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Note: 1 or 1.5 lectures, §5.4 in [EP], §7.8 in [BD]

It may happen that a matrix has some “repeated” eigenvalues. That is, the characteristic equation may have repeated roots. As we said before, this is actually unlikely to happen for a random matrix. If we take a small perturbation of (we change the entries of slightly), we get a matrix with distinct eigenvalues. As any system we want to solve in practice is an approximation to reality anyway, it is not absolutely indispensable to know how to solve these corner cases. On the other hand, these cases do come up in applications from time to time. Furthermore, if we have distinct but very close eigenvalues, the behavior is similar to that of repeated eigenvalues, and so understanding that case will give us insight into what is going on.

Take the diagonal matrix

has an eigenvalue 3 of multiplicity 2. We call the multiplicity of the eigenvalue in the characteristic equation the algebraic multiplicity. In this case, there also exist 2 linearly independent eigenvectors, and corresponding to the eigenvalue 3. This means that the so-called geometric multiplicity of this eigenvalue is also 2.

In all the theorems where we required a matrix to have distinct eigenvalues, we only really needed to have linearly independent eigenvectors. For example, has the general solution

Let us restate the theorem about real eigenvalues. In the following theorem we will repeat eigenvalues according to (algebraic) multiplicity. So for the above matrix , we would say that it has eigenvalues 3 and 3.

Theorem 3.7.1. Suppose the matrix has real eigenvalues (not necessarily distinct), , , …, , and there are linearly independent corresponding eigenvectors , , …, . Then the general solution to can be written as

The geometric multiplicity of an eigenvalue of algebraic multiplicity is equal to the number of corresponding linearly independent eigenvectors. The geometric multiplicity is always less than or equal to the algebraic multiplicity. The theorem handles the case when these two multiplicities are equal for all eigenvalues. If for an eigenvalue the geometric multiplicity is equal to the algebraic multiplicity, then we say the eigenvalue is complete.

In other words, the hypothesis of the theorem could be stated as saying that if all the eigenvalues of are complete, then there are linearly independent eigenvectors and thus we have the given general solution.

If the geometric multiplicity of an eigenvalue is 2 or greater, then the set of linearly independent eigenvectors is not unique up to multiples as it was before. For example, for the diagonal matrix we could also pick eigenvectors and , or in fact any pair of two linearly independent vectors. The number of linearly independent eigenvectors corresponding to is the number of free variables we obtain when solving . We pick speciﬁc values for those free variables to obtain eigenvectors. If you pick diﬀerent values, you may get diﬀerent eigenvectors.

If an matrix has less than linearly independent eigenvectors, it is said to be deﬁcient. Then there is at least one eigenvalue with an algebraic multiplicity that is higher than its geometric multiplicity. We call this eigenvalue defective and the diﬀerence between the two multiplicities we call the defect.

has an eigenvalue 3 of algebraic multiplicity 2. Let us try to compute eigenvectors.

We must have that . Hence any eigenvector is of the form . Any two such vectors are linearly dependent, and hence the geometric multiplicity of the eigenvalue is 1. Therefore, the defect is 1, and we can no longer apply the eigenvalue method directly to a system of ODEs with such a coeﬃcient matrix.

The key observation we use here is that if is an eigenvalue of of algebraic multiplicity , then we can ﬁnd certain linearly independent vectors solving the equation . We will call these generalized eigenvectors.

Let us continue with the example and the equation . We have an eigenvalue of (algebraic) multiplicity 2 and defect 1. We have found one eigenvector . We have the solution

In this case, let us try (in the spirit of repeated roots of the characteristic equation for a single equation) another solution of the form

We diﬀerentiate to get

As we are assuming that is a solution, must equal , and

By looking at the coeﬃcients of and we see and . This means that

Therefore, is a solution if these two equations are satisﬁed. The second equation is satisﬁed as is an eigenvector. So, if we can ﬁnd a that solves , then we are done. This is just a bunch of linear equations to solve and we are by now very good at that. Let us solve . Write

By inspection we see that letting ( could be anything in fact) and does the job. Hence we can take . Our general solution to is

Let us check that we really do have the solution. First . Good. Now . Good.

In the example, if we plug into we ﬁnd

Furthermore, if , then is an eigenvector, a multiple of . In this case is just the zero matrix (exercise). So any vector solves and we just need a such that . Then we could use for , and for .

Note that the system has a simpler solution since is a so-called upper triangular matrix, that is every entry below the diagonal is zero. In particular, the equation for does not depend on . Mind you, not every defective matrix is triangular.

Exercise 3.7.1: Solve by ﬁrst solving for and then for independently. Check that you got the same solution as we did above.

Let us describe the general algorithm. Suppose that is an eigenvalue of multiplicity 2, defect 1. First ﬁnd an eigenvector of . That is, solves . Then, ﬁnd a vector such that

This gives us two linearly independent solutions

This machinery can also be generalized to higher multiplicities and higher defects. We will not go over this method in detail, but let us just sketch the ideas. Suppose that has an eigenvalue of multiplicity . We ﬁnd vectors such that

Such vectors are called generalized eigenvectors (then is an eigenvector). For every eigenvector we ﬁnd a chain of generalized eigenvectors through such that:

We form the linearly independent solutions Recall that is the factorial. We proceed to ﬁnd chains until we form linearly independent solutions ( is the multiplicity). You may need to ﬁnd several chains for every eigenvalue.Exercise 3.7.3: Let . a) What are the eigenvalues? b) What is/are the defect(s) of the eigenvalue(s)? c) Find the general solution of .

Exercise 3.7.4: Let . a) What are the eigenvalues? b) What is/are the defect(s) of the eigenvalue(s)? c) Find the general solution of in two diﬀerent ways and verify you get the same answer.

Exercise 3.7.5: Let . a) What are the eigenvalues? b) What is/are the defect(s) of the eigenvalue(s)? c) Find the general solution of .

Exercise 3.7.6: Let . a) What are the eigenvalues? b) What is/are the defect(s) of the eigenvalue(s)? c) Find the general solution of .

Exercise 3.7.7: Let . a) What are the eigenvalues? b) What is/are the defect(s) of the eigenvalue(s)? c) Find the general solution of .

Exercise 3.7.8: Suppose that is a matrix with a repeated eigenvalue . Suppose that there are two linearly independent eigenvectors. Show that .

Exercise 3.7.101: Let . a) What are the eigenvalues? b) What is/are the defect(s) of the eigenvalue(s)? c) Find the general solution of .

Exercise 3.7.102: Let . a) What are the eigenvalues? b) What is/are the defect(s) of the eigenvalue(s)? c) Find the general solution of .