General solution of ODE of a constant $3\times 3$ matrix

Question

Determine the general solution of the system $y'=Ay$ , where $A$ is a constant matrix, defined by

$$A = \begin{pmatrix}-5&-8&4\\2&3&-2\\6&14&-5\end{pmatrix}$$

After attempting to find the eigenvalues of the system, I end up with eigenvalues $\lambda=-1,-3,-3$, where $-3$ has a multiplicity of $2$. Then, finding the corresponding vector for $\lambda=-1$, $$ u= \begin{pmatrix}3\\-1\\1\end{pmatrix}, $$ for $\lambda=-3$, $$v= \begin{pmatrix}-2\\1\\1\end{pmatrix}, $$

and for the second $ \lambda =-3$ the vector $$w=\begin{pmatrix}-1-2t\\\frac{1}{2}+t\\t\end{pmatrix}.$$

I understand how to get the first two eigevectors $u$ and $v$, but how did they get the third eigenvector to be in terms of $t$? Is it because it has a multiplicity of $2$, so there is a second solution that can be described in terms of a variable? If so, what is the process to finding that third eigenvector? And does this process generalize for an eigenvalue that would have perhaps a multiplicity of $3$?

The general solution of the system is

$$y(t) = C_{1}e^{-t}u + C_{2}e^{-3t}v + C_{3}e^{-3t}w$$

where $u$, $v$, and $w$ are defined above.

Answer 1

To find the generalized eigenvector $w$ corresponding to the eigenvalue $\lambda=-3$, you have to solve the linear system of equations in $(A+3I)w=v$.

This amounts to row reducing the augmented matrix: $$ \left( \begin{array}{rrr|r} -2 & -8 & 4 & -2 \\ 2 & 6 & -2 & 1 \\ 6 & 14 & -2 & 1 \\ \end{array} \right) \stackrel{RREF}{\rightsquigarrow} \left( \begin{array}{rrr|r} 1 & 0 & 2 & -1 \\ 0 & 1 & -1 & \frac{1}{2} \\ 0 & 0 & 0 & 0 \\ \end{array} \right). $$ Here, we put the augmented matrix into reduced row echelon form (RREF).

So $w=(x_1,x_2,x_3)$ must satisfy $$ x_1 + 2 x_3 =-1 \qquad \mbox{ and } \qquad x_2 -x_3 =\frac{1}{2}, $$ or $$ x_1 = -2 x_3 -1, \qquad x_2 = x_3 + \frac{1}{2}, \qquad \mbox{ and } \qquad x_3 = x_3. $$ Since $x_3$ is a free variable, let's use $t$ as the free variable instead.

So the generalized eigenvector $w$ must be of the form $$ \boxed{ w= t \begin{pmatrix} -2 \\ 1 \\ 1 \\ \end{pmatrix} + \begin{pmatrix} -1 \\ \frac{1}{2} \\ 0 \\ \end{pmatrix}, \qquad \mbox{ where } t\in \mathbb{R}. } $$ Note that if $w$ is a generalized eigenvector for the eigenvalue $\lambda=-3$, then $w$ must satisfy $$ (A+3I)w=v $$ because $(A+3I)v=0$. So multiply $A+3I$ on the left to both sides of $(A+3I)w=v$ to obtain $(A+3I)^2 w = 0$, which is precisely what it means for a vector to be a generalized eigenvector.

Answer 2

We have a defective matrix and have to resort to finding a generalized eigenvector.

Following Example 1, we can try

$$[A- \lambda I]v_2 = [A + 3I]v_2 = v_1$$

This gives us the RREF of

$$\DeclareMathOperator{RREF}{RREF}\RREF \left[\begin{array}{rrr|r} -2 & -8 & 4 & -2 \\ 2 & 6 & -2 & 1 \\ 6 & 14 & -2 & 1 \end{array}\right]= \left[\begin{array}{rrr|r} 1 & 0 & 2 & -1 \\ 0 & 1 & -1 & \dfrac{1}{2}\\ 0 & 0 & 0 & 0 \end{array}\right] \implies v_2 = \begin{bmatrix}-1-2t\\\dfrac{1}{2}+t\\t\end{bmatrix}$$

You will learn about the Jordan Normal Form and Eigenvector Chaining.