In this section, all of the matrices will be real or complex matrices.

The solution to the exponential growth equation

A constant coefficient linear system has a similar form, but we have vectors and a matrix instead of scalars:

Thus, if A is an real matrix, then .

It's natural to ask whether you can solve a constant coefficient linear system using some kind of exponential, as with the exponential growth equation.

If a solution to the system is to have the same form as the growth equation solution, it should look like

But " " seems to be e raised to a matrix
power! How does that make any sense? It turns out that the * matrix exponential* can be defined in a reasonable way.

From calculus, the Taylor series for is

It converges absolutely for all z.

It A is an matrix with real entries, define

The powers make sense, since A is a square matrix. It is possible to show that this series converges for all t and every matrix A.

As a consequence, I can differentiate the series term-by-term:

This shows that solves the differential equation . The initial condition vector yields the particular solution

This works, because (by setting in the power series).

Another familiar property of ordinary exponentials holds for the matrix exponential: If A and B commute (that is, ), then

You can prove this by multiplying the power series for the exponentials on the left. ( is just with .)

* Example.* Compute if

Compute the successive powers of A:

Therefore,

You can compute the exponential of an arbitrary diagonal matrix in the same way:

For example, if

Using this idea, we can compute when A is diagonalizable. First, I'll give examples where we can compute : First, using a "lucky" pattern, and second, using our knowledge of how to solve a system of differential equations.

* Example.* Compute if

Compute the successive powers of A:

Hence,

Here's where the last equality came from:

In the last example, we got lucky in being able to recognize a pattern in the terms of the power series. Here is what happens if we aren't so lucky.

* Example.* Compute , if

If you compute powers of A as in the last two examples, there is no evident pattern:

It looks like it would be difficult to compute the matrix exponential using the power series.

I'll use the fact that is the solution to a linear system. The system's coefficient matrix is A, so the system is

You can solve this system by hand. For instance, the first equation gives

Plugging these expressions for y and into , I get

After some simplification, this becomes

The solution is

Plugging this into the expression for y above and doing some ugly algebra gives

Next, remember that if B is a matrix,

Try this out with a particular B to see how it works.

In particular, this is true for . Now is the solution satisfying , but

Set to get the first column of :

Solving this system of equations for and , I get , . So

Set to get the second column of :

Therefore, , . Hence,

Therefore,

So I found , but this was a lot of work (not all of which I wrote out!), and A was just a matrix.

We noted earlier that we can compute fairly easily in case A is diagonalizable. Recall
that an matrix A is *
diagonalizable* if it has n independent eigenvectors. (This is
true, for example, if A has n distinct eigenvalues.)

Suppose A is diagonalizable with independent eigenvectors and corresponding eigenvalues . Let S be the matrix whose columns are the eigenvectors:

Then

We saw earlier how to compute the exponential for the diagonal matrix D:

But note that

Notice how each " " pair cancels. Continuing in this way --- you can give a formal proof using induction --- we have . Therefore,

Then

Notice that S and have "switched places" from the original diagonalization equation.

Hence,

Thus, if A is diagonalizable, find the eigenvalues and use them to construct the diagonal matrix with the exponentials in the middle. Find a set of independent eigenvectors and use them to construct S and . Putting everything into the equation above gives .

* Example.* Compute if

The characteristic polynomial is and the eigenvalues are , . Since there are two different eigenvalues and A is a matrix, A is diagonalizable. The corresponding eigenvectors are and . Thus,

Hence,

* Example.* Compute if

The characteristic polynomial is and the eigenvalues are and (a double root). The corresponding eigenvectors are for , and and for . Since I have 3 independent eigenvectors, the matrix is diagonalizable.

I have

From this, it follows that

Here's a quick check you can use when computing . Plugging into gives , the identity matrix. For instance, in the last example, if you set in the right side, it checks:

However, this check isn't foolproof --- just because you get I by
setting doesn't mean your answer is right. However,
if you *don't* get I, your answer is surely wrong!

A better check that is a little more work is to compute the derivative of , and then set . You should get A. To see this, note that

Evaluating both sides of this equation at gives . Since the theory of differential
equations tells us that the solution to an initial value problem of
this kind is unique, if your answer passes these checks then it
*is* . I think it's good to do the
first (easier) check, even if you don't do the second.

If you try this in the previous example, you'll find that the second check works as well.

Unfortunately, not every matrix is diagonalizable. How do we compute
for an *arbitrary* real matrix?

One approach is to use the * Jordan canonical
form* for a matrix, but this would require a discussion of
canonical forms, a large subject in itself.

Note that any method for finding requires finding
the eigenvalues of A which is, in general, a difficult problem. For
instance, there are methods from numerical analysis for
*approximating* the eigenvalues of a matrix.

I'll describe an iterative algorithm for computing that only requires that one know the eigenvalues of
A. There are various such algorithms for computing the matrix
exponential; this one, which is due to Richard Williamson [1], seems
to me to be the easiest for hand computation. It's also possible to
implement this method using a computer algebra system like
*maxima* or *Mathematica*.

Let A be an matrix. Let be a list of the eigenvalues, with multiple eigenvalues repeated according to their multiplicity.

The last phrase means that if the characteristic polynomial is , the eigenvalue 1 is listed 3 times. So your list of eigenvalues might be . But you can list them in any order; if you wanted to show off, you could make your list . It will probably make the computations easier and less error-prone if you list the eigenvalues in some "nice" way (so either or ).

Let

Then

* Remark.* If you've seen *
convolutions* before, you might recognize that the expression for
is a convolution:

In general, the convolution of f and g is

If you haven't seen this before, don't worry: you do not need to know this! The important thing (which gives the definition of ) is the integral on the right side.

To prove that this algorithm works, I'll show that the expression on the right satisfies the differential equation . To do this, I'll need two facts about the characteristic polynomial .

1. .

2. (* Cayley-Hamilton Theorem*) .

Observe that if is the characteristic polynomial, then using the first fact and the definition of the B's,

By the Cayley-Hamilton Theorem,

I will use this fact in the proof below. First, let's see an example of the Cayley-Hamilton theorem. Let

The characteristic polynomial is . The Cayley-Hamilton theorem asserts that if you plug A into , you'll get the zero matrix.

First,

Therefore,

We've verified the Cayley-Hamilton theorem for this matrix. We'll provide a proof elsewhere. Let's give a proof of the algorithm for .

* Proof of the algorithm.* First,

Recall that the Fundamental Theorem of Calculus says that

Applying this and the Product Rule, I can differentiate to obtain

Therefore,

Expand the terms using

Making this substitution and telescoping the sum, I have

(The result (*) proved above was used in the next-to-the-last equality.) Combining the results above, I've shown that

This shows that satisfies .

Using the power series expansion, I have . So

(Remember that matrix multiplication is not commutative in general!) It follows that is a constant matrix.

Set . Since , it follows that . In addition, . Therefore, , and hence .

* Example.* Use the matrix exponential to solve

is the solution vector.

The characteristic polynomial is . You can check
that there is only one independent eigenvector, so I can't solve the
system by diagonalizing. I could use * generalized
eigenvectors* to solve the system, but I will use the matrix
exponential to illustrate the algorithm.

First, list the eigenvalues: . Since is a double root, it is listed twice.

First, I'll compute the 's:

Here are the 's:

Therefore,

As a rough check, note that setting produces the identity.)

The solution to the given initial value problem is

You can get the general solution by replacing with .

* Example.* Find if

The eigenvalues are obviously (double) and .

First, I'll compute the 's. I have , and

Next, I'll compute the 's. , and

Therefore,

* Example.* Use the matrix exponential to solve

is the solution vector.

This example will demonstrate how the algorithm for works when the eigenvalues are complex.

The characteristic polynomial is . The eigenvalues are . I will list them as .

First, I'll compute the 's. , and

Next, I'll compute the 's. , and

Therefore,

I want a *real* solution, so I'll use DeMoivre's Formula to
simplify:

Plugging these into the expression for above, I have

Notice that all the i's have dropped out! This reflects the obvious fact that the exponential of a real matrix must be a real matrix.

Finally, the general solution to the original system is

* Example.* Solve the following system using both
the matrix exponential and the eigenvector methods.

is the solution vector.

The characteristic polynomial is . The eigenvalues are .

Consider :

As this is an eigenvector matrix, it must be singular, and hence the rows must be multiples. So ignore the second row. I want a vector such that . To get such a vector, switch the and -1 and negate one of them: , . Thus, is an eigenvector.

The corresponding solution is

Take the real and imaginary parts:

The solution is

Now I'll solve the equation using the matrix exponential. The eigenvalues are . Compute the 's. , and

(Here and below, I'm cheating a little in the comparison by not showing all the algebra involved in the simplification. You need to use DeMoivre's Formula to eliminate the complex exponentials.)

Next, compute the 's. , and

Therefore,

The solution is

Taking into account some of the algebra I didn't show for the matrix exponential, I think the eigenvector approach is easier.

* Example.* Solve the following system using both
the matrix exponential and the (generalized) eigenvector methods.

is the solution vector.

I'll do this first using the generalized eigenvector method, then using the matrix exponential.

The characteristic polynomial is . The eigenvalue is (double).

Ignore the first row, and divide the second row by 2, obtaining the vector . I want such that . Swap 1 and -2 and negate the -2: I get . This is an eigenvector for .

Since I only have one eigenvector, I need a generalized eigenvector. This means I need such that

Row reduce:

This means that . Setting yields . The generalized eigenvector is .

The solution is

Next, I'll solve the system using the matrix exponential. The eigenvalues are . First, I'll compute the 's. , and

Next, compute the 's. , and

Therefore,

The solution is

In this case, finding the solution using the matrix exponential may be a little bit easier.

[1] Richard Williamson, *Introduction to differential
equations*. Englewood Cliffs, NJ: Prentice-Hall, 1986.

Copyright 2023 by Bruce Ikenaga