MATH 2090  Systems with Defective Coefficient Matrix  Fall 2001

These examples illustrate the determination of a fundamental set of solutions for the system x⃗′ = Ax⃗ in instances where the matrix A is defective. We will consider only the case of an eigenvalue of algebraic multiplicity 3 here. A more general discussion may be found in §8.6 of the text. If λ is an eigenvalue of A of algebraic multiplicity 3, there are three possibilities:

(a) The geometric multiplicity of λ may also be 3. In this case, we can find three linearly independent eigenvectors corresponding to the eigenvalue λ. If v⃗_{1}, v⃗_{2}, and v⃗_{3} are three such eigenvectors, we obtain three linearly independent solutions, x⃗_{1}(t) = e^{λt}v⃗_{1}, x⃗_{2}(t) = e^{λt}v⃗_{2}, and x⃗_{3}(t) = e^{λt}v⃗_{3} of the system x⃗′ = Ax⃗.

(b) The geometric multiplicity of λ may be 2. In this case, we can find two linearly independent eigenvectors, say v⃗_{1} and v⃗_{2}, corresponding to the eigenvalue λ. From these, we obtain two linearly independent solutions, x⃗_{1}(t) = e^{λt}v⃗_{1} and x⃗_{2}(t) = e^{λt}v⃗_{2}, of the system x⃗′ = Ax⃗. A third solution of this system is of the form x⃗(t) = te^{λt}v⃗ + e^{λt}w⃗, where (A−λI)v⃗ = 0 and (A−λI)w⃗ = v⃗. [Check this.] These conditions say that v⃗ is an eigenvector corresponding to λ, and w⃗ is an associated generalized eigenvector. The vector v⃗ may be expressed as a linear combination, v⃗ = c_{1}v⃗_{1} + c_{2}v⃗_{2} of the two eigenvectors we started with. Some care must be taken here: we must choose v⃗ so that the system of equations (A − λI)w⃗ = v⃗ is consistent.

§8.6 #6 A = [ \begin{matrix}15 & −32 & 12 \\ 8 & −17 & 6 \\ 0 & 0 & −1 \end{matrix} ] has eigenvalues λ_{1} = λ_{2} = λ_{3} = −1 and corresponding eigenvectors v⃗_{1} = [ \begin{matrix}2 \\ 1 \\ 0 \end{matrix} ] and v⃗_{2} = [\begin{matrix}−3 \\ 0 \\ 4 \end{matrix} ]

Solutions of x⃗′ = Ax⃗ are x⃗_{1}(t) = e^{−t}v⃗_{1} = e^{−t} [ \begin{matrix}2 \\ 1 \\ 0 \end{matrix} ], x⃗_{2}(t) = e^{−t}v⃗_{2} = e^{−t} [\begin{matrix}−3 \\ 0 \\ 4 \end{matrix} ], and x⃗_{3}(t) = te^{−t}v⃗ + e^{−t}w⃗. We must

choose v⃗ = c_{1}v⃗_{1} + c_{2}v⃗_{2} = [ \begin{matrix}2c_{1} − 3c_{2} \\ c_{1} \\ 4c_{2} \end{matrix} ] so that (A + I)w⃗ = v⃗ is consistent. The augmented matrix of this system is

[\begin{matrix}16 & −32 & 12 & 2c_{1} − 3c_{2} \\ 8 & −16 & 6 & c_{1} \\ 0 & 0 & 0 & 4c_{2} \end{matrix}] ⟶ [\begin{matrix}8 & −16 & 6 & c_{1} \\ 0 & 0 & 0 & −3c_{2} \\ 0 & 0 & 0 & 4c_{2} \end{matrix}]. So the system is consistent if c_{2} = 0. If we take c_{1} = 1, then

v⃗ = v⃗_{1} = [ \begin{matrix}2 \\ 1 \\ 0 \end{matrix} ] and one solution of (A + I)w⃗ = v⃗ is w⃗ = [ \frac{1}{8} 0 0 ]. Thus x⃗_{3}(t) = te^{−t}v⃗ + e^{−t}w⃗ = te^{−t} [ \begin{matrix}2 \\ 1 \\ 0 \end{matrix} ] + e^{−t} [ \frac{1}{8} 0 0 ] is

a third solution of x⃗′ = Ax⃗. Checking that W [x⃗_{1}, x⃗_{2}, x⃗_{3}] = e^{−3t}/2 ≠ 0, we have found a fundamental set of solutions.

(c) The geometric multiplicity of λ may be 1. In this case, we can find only one (independent) eigenvector, say v⃗, corresponding to the eigenvalue λ. From this, we obtain one solution, x⃗_{1}(t) = e^{λt}v⃗, of the system x⃗′ = Ax⃗. Two additional solutions of this system are of the form x⃗_{2}(t) = te^{λt}v⃗ + e^{λt}w⃗ and x⃗_{3}(t) = \frac{t^{2}}{2} e^{λt}v⃗ + te^{λt}w⃗ + e^{λt}z⃗, where (A − λI)v⃗ = 0, (A − λI)w⃗ = v⃗, and (A − λI)z⃗ = w⃗. [Check this.] These conditions say that v⃗ is an eigenvector corresponding to λ, and that w⃗ and z⃗ are associated generalized eigenvectors.

§8.6 #8 A = [ \begin{matrix}1 & 0 & 0 \\ 0 & 3 & 2 \\ 2 & −2 & −1 \end{matrix} ] has eigenvalues λ_{1} = λ_{2} = λ_{3} = 1 and corresponding eigenvector v⃗ = [\begin{matrix}0 \\ −1 \\ 1 \end{matrix} ]

Solutions of x⃗′ = Ax⃗ are x⃗_{1}(t) = e^{t}v⃗ = e^{−t} [\begin{matrix}0 \\ −1 \\ 1 \end{matrix} ], x⃗_{2}(t) = te^{t}v⃗ + e^{t}w⃗, and x⃗_{3}(t) = \frac{t^{2}}{2} e^{t}v⃗ + te^{t}w⃗ + e^{t}z⃗.

For the eigenvector v⃗ above, one solution of the system (A−I)w⃗ = v⃗ is w⃗ = [\begin{matrix}0 \\ −1 \\ 0 \end{matrix} ]. So x⃗_{2}(t) = te^{t} [\begin{matrix}0 \\ −1 \\ 1 \end{matrix} ] + e^{t} [\begin{matrix}0 \\ −1 \\ 0 \end{matrix} ].

For w⃗ as above, one solution of the system (A−I)z⃗ = w⃗ is z⃗ = [\begin{matrix}−\frac{1}{2} \\ −\frac{1}{2} \\ 0 \end{matrix} ]. So x⃗_{3}(t) = \frac{t^{2}}{2} e^{t} [\begin{matrix}0 \\ −1 \\ 1 \end{matrix} ] + te^{t} [\begin{matrix}0 \\ −1 \\ 0 \end{matrix} ] + e^{t} [\begin{matrix}−\frac{1}{2} \\ −\frac{1}{2} \\ 0 \end{matrix} ].

Checking that W [x⃗_{1}, x⃗_{2}, x⃗_{3}] = −e^{3t}/2 ≠ 0, we have found a fundamental set of solutions.