Section 3.5 Repeated Eigenvalues
Objectives
To understand and be able to solve systems \({\mathbf x}' = A {\mathbf x}\text{,}\) where \(A\) is a \(2 \times 2\) matrix with a single eigenvalue \(\lambda\text{.}\)
Consider the following system
The characteristic polynomial of the system (3.5.1) is \(\lambda^2  6\lambda + 9\) and \(\lambda^2  6 \lambda + 9 = (\lambda  3)^2\text{.}\) This polynomial has a single root \(\lambda = 3\) with eigenvector \(\mathbf v = (1, 1)\text{.}\) There is a single straightline solution for this system (FigureÂ 3.5.1). The strategy that we used to find the general solution to a system with distinct real eigenvalues will clearly have to be modified if we are to find a general solution to a system with a single eigenvalue.
Subsection 3.5.1 Repeated Eigenvalues
The remaining case that we must consider is when the characteristic equation of a matrix \(A\) has repeated roots. The simplest such case is
The eigenvalues of \(A\) are both \(\lambda\text{.}\) Since \(A{\mathbf v} = \lambda {\mathbf v}\text{,}\) any nonzero vector in \({\mathbb R}^2\) is an eigenvector for \(\lambda\text{.}\) Thus, solutions to this system are of the form
Each solution to our system lies on a straight line through the origin and either tends to the origin if \(\lambda \lt 0\) or away from zero if \(\lambda \gt 0\text{.}\)
A more interesting case occurs if
Again, both eigenvalues are \(\lambda\text{;}\) however, there is only one linearly independent eigenvector, which we can take to be \((1, 0)\text{.}\) Therefore, we have a single straightline solution
To find other solutions, we will rewrite the system as
This is a partially coupled system (SubsectionÂ 2.4.1). If \(y \neq 0\text{,}\) the solution of the second equation is
Therefore, the first equation becomes
which is a firstorder linear differential equation with solution
Consequently, a solution to our system is
Example 3.5.2.
Consider the linear system
The matrix that corresponds to this system is
has a single eigenvalue, \(\lambda = 1\text{.}\) An eigenvector for \(\lambda\) is \(\mathbf v = (1, 0)\text{.}\) The general solution to our system is
Applying the initial conditions \(x(0) = 1\) and \(y(0) = 3\text{,}\) the solution to our initial value problem is
Notice that we have only one straightline solution (FigureÂ 3.5.3).
Activity 3.5.1. Systems with Repeated Eigenvalues.
Consider the system \(d\mathbf x/dt = A \mathbf x\text{,}\) where
(a)
Find the eigenvalues of \(A\text{.}\) There should be a single real eigenvalue \(\lambda\text{.}\)
(b)
Find the eigenvectors \(\mathbf v\) for the eigenvalues \(\lambda\text{.}\)
(c)
Find the straightline solution of \(d\mathbf x/dt = A \mathbf x\text{.}\) Plot the solution in the \(xy\)plane.
(d)
Find the general solution of \(d\mathbf x/dt = A \mathbf x\text{.}\)
(e)
Sketch several solution curves for the system \(d\mathbf x/dt = A \mathbf x\text{.}\) What do you notice about the solution curves, especially with respect to the straightline solution?
Subsection 3.5.2 Solving Systems with Repeated Eigenvalues
If the characteristic equation has only a single repeated root, there is a single eigenvalue. If this is the situation, then we actually have two separate cases to examine, depending on whether or not we can find two linearly independent eigenvectors.
Example 3.5.4.
Suppose we have the system \(\mathbf x' = A \mathbf x\text{,}\) where
The single eigenvalue is \(\lambda = 2\text{,}\) but there are two linearly independent eigenvectors, \(\mathbf v_1 = (1,0)\) and \(\mathbf v_2 = (0,1)\text{.}\) In this case our solution is
This is not too surprising since the system
is uncoupled and each equation can be solved separately.
Example 3.5.5.
Now let us consider the example \(\mathbf x' = A \mathbf x\text{,}\) where
Since the characteristic polynomial of \(A\) is \(\lambda^2  6 \lambda + 9 = (\lambda  3)^2\text{,}\) we have only a single eigenvalue \(\lambda = 3\) with eigenvector \(\mathbf v_1 = (1, 2)\text{.}\) This gives us one solution to our system, \(\mathbf x_1(t) = e^{3t}\mathbf v_1\text{;}\) however, we still need a second solution.
Since all other eigenvectors of \(A\) are a multiple of \(\mathbf v\text{,}\) we cannot find a second linearly independent eigenvector, and we need to obtain the second solution in a different manner. Furthermore, since this system is not partially coupled, we will need a more general strategy.
First, we must find a vector \({\mathbf v}_2\) such that \((A  \lambda I){\mathbf v}_2 = {\mathbf v}_1\text{.}\) To do this we can start with any nonzero vector \({\mathbf w}\) that is not a multiple of \({\mathbf v}_1\text{,}\) say \({\mathbf w} = (1, 0)\text{.}\) We then compute
Thus, we can take \({\mathbf v}_2 = (1/2)\mathbf w = (1/2, 0)\text{,}\) and our second solution is
Thus, our general solution is
If the eigenvalue is positive, we will have a nodal source. If it is negative, we will have a nodal sink. Notice that we have only given a recipe for finding a solution to \(\mathbf x' = A \mathbf x\text{,}\) where \(A\) has a repeated eigenvalue and any two eigenvectors are linearly dependent. We will justify our procedure in the next section (SubsectionÂ 3.6.6).
Activity 3.5.2. Systems with Repeated Eigenvaluesâ€”Finding a Second Solution.
Consider the system \(d\mathbf x/dt = A \mathbf x\text{,}\) where
(a)
Find the eigenvalues of \(A\text{.}\) There should be a single real eigenvalue \(\lambda\text{.}\)
(b)
Find the eigenvectors \(\mathbf v_1\) for the eigenvalues \(\lambda\text{.}\)
(c)
Find the straightline solution of \(d\mathbf x/dt = A \mathbf x\text{.}\) Plot the solution in the \(xy\)plane.
(d)
Find one solution, \(\mathbf x_1\text{,}\) of \(d\mathbf x/dt = A \mathbf x\text{.}\)
(e)
To find a second solution of \(d\mathbf x/dt = A \mathbf x\text{,}\) choose a vector \(\mathbf w\) that is not a multiple of \(\mathbf v_1\) and compute \((A  \lambda I) {\mathbf w}\text{.}\) This should give you a vector of the form \(\alpha \mathbf v_1\text{.}\) Let \(\mathbf v_2 = (1/\alpha) \mathbf w\text{.}\) The second solution is \({\mathbf x}_2 = e^{\lambda t} ({\mathbf v}_2 + t {\mathbf v}_1)\text{.}\) What is the general solution?
(f)
Sketch several solution curves for the system \(d\mathbf x/dt = A \mathbf x\text{.}\) What do you notice about the solution curves, especially with respect to the straightline solution?
Subsection 3.5.3 Important Lessons

If
\begin{equation*} A = \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}, \end{equation*}then \(A\) has one repeated real eigenvalue. The general solution to the system \({\mathbf x}' = A {\mathbf x}\) is
\begin{equation*} {\mathbf x}(t) = \alpha e^{\lambda t} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \beta e^{\lambda t} \begin{pmatrix} t \\ 1 \end{pmatrix}. \end{equation*}If \(\lambda \lt 0\text{,}\) then the solutions tend towards the origin as \(t \to \infty\text{.}\) For \(\lambda \gt 0\text{,}\) the solutions tend away from the origin.

Suppose that a system \(d\mathbf x/dt = A \mathbf x\) has a single eigenvalue with an \(\mathbf v_1\) and that all other eigenvectors are multiples of \(\mathbf v_1\text{.}\) Then one solution is \(\mathbf x_1 = e^{\lambda t} {\mathbf v}_1\text{.}\) To find a second linearly independent solution of \(d\mathbf x/dt = A \mathbf x\text{,}\) choose a vector \(\mathbf w\) that is not a multiple of \(\mathbf v_1\) and compute \((A  \lambda I) {\mathbf w}\text{.}\) This should give you a vector of the form \(\alpha \mathbf v_1\text{.}\) Let \(\mathbf v_2 = (1/\alpha) \mathbf w\text{.}\) The second solution is \({\mathbf x}_2 = e^{\lambda t} ({\mathbf v}_2 + t {\mathbf v}_1)\text{.}\) The general solution of \(d\mathbf x/dt = A \mathbf x\) will be
\begin{equation*} \mathbf x(t) = c_1 e^{\lambda t} {\mathbf v}_1+ c_2 e^{\lambda t} ({\mathbf v}_2 + t {\mathbf v}_1) \text{.} \end{equation*}
Reading Questions 3.5.4 Reading Questions
1.
Given a \(2 \times 2\) system with repeated eigenvalues, how many straightline solutions are there?
2.
Given a \(2 \times 2\) system with repeated eigenvalues, explain why it is necessary to find a second linearly independent solution.
Exercises 3.5.5 Exercises
Solving Linear Systems with Repeated Eigenvalues.
Find the general solution of each of the linear systems in Exercise GroupÂ 3.5.5.1â€“4.
1.
2.
3.
4.
Solving Initial Value Problems.
Solve each of the following linear systems for the given initial values in Exercise GroupÂ 3.5.5.5â€“8.
5.
6.
7.
8.
9.
Consider the linear system \(d \mathbf x/dt = A \mathbf x\text{,}\) where
Suppose the initial conditions for the solution curve are \(x(0) = 2\) and \(y(0) = 5\text{.}\) We can use the following Sage code to plot the phase portrait of this system, including a solution curve and the straightline solution.
Use Sage to graph the direction field for the system linear systems \(d\mathbf x/dt = A \mathbf x\) in Exercise GroupÂ 3.5.5.5â€“8. Plot the straightline solutions and the solution curve for the given initial condition.