3407 Repeated Eigenvalues
Repeated Eigenvalues
Again, we start with the real system
We say an eigenvalue of is repeated if it is a multiple root of the characteristic equation of ; in our case, as this is a quadratic equation, the only possible case is when is a double real root.
We need to find two linearly independent solutions to the system . We can get one solution in the usual way. Let be an eigenvector corresponding to . This is found by solving the system
This gives the solution to the system . Our problem is to find a second solution. To do this we have to distinguish two cases, called complete and defective. The first one is easier, especially in the case.
A. The complete case.
Still assuming is a real double root of the characteristic equation of , we say is a complete eigenvalue if there are two linearly independent eigenvectors and corresponding to ; i.e., if these two vectors are two linearly independent solutions to the system .
In the case, this only occurs when is a scalar matrix that is, when . In this case, , and every vector is an eigenvector. It is easy to find two independent solutions; the usual choices are
So the general solution is
Of course, we could choose any other pair of independent eigenvectors to generate the solutions, e.g.,
Remark. For and above the situation is more complicated. We will not discuss it here. The interested reader can consult, for instance, the textbook by Edwards and Penney.
B. The defective case.
If the eigenvalue is a double root of the characteristic equation, but the system has only one non-zero solution (up to constant multiples), then the eigenvalue is said to be incomplete or defective and is the unique normal mode. However, a second order system needs two independent solutions. Our experience with repeated roots in second order ODE's suggests we try multiplying our normal solution by . It turns out this doesn't quite work, but it can be fixed as follows: a second independent solution is given by
where is any vector satisfying
Fact. The equation for is guaranteed to have a solution, provided that the eigenvalue really is defective. When solving for , try setting , and solving for . If that does not work, try setting and solving for .
Remarks
1. Some people do not bother with . When they encounter the defective case (at least when ), they give up on eigenvalues, and simply solve the original system by elimination.
2. Although we will not go into it in this course, there is a well developed theory of defective matrices which gives insight into where this formula comes from. You will learn about all this when you study linear algebra.
Worked example: Defective Repeated Eigenvalue
Problem. Solve , where
.
Comments are given in italics.
Solution.
Step 0. Write down :
Step 1. Find the characteristic equation of :
Thus
Step 2. Find the eigenvalues of .
The characteristic polynomial factors: . This has a repeated root, .
As the matrix is not the identity matrix, we must be in the defective repeated root case.
Step 3. Find an eigenvector.
This is vector that must satisfy:
Check: this gives two identical equations, which is good.
The equation is . Setting gives . Thus, one eigenvector for is . All other eigenvectors for are multiples of this.
Step 4. Find : This vector must satisfy
Setting gives ; so is suitable.
Step 5. General solution.
The general solution is