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Home / Finding General Solution to a Given Matrix 2x2

Finding General Solution to a Given Matrix 2x2

The Exponential of a Matrix

The solution to the exponential growth equation

$$\der x t = kx \quad \hbox{is given by} \quad x = c_0 e^{kt}.$$

It is natural to ask whether you can solve a constant coefficient linear system

$$\vec x\,' = A\vec x$$

in a similar way.

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

$$\vec x = e^{At} \vec x_0.$$

The first thing I need to do is to make sense of the matrix exponential $e^{At}$ .

The Taylor series for $e^z$ is

$$e^z = \sum_{n=0}^\infty \dfrac{z^n}{n!}.$$

It converges absolutely for all z.

It A is an $n \times n$ matrix with real entries, define

$$e^{At} = \sum_{n=0}^\infty \dfrac{t^n A^n}{n!}.$$

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

Differentiating the series term-by-term,

$$\der {} t e^{At} = \sum_{n=0}^\infty n \dfrac{t^{n-1} A^n}{n!} = \sum_{n=1}^\infty \dfrac{t^{n-1} A^n}{(n - 1)!} = A \sum_{n=1}^\infty \dfrac{t^{n-1} A^{n-1}}{(n - 1)!} = A \sum_{m=0}^\infty \dfrac{t^m A^m}{m!} = A e^{At}.$$

This shows that $e^{At}$ solves the differential equation $\vec x\,' = A\vec x$ . The initial condition vector $\vec x(0) = \vec x_0$ yields the particular solution

$$\vec x = e^{At} \vec x_0.$$

This works, because $e^{0\cdot A} = I$ (by setting $t = 0$ in the power series).

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

$$e^{A} e^{B} = e^{A+B}.$$

You can prove this by multiplying the power series for the exponentials on the left. ($e^A$ is just $e^{At}$ with $t = 1$ .)

Example.Compute $e^{At}$ if

$$A = \left[\matrix{2 & 0 \cr 0 & 3 \cr}\right].$$

Compute the successive powers of A:

$$A = \left[\matrix{2 & 0 \cr 0 & 3 \cr}\right], \quad A^2 = \left[\matrix{4 & 0 \cr 0 & 9 \cr}\right], \quad \ldots , A^n = \left[\matrix{2^n & 0 \cr 0 & 3^n \cr}\right].$$

Therefore,

$$e^{At} = \sum_{n=0}^\infty \dfrac{t^n}{n!} \left[\matrix{2^n & 0 \cr 0 & 3^n \cr}\right] = \left[\matrix{\sum_{n=0}^\infty \dfrac{(2t)^n}{n!} & 0 \cr \noalign{\vskip2pt} 0 & \sum_{n=0}^\infty \dfrac{(3t)^n}{n!} \cr}\right] = \left[\matrix{e^{2t} & 0 \cr 0 & e^{3t} \cr}\right].\quad\halmos$$

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

$$A = \left[\matrix{\lambda_1 & 0 & \cdots & 0 \cr 0 & \lambda_2 & \cdots & 0 \cr \vdots & \vdots & & \vdots \cr 0 & 0 & \cdots & \lambda_n \cr}\right], \quad e^{At} = \left[\matrix{e^{\lambda_1 t} & 0 & \cdots & 0 \cr 0 & e^{\lambda_2 t} & \cdots & 0 \cr \vdots & \vdots & & \vdots \cr 0 & 0 & \cdots & e^{\lambda_n t} \cr}\right].\quad\halmos$$

Example. Compute $e^{At}$ if

$$A = \left[\matrix{1 & 2 \cr 0 & 1 \cr}\right].$$

Compute the successive powers of A:

$$A = \left[\matrix{1 & 2 \cr 0 & 1 \cr}\right], \quad A^2 = \left[\matrix{1 & 4 \cr 0 & 1 \cr}\right], \quad A^3 = \left[\matrix{1 & 6 \cr 0 & 1 \cr}\right], \quad \ldots , A^n = \left[\matrix{1 & 2n \cr 0 & 1 \cr}\right].$$

Hence,

$$e^{A} = \sum_{n=0}^\infty \dfrac{t^n}{n!} \left[\matrix{1 & 2n \cr 0 & 1 \cr}\right] = \left[\matrix{\sum_{n=0}^\infty \dfrac{t^n}{n!} & \sum_{n=0}^\infty \dfrac{2n t^n}{n!} \cr \noalign{\vskip2pt} 0 & \sum_{n=0}^\infty \dfrac{t^n}{n!} \cr}\right] = \left[\matrix{e^t & 2t e^t \cr 0 & e^t \cr}\right].$$

Here's where the last equality came from:

$$\sum_{n=0}^\infty \dfrac{t^n}{n!} = e^t,$$

$$\sum_{n=0}^\infty \dfrac{2n t^n}{n!} = 2t \sum_{n=1}^\infty \dfrac{t^{n-1}}{(n - 1)!} = 2t \sum_{m=0}^\infty \dfrac{t^m}{m!} = 2t e^t.\quad\halmos$$

Example. Compute $e^{At}$ , if

$$A = \left[\matrix{3 & -10 \cr 1 & -4 \cr}\right].$$

If you compute powers of A as in the last two examples, there is no evident pattern. Therefore, it would be difficult to compute the exponential using the power series.

Instead, set up the system whose coefficient matrix is A:

$$x' = 3x - 10y,$$

$$y' = x - 4y.$$

The solution is

$$x = c_1 e^t + c_2 e^{-2t}, \quad y = \dfrac{1}{5} c_1 e^t + \dfrac{1}{2} c_2 e^{-2t}.$$

Next, note that if B is a $2 \times 2$ matrix,

$$B \left[\matrix{1 \cr 0 \cr}\right] = \hbox{first column of B} \quad\hbox{and}\quad B \left[\matrix{0 \cr 1 \cr}\right] = \hbox{second column of B}.$$

In particular, this is true for $e^{At}$ . Now

$$\vec x = e^{At} \vec x_0$$

is the solution satisfying $\vec x(0) = \vec x_0$ , but

$$\vec x = \left[\matrix{c_1 e^t + c_2 e^{-2t} \cr \noalign{\vskip2pt} \dfrac{1}{5} c_1 e^t + \dfrac{1}{2} c_2 e^{-2t} \cr}\right].$$

Set $\vec x(0) = (1, 0)$ to get the first column of $e^{At}$ :

$$\left[\matrix{1 \cr 0 \cr}\right] = \left[\matrix{c_1 + c_2 \cr \noalign{\vskip2pt} \dfrac{1}{5} c_1 + \dfrac{1}{2} c_2 \cr}\right].$$

Hence, $c_1 = \dfrac{5}{3}$ , $c_2 = -\dfrac{2}{3}$ . So

$$\left[\matrix{x \cr y \cr}\right] = \left[\matrix{\dfrac{5}{3} e^t - \dfrac{2}{3} e^{-2t} \cr \noalign{\vskip2pt} \dfrac{1}{3} e^t - \dfrac{1}{3} e^{-2t} \cr}\right].$$

Set $\vec x(0) = (0, 1)$ to get the second column of $e^{At}$ :

$$\left[\matrix{0 \cr 1 \cr}\right] = \left[\matrix{c_1 + c_2 \cr \noalign{\vskip2pt} \dfrac{1}{5} c_1 + \dfrac{1}{2} c_2 \cr}\right].$$

Therefore, $c_1 = -\dfrac{10}{3}$ , $c_2 = \dfrac{10}{3}$ . Hence,

$$\left[\matrix{x \cr y \cr}\right] = \left[\matrix{-\dfrac{10}{3} e^t + \dfrac{10}{3} e^{-2t} \cr \noalign{\vskip2pt} -\dfrac{2}{3} e^t + \dfrac{5}{3} e^{-2t} \cr}\right].$$

Therefore,

$$e^{At} = \left[\matrix{\dfrac{5}{3} e^t - \dfrac{2}{3} e^{-2t} & -\dfrac{10}{3} e^t + \dfrac{10}{3} e^{-2t} \cr \noalign{\vskip2pt} \dfrac{1}{3} e^t - \dfrac{1}{3} e^{-2t} & -\dfrac{2}{3} e^t + \dfrac{5}{3} e^{-2t} \cr}\right].$$

I found $e^{A t}$ , but I had to solve a system of differential equations in order to do it.

In some cases, it's possible to use linear algebra to compute the exponential of a matrix. An $n \times n$ 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 $\vec v_1, \ldots, \vec v_n$ and corresponding eigenvalues $\lambda_1, \ldots, \lambda_n$ . Let S be the matrix whose columns are the eigenvectors:

$$S = \left[\matrix{ \uparrow & \uparrow & & \uparrow \cr \vec v_1 & \vec v_2 & \cdots & \vec v_n \cr \downarrow & \downarrow & & \downarrow \cr}\right].$$

Then

$$S^{-1} A S = \left[\matrix{ \lambda_1 & 0 & \cdots & 0 \cr 0 & \lambda_2 & \cdots & 0 \cr \vdots & \vdots & & \vdots \cr 0 & 0 & \cdots & \lambda_n \cr}\right] = D.$$

As I observed above,

$$e^{D t} = \left[\matrix{ e^{\lambda_1 t} & 0 & \cdots & 0 \cr 0 & e^{\lambda_2 t} & \cdots & 0 \cr \vdots & \vdots & & \vdots \cr 0 & 0 & \cdots & e^{\lambda_n t} \cr}\right].$$

On the other hand, since $(S^{-1} A S)^n = S^{-1} A^n S$ ,

$$e^{D t} = \sum_{n=0}^\infty \dfrac{t^n (S^{-1} A S)^n}{n!} = S^{-1} \left(\sum_{n=0}^\infty \dfrac{t^n A^n}{n!}\right) S = S^{-1} e^{At} S.$$

Hence,

$$e^{A t} = S \left[\matrix{ e^{\lambda_1 t} & 0 & \cdots & 0 \cr 0 & e^{\lambda_2 t} & \cdots & 0 \cr \vdots & \vdots & & \vdots \cr 0 & 0 & \cdots & e^{\lambda_n t} \cr}\right] S^{-1}.$$

I can use this approach to compute $e^{A t}$ in case A is diagonalizable.

Example. Compute $e^{At}$ if

$$A = \left[\matrix{3 & 5 \cr 1 & -1 \cr}\right].$$

The eigenvalues are $\lambda = $ , $\lambda = -2$ . Since there are two different eigenvalues and A is a $2 \time 2$ matrix, A is diagonalizable. The corresponding eigenvectors are $(5, 1)$ and $(-1, 1)$ . Thus,

$$S = \left[\matrix{ 5 & -1 \cr 1 & 1 \cr}\right], \quad S^{-1} = \dfrac{1}{6} \left[\matrix{ 1 & 1 \cr -1 & 5 \cr}\right].$$

Hence,

$$e^{A t} = \left[\matrix{5 & -1 \cr 1 & 1 \cr}\right] \left[\matrix{e^{4 t} & 0 \cr 0 & e^{-2 t} \cr}\right] \left(\dfrac{1}{6}\right) \left[\matrix{1 & 1 \cr -1 & 5 \cr}\right] = \dfrac{1}{6} \left[\matrix{5e^{4 t} + e^{-2 t} & 5e^{4 t} - 5e^{-2 t} \cr e^{4 t} - e^{-2 t} & e^{4 t} + 5e^{-2 t} \cr}\right].\quad\halmos$$

Example. Compute $e^{A t}$ if

$$A = \left[\matrix{ 5 & -6 & -6 \cr -1 & 4 & 2 \cr 3 & -6 & -4 \cr}\right].$$

The eigenvalues are $\lambda = 1$ and $\lambda = 2$ (double). The corresponding eigenvectors are $(3, -1, 3)$ for $\lambda = 1$ , and $(2, 1, 0)$ and $(2, 0, 1)$ for $\lambda = 2$ . Since I have 3 independent eigenvectors, the matrix is diagonalizable.

I have

$$S = \left[\matrix{ 3 & 2 & 2 \cr -1 & 1 & 0 \cr 3 & 0 & 1 \cr}\right], \quad S^{-1} = \left[\matrix{ -1 & 2 & 2 \cr -1 & 3 & 2 \cr 3 & -6 & -5 \cr}\right].$$

From this, it follows that

$$e^{A t} = \left[\matrix{ -3e^t + 4e^{2t} & 6e^t - 6e^{2t} & 6e^t - 6e^{2t} \cr e^t - e^{2t} & -2e^t + 3e^{2t} & -2e^t + 2e^{2t} \cr -3e^t + 3e^{2t} & 6e^t - 6e^{2t} & 6e^t - 5e^{2t} \cr}\right].$$

Here's a quick check on the computation: If you set $t = 0$ in the right side, you get

$$\left[\matrix{1 & 0 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 \cr}\right].$$

This checks, since $e^{A \cdot 0} = I$ .

Note that this check isn't foolproof --- just because you get I by setting $t = 0$ doesn't mean your answer is right. However, if you don't get I, your answer is surely wrong!

How do you compute $e^{A t}$ is A is not diagonalizable?

I'll describe an iterative algorithm for computing $e^{At}$ that only requires that one know the eigenvalues of A. There are various algorithms for computing the matrix exponential; this one, which is due to Williamson [1], seems to me to be the easiest for hand computation.

(Note that finding the eigenvalues of a matrix is, in general, a difficult problem: Any method for finding $e^{A t}$ will have to deal with it.)

Let A be an $n \times n$ matrix. Let $\{\lambda_1, \lambda_2, \ldots, \lambda_n\}$ be a list of the eigenvalues, with multiple eigenvalues repeated according to their multiplicity.

Let

$$\eqalign{ a_1 &= e^{\lambda_1 t},\cr a_k = e^{\lambda_k t} \star a_{k-1}(t) = \int_0^t & e^{\lambda_k (t-u)} a_{k-1}(u)\,du, \quad k = 2, \ldots, n,\cr & \cr B_1 &= I,\cr B_k = (A - \lambda_{k-1} I) & \cdot B_{k-1}, \quad k = 2, \ldots, n,\cr}$$

Then

$$e^{A t} = a_1 B_1 + a_2 B_2 + \ldots + a_n B_n.$$

To prove this, I'll show that the expression on the right satisfies the differential equation $\vec x\,' = A\vec x$ . To do this, I'll need two facts about the characteristic polynomial $p(x)$ .

1. $(x - \lambda_1)(x - \lambda_2)\cdots (x - \lambda_n) = \pm p(x)$ .

2. ( Cayley-Hamilton Theorem) $p(A) = 0$ .

Observe that if $p(x)$ is the characteristic polynomial, then using the first fact and the definition of the B's,

$$\eqalign{ p(x) & = \pm (x - \lambda_1)(x - \lambda_2) \cdots (x - \lambda_n) \cr p(A) & = \pm (A - \lambda_1 I)(A - \lambda_2 I) \cdots (A - \lambda_n I) \cr & = \pm I(A - \lambda_1 I)(A - \lambda_2 I) \cdots (A - \lambda_n I) \cr & = \pm B_1 (A - \lambda_1 I)(A - \lambda_2 I) \cdots (A - \lambda_n I) \cr & = \pm B_2 (A - \lambda_2 I) \cdots (A - \lambda_n I) \cr & \vdots \cr & = \pm B_n (A - \lambda_n I) \cr}$$

By the Cayley-Hamilton Theorem,

$$\pm B_n (A - \lambda_n I) = 0. \eqno{(*)}$$

I will use this fact in the proof below.

Example. I'll illustrate the Cayley-Hamilton theorem with the matrix

$$A = \left[\matrix{2 & 3 \cr 2 & 1 \cr}\right].$$

The characteristic polynomial is $(2 - \lambda)(1 - \lambda) - 6 = \lambda^2 - 3\lambda - 4$ . The Cayley-Hamilton theorem asserts that if you plug A into $\lambda^2 - 3\lambda - 4$ , you'll get the zero matrix.

First,

$$A^2 = \left[\matrix{2 & 3 \cr 2 & 1 \cr}\right] \left[\matrix{2 & 3 \cr 2 & 1 \cr}\right] = \left[\matrix{10 & 9 \cr 6 & 7 \cr}\right].$$

Therefore,

$$A^2 - A - 4I = \left[\matrix{10 & 9 \cr 6 & 7 \cr}\right] - \left[\matrix{6 & 9 \cr 6 & 3 \cr}\right] - \left[\matrix{4 & 0 \cr 0 & 4 \cr}\right] = \left[\matrix{0 & 0 \cr 0 & 0 \cr}\right].\quad\halmos$$

Proof of the algorithm. First,

$$a_k = \int_0^t e^{\lambda_k (t-u)} a_{k-1}(u)\,du = e^{\lambda_k t} \int_0^t e^{-\lambda_k u} a_{k-1}(u)\,du.$$

Recall that the Fundamental Theorem of Calculus says that

$$\der {} t \int _0^t f(u)\,du = f(t).$$

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

$$a_k' = \lambda_k e^{\lambda_k t} \int_0^t e^{-\lambda_k u} a_{k-1}(u)\,du + e^{\lambda_k t} e^{-\lambda_k t} a_{k-1}(t),$$

$$a_k' = \lambda_k a_k + a_{k-1}.$$

Therefore,

$$\eqalign{ (a_1 B_1 + a_2 B_2 + &\ldots + a_n B_n)' = \cr \lambda_1 a_1 B_1 &+ \cr \lambda_2 a_2 B_2 &+ a_1 B_2 + \cr \lambda_3 a_3 B_3 &+ a_2 B_3 + \cr &\vdots \cr \lambda_n a_n B_n &+ a_{n-1} B_n.}$$

Expand the $a_{i-1} B_i$ terms using

$$a_{i-1} B_i = a_{i-1}(A - \lambda_{i-1}I)B_{i-1} = a_{i-1} A B_{i-1} - \lambda_{i-1} a_{i-1} B_{i-1}.$$

Making this substitution and telescoping the sum, I have

$$\eqalign{ \lambda_1 a_1 B_1 &+ \cr \lambda_2 a_2 B_2 &+ a_1 A B_1 - \lambda_1 a_1 B_1 + \cr \lambda_3 a_3 B_3 &+ a_2 A B_2 - \lambda_2 a_2 B_2 + \cr &\vdots \cr \lambda_n a_n B_n &+ a_{n-1} A B_{n-1} - \lambda_{n-1} a_{n-1} B_{n-1} = \cr \lambda_n a_n B_n &+ A(a_1 B_1 + a_2 B_2 + \ldots + a_{n-1} B_{n-1}) = \cr \lambda_n a_n B_n - A a_n B_n &+ A(a_1 B_1 + a_2 B_2 + \ldots + a_n B_n) = \cr -a_n(A - \lambda_nI)B_n &+ A(a_1 B_1 + a_2 B_2 + \ldots + a_n B_n) \cr -a_n \cdot 0 &+ A(a_1 B_1 + a_2 B_2 + \ldots + a_n B_n) = \cr & A(a_1 B_1 + a_2 B_2 + \ldots + a_n B_n) \cr}$$

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

$$(a_1 B_1 + a_2 B_2 + \ldots + a_n B_n)' = A(a_1 B_1 + a_2 B_2 + \ldots + a_n B_n).$$

This shows that $M = a_1 B_1 + a_2 B_2 + \ldots + a_n B_n$ satisfies $M' = AM$ .

Using the power series expansion, I have $e^{-t A} A = A e^{-t A}$ . So

$$(e^{-tA} M)' = -A e^{-t A} M + e^{-t A} A M = -e^{-t A} A M + e^{-t A} A M = 0.$$

(Remember that matrix multiplication is not commutative in general!) It follows that $e^{-t A} M$ is a constant matrix.

Set $t = 0$ . Since $a_2 = \cdots = a_n = 0$ , it follows that $M(0) = I$ . In addition, $e^{-0\cdot A} = I$ . Therefore, $e^{-tA} M = I$ , and hence $M = e^{A t}$ .

Example. Use the matrix exponential to solve

$$\vec x\,' = \left[\matrix{3 & -1 \cr 1 & 1 \cr}\right] \vec x, \qquad \vec x(0) = \left[\matrix{3 \cr 4 \cr}\right].$$

The characteristic polynomial is $(\lambda - 2)^2$ . 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: $\{2, 2\}$ . Since $\lambda = 2$ is a double root, it is listed twice.

First, I'll compute the $a_k$ 's:

$$a_1 = e^{2t},$$

$$a_2 = e^{2t} \star a_1(t) = \int_0^t e^{2(t-u)} e^{2u}\,du = e^{2t} \int_0^t du = te^{2t}.$$

Here are the $B_k$ 's:

$$B_1 = I, \quad B_2 = (A - 2I) B_1 = A - 2I = \left[\matrix{1 & -1 \cr 1 & -1 \cr}\right].$$

Therefore,

$$e^{At} = e^{2t} \left[\matrix{1 & 0 \cr 0 & 1 \cr}\right] + te^{2t} \left[\matrix{1 & -1 \cr 1 & -1 \cr}\right] = \left[\matrix{e^{2t} + te^{2t} & -te^{2t} \cr te^{2t} & e^{2t} - te^{2t} \cr}\right].$$

As a check, note that setting $t = 0$ produces the identity.)

The solution to the given initial value problem is

$$\vec x = \left[\matrix{e^{2t} + te^{2t} & -te^{2t} \cr te^{2t} & e^{2t} - te^{2t} \cr}\right] \left[\matrix{3 \cr 4 \cr}\right].$$

You can get the general solution by replacing $(3, 4)$ with $(c_1, c_2)$ .

Example. Find $e^{At}$ if

$$A = \left[\matrix{ 1 & 0 & 0 \cr 1 & 1 & 0 \cr -1 & -1 & 2 \cr}\right].$$

The eigenvalues are obviously $\lambda = 1$ (double) and $\lambda = 2$ .

First, I'll compute the $a_k$ 's. I have $a_1 = e^t$ , and

$$a_2 = \int_0^t e^{t-u} e^u\,du = e^t \int_0^t\,du = te^t,$$

$$a_3 = \int_0^t e^{2(t-u)} ue^u\,du = -te^t - e^t + e^{2t}.$$

Next, I'll compute the $B_k$ 's. $B_1 = I$ , and

$$B_2 = A - I = \left[\matrix{0 & 0 & 0 \cr 1 & 0 & 0 \cr -1 & -1 & 1 \cr}\right],$$

$$B_3 = (A - I)B_2 = \left[\matrix{0 & 0 & 0 \cr 0 & 0 & 0 \cr -2 & -1 & 1 \cr}\right].$$

Therefore,

$$e^{At} = \left[\matrix{e^t & 0 & 0 \cr te^t & e^t & 0 \cr te^t + 2e^t - 2e^{2t} & e^t - e^{2t} & e^{2t} \cr}\right].\quad\halmos$$

Example. Use the matrix exponential to solve

$$\vec x\,' = \left[\matrix{2 & -5 \cr 2 & -4 \cr}\right]\vec x.$$

This example will demonstrate how the algorithm for $e^{A t}$ works when the eigenvalues are complex.

The characteristic polynomial is $\lambda^2 + 2\lambda + 2$ . The eigenvalues are $\lambda = -1 \pm i$ . I will list them as $\{-1 + i, -1 - i\}$ .

First, I'll compute the $a_k$ 's. $a_1 = e^{(-1+i)t}$ , and

$$a_2 = \int_0^t e^{(-1+i)(t-u)} e^{(-1-i)u}\,du = e^{(-1+i)t} \int_0^t e^{(1-i)u} e^{(-1-i)u}\,du =$$

$$e^{(-1+i)t} \int_0^t e^{-2iu}\,du = e^{(-1+i)t} \dfrac{i}{2} \left(e^{-2it} - 1\right) = \dfrac{i}{2} \left(e^{(-1-i)t} - e^{(-1+i)t}\right).$$

Next, I'll compute the $B_k$ 's. $B_1 = I$ , and

$$B_2 = A - (-1 + i)I = \left[\matrix{3 - i & -5 \cr 2 & -3 - i \cr}\right].$$

Therefore,

$$e^{At} = e^{(-1+i)t} \left[\matrix{ 1 & 0 \cr 0 & 1 \cr}\right] + \dfrac{i}{2} \left(e^{(-1-i)t} - e^{(-1+i)t}\right) \left[\matrix{ 3 - i & -5 \cr 2 & -3 - i \cr}\right].$$

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

$$\eqalign{ e^{(-1+i)t} & = e^{-t} (\cos t + i \sin t) \cr e^{(-1-i)t} - e^{(-1+i)t} & = e^{-t} (\cos t - i \sin t) - e^{-t} (\cos t + i \sin t) \cr & = - 2 i e^{-t} \sin t \cr \dfrac{i}{2} \left(e^{(-1-i)t} - e^{(-1+i)t}\right) & = e^{-t} \sin t \cr}$$

Plugging these into the expression for $e^{A t}$ above, I have

$$e^{A t} = e^{-t} (\cos t + i \sin t) \left[\matrix{ 1 & 0 \cr 0 & 1 \cr}\right] + e^{-t} \sin t \left[\matrix{ 3 - i & -5 \cr 2 & -3 - i \cr}\right] = e^{-t} \left[\matrix{ \cos t + 3 \sin t & -5 \sin t \cr 2 \sin t & \cos t - 3 \sin t \cr}\right].$$

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

$$\left[\matrix{x \cr y \cr}\right] = e^{-t}\left[\matrix{\cos t + 3 \sin t & -5 \sin t \cr 2 \sin t & \cos t - 3 \sin t \cr}\right] \left[\matrix{c_1 \cr c_2 \cr}\right].\quad\halmos$$

Example. I'll compare the matrix exponential and the eigenvector solution methods by solving the following system both ways:

$$\vec x\,' = \left[\matrix{ 2 & -1 \cr 1 & 2 \cr}\right] \vec x.$$

The characteristic polynomial is $\lambda^2 - 4\lambda + 5$ . The eigenvalues are $\lambda = 2 \pm i$ .

Consider $\lambda = 2 + i$ :

$$A - (2 + i)I = \left[\matrix{ -i & -1 \cr 1 & -i \cr}\right].$$

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 $(a, b)$ such that $(-i)a + (-1)b = 0$ . To get such a vector, switch the $-i$ and -1 and negate one of them: $a = 1$ , $b = -i$ . Thus, $(1, -i)$ is an eigenvector.

The corresponding solution is

$$e^{(2+i)t} \left[\matrix{1 \cr -i \cr}\right] = e^{2t} \left[\matrix{\cos t + i \sin t \cr \sin t - i \cos t \cr}\right].$$

Take the real and imaginary parts:

$$\re e^{(2+i)t} \left[\matrix{1 \cr -i \cr}\right] = e^{2t} \left[\matrix{\cos t \cr \sin t \cr}\right],$$

$$\im e^{(2+i)t} \left[\matrix{1 \cr -i \cr}\right] = e^{2t} \left[\matrix{\sin t \cr -\cos t \cr}\right].$$

The solution is

$$\vec x = e^{2t} \left(c_1 \left[\matrix{\cos t \cr \sin t \cr}\right] + c_2 \left[\matrix{\sin t \cr -\cos t \cr}\right]\right).$$

Now I'll solve the equation using the exponential. The eigenvalues are $\{2 + i, 2 - i\}$ . Compute the $a_k$ 's. $a_1 = e^{(2+i)t}$ , and

$$a_2 = e^{(2-i)t}\star e^{(2+i)t} = \int_0^t e^{(2-i)(t-u)} e^{(2+i)u}\,du = e^{(2-i)t} \int_0^t e^{2iu}\,du =$$

$$e^{(2-i)t} \left[-\dfrac{i}{2} e^{2iu}\right]_0^t = \dfrac{i}{2} e^{(2-i)t} \left(1 - e^{2it}\right) = \dfrac{i}{2} e^{2t}\left(e^{-it} - e^{-it}\right) = e^{2t} \sin t.$$

(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 $B_k$ 's. $B_1 = I$ , and

$$B_2 = A - (2 + i)I = \left[\matrix{ -i & -1 \cr 1 & -i \cr}\right].$$

Therefore,

$$e^{At} = e^{(2+i)t} \left[\matrix{ 1 & 0 \cr 0 & 1 \cr}\right] + e^{2t} \sin t \left[\matrix{ -i & -1 \cr 1 & -i \cr}\right] = e^{2t} \left[\matrix{ \cos t & -\sin t \cr \sin t & \cos t \cr}\right].$$

The solution is

$$\vec x = e^{2t} \left[\matrix{ \cos t & -\sin t \cr \sin t & \cos t \cr}\right] \left[\matrix{c_1 \cr c_2 \cr}\right].$$

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 system

$$\vec x\,' = \left[\matrix{5 & -8 \cr 2 & -3 \cr}\right]\vec x.$$

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

The characteristic polynomial is $\lambda^2 - 2\lambda + 1$ . The eigenvalue is $\lambda = 1$ (double).

$$A - I = \left[\matrix{4 & -8 \cr 2 & -4 \cr}\right].$$

Ignore the first row, and divide the second row by 2, obtaining the vector $(1, -2)$ . I want $(a, b)$ such that $(1)a + (-2)b = 0$ . Swap 1 and -2 and negate the -2: I get $(a, b) = (2, 1)$ . This is an eigenvector for $\lambda = -1$ .

Since I only have one eigenvector, I need a generalized eigenvector. This means I need $(a', b')$ such that

$$\left[\matrix{4 & -8 \cr 2 & -4 \cr}\right] \left[\matrix{a' \cr b' \cr}\right] = \left[\matrix{2 \cr 1 \cr}\right].$$

Row reduce:

$$\left[\matrix{4 & -8 & 2 \cr 2 & -4 & 1 \cr}\right] \quad \to \quad \left[\matrix{1 & -2 & \dfrac{1}{2} \cr \noalign{\vskip2pt} 0 & 0 & 0 \cr}\right]$$

This means that $a' = 2b' + \dfrac{1}{2}$ . Setting $b' = 0$ yields $a' = \dfrac{1}{2}$ . The generalized eigenvector is $\left(\dfrac{1}{2}, 0\right)$ .

The solution is

$$\vec x = c_1 e^t \left[\matrix{2 \cr 1 \cr}\right] + c_2\left(te^t \left[\matrix{2 \cr 1 \cr}\right] + e^t \left[\matrix{\dfrac{1}{2} \cr \noalign{\vskip2pt} 0 \cr}\right]\right).$$

Next, I'll solve the system using the matrix exponential. The eigenvalues are $\{1, 1\}$ . First, I'll compute the $a_k$ 's. $a_1 = e^t$ , and

$$a_2 = e^t \star e^t = \int_0^t e^{t-u} e^u\,du = e^t \int_0^t\,du = te^t.$$

Next, compute the $B_k$ 's. $B_1 = I$ , and

$$B_2 = A - I = \left[\matrix{4 & -8 \cr 2 & -4 \cr}\right].$$

Therefore,

$$e^{At} = e^t \left[\matrix{1 & 0 \cr 0 & 1 \cr}\right] + te^t \left[\matrix{4 & -8 \cr 2 & -4 \cr}\right] = \left[\matrix{e^t + 4te^t & -8te^t \cr 2te^t & e^t - 4te^t \cr}\right].$$

The solution is

$$\vec x = \left[\matrix{e^t + 4te^t & -8te^t \cr 2te^t & e^t - 4te^t \cr}\right] \left[\matrix{c_1 \cr c_2 \cr}\right].$$

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.

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Finding General Solution to a Given Matrix 2x2

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