In Section ?? we discussed solutions to differential equations for three classes of matrices . See Table ??. We stated that in a certain sense every matrix can be thought of as a member of one of these families. In this section we show that every matrix is similar to one of the matrices in that table (see Theorem ??), where similarity is defined as follows.

Our present interest in similar matrices stems from the fact that if we know the solutions to the system of differential equations in closed form, then we know the solutions to the system of differential equations in closed form. More precisely,

Proof
Since the entries in the matrix are constants, it follows that Since is a solution to the equation, it follows that Since and , Thus is a solution to , as claimed.

Invariants of Similarity

Proof
The determinant is a function on matrices that has several important properties. Recall, in particular, from Chapter ??, Theorem ?? that for any pair of matrices and : and for any invertible matrix

Let be an invertible matrix so that . Using (??) and (??) we see that

Hence the eigenvalues of and are the same. It follows from (??) and (??) of Section ?? that the determinants and traces of and are equal.

For example, if then and A calculation shows that as stated in Lemma ??.

Similarity and Matrix Exponentials

We introduce similarity at this juncture for the following reason: if is a matrix that is similar to , then can be computed from . More precisely:

Proof
Note that for all powers of we have Next verify (??) by computing

Classification of Matrices

We now classify all matrices up to similarity.

Proof
The strategy in the proof of this theorem is to determine the and columns of by computing (in each case) for and . Note from the definition of that In addition, if is invertible, then Note that if and are linearly independent, then is invertible.

(a) Since and are assumed to be linearly independent, is invertible. So we can compute It follows that the column of is Similarly, the column of is thus verifying (a).

(b) Lemma ?? implies that and are linearly independent and hence that is invertible. Using (??), with replaced by , replaced by , and replaced by , we calculate and Thus the columns of are as desired.

(c) Let be an eigenvector and assume that is a generalized eigenvector satisfying (??). By Lemma ?? the vectors and exist and are linearly independent.

For this choice of and , compute and Thus the two columns of are:

Closed Form Solutions Using Similarity

We now use Lemma ??, Theorem ??, and the explicit solutions to the normal form equations Table ?? to find solutions for where is any matrix. The idea behind the use of similarity to solve systems of ODEs is to transform a given system into another normal form system whose solution is already known. This method is very much like the technique of change of variables used when finding indefinite integrals in calculus.

We suppose that we are given a system of differential equations and use Theorem ?? to transform by similarity to one of the normal form matrices listed in that theorem. We then solve the transformed equation, as we did in Section ?? (see Table ??), and use Lemma ?? to transform the solution back to the given system.

For example, suppose that has a complex eigenvalue with corresponding eigenvector . Then Theorem ?? states that where is an invertible matrix. Using Table ?? the general solution to the system of equations is: Lemma ?? states that is the general solution to the system. Moreover, we can solve the initial value problem by solving for and . In particular, Putting these steps together implies that

is the solution to the initial value problem.
The Example with Complex Eigenvalues Revisited

Recall the example in (??) with initial values This linear system has a complex eigenvalue with corresponding eigenvector Thus the matrix that transforms into normal form is It follows from (??) that the solution to the initial value problem is

A calculation gives
Thus the solution to (??) that we have found using similarity of matrices is identical to the solution (??) that we found by the direct method.

Solving systems with either distinct real eigenvalues or equal eigenvalues works in a similar fashion.

Exercises

Suppose that the matrices and are similar and the matrices and are similar. Show that and are also similar matrices.
Use (??) in Chapter ?? to verify that the traces of similar matrices are equal.

In Exercises ?? – ?? determine whether or not the given matrices are similar, and why.

.
.
Let so that and are similar matrices. Suppose that is an eigenvector of with eigenvalue . Show that is an eigenvector of with eigenvalue .
Which matrices are similar to ?
Compute where Check your answer using MATLAB.