T} , for instance, we calculate

v_{2} = (A + I)v_{3} =\left [ \begin{matrix} 1 & 1 & 2 \\ -5 & -2 & -7 \\ 1 & 0 & 1 \end{matrix} \right ] \left [ \begin{matrix} 1 \\ 0 \\ 0 \end{matrix} \right ] = \left [ \begin{matrix} 1 \\ -5 \\ 1 \end{matrix} \right ] ,

v_{1} = (A + I)v_{2}=\left [ \begin{matrix} 1 & 1 & 2 \\ -5 & -2 & -7 \\ 1 & 0 & 1 \end{matrix} \right ] \left [ \begin{matrix} 1 \\ -5 \\ 1 \end{matrix} \right ] = \left [ \begin{matrix} -2 \\ -2 \\ 2 \end{matrix} \right ].

Note that v_{1} is the previously found eigenvector v with a = -2; this agreement serves as a check of the accuracy of our matrix computations. Thus we have found a length 3 chain {v_{1}, v_{2}, v_{3}} of generalized eigenvectors associated with the triple eigenvalue λ = -1. Substitution in (24) now yields the linearly independent solutions

x_{1}(t) = v_{1} e^{-t} = \left [ \begin{matrix} -2 \\ -2 \\ 2 \end{matrix} \right ]e^{-t},

x_{2}(t) = (v_{1} t + v_{2}) e^{-t} = \left [ \begin{matrix} -2t + 1 \\ -2t – 5 \\ 2t + 1 \end{matrix} \right ]e^{-t},

x_{3}(t) = (\frac{1}{2} v_{1} t^{2} + v_{2}t + v_{3})e^{-t} = \left [ \begin{matrix} -t^{2} + t + 1 \\ -t^{2} – 5t \\ t^{2} + t \end{matrix} \right ]e^{-t}

of the system x^{′} = Ax.