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8 Matrix Exponential Methods

8.1 Matrix Exponentials and Linear Systems

The solution vectors of an n×n $n \times n$ homogeneous linear system

x' = A x

$x^{′} = Ax$ (1)

can be used to construct a square matrix X=Φ(t) $X = Φ (t)$ that satisfies the matrix differential equation

X' = A X

$X^{′} = A X$ (1′)

associated with Eq. (1). Suppose that x1(t), x2(t),…, xn(t) $x_{1} (t), x_{2} (t), \dots, x_{n} (t)$ are n linearly independent solutions of Eq. (1). Then the n×n $n \times n$ matrix

Φ (t) = ⎡ ⎣ ⎢ | x 1 (t) | | x 2 (t) | \dots | x n (t) | ⎤ ⎦ ⎥

$Φ (t) = [\begin{matrix} | & | & | \\ x_{1} (t) & x_{2} (t) & \dots & x_{n} (t) \\ | & | & | \end{matrix}]$ (2)

having these solution vectors as its column vectors, is called a fundamental matrix for the system in (1).

Fundamental Matrix Solutions

Because the column vector x=xj(t) $x = x_{j} (t)$ of the fundamental matrix Φ(t) $Φ (t)$ in (2) satisfies the differential equation x'=Ax, $x^{′} = Ax,$ it follows (from the definition of matrix multiplication) that the matrix X=Φ(t) $X = Φ (t)$ itself satisfies the matrix differential equation X'=AX. $X^{′} = A X .$ Because its column vectors are linearly independent, it also follows that the fundamental matrix Φ(t) $Φ (t)$ is nonsingular, and therefore has an inverse matrix Φ(t)−1. $Φ {(t)}^{- 1} .$ Conversely, any nonsingular matrix solution Ψ(t) $Ψ (t)$ of Eq. (1′) has linearly independent column vectors that satisfy Eq. (1), so Ψ(t) $Ψ (t)$ is a fundamental matrix for the system in (1).

In terms of the fundamental matrix Φ(t) $Φ (t)$ in (2), the general solution

x (t) = c 1 x 1 (t) + c 2 x 2 (t) + \dots + c n x n (t)

$x (t) = c_{1} x_{1} (t) + c_{2} x_{2} (t) + \dots + c_{n} x_{n} (t)$ (3)

of the system x'=Ax $x^{′} = Ax$ can be written in the form

x (t) = Φ (t) c

$x (t) = Φ (t) c$ (4)

where c=[c1c2⋯cn]T $c = {[\begin{array}{l} c_{1} & c_{2} & \dots & c_{n} \end{array}]}^{T}$ is an arbitrary constant vector. If Ψ(t) $Ψ (t)$ is any other fundamental matrix for (1), then each column vector of Ψ(t) $Ψ (t)$ is a linear combination of the column vectors of Φ(t), $Φ (t),$ so it follows from Eq. (4) that

Ψ (t) = Φ (t) C

$Ψ (t) = Φ (t) C$ (4′)

for some n×n $n \times n$ matrix C of constants.

In order that the solution x(t) in (3) satisfy a given initial condition

x (0) = x 0,

$x (0) = x_{0},$ (5)

it suffices that the coefficient vector c in (4) be such that Φ(0)c=x0; $Φ (0) c = x_{0};$ that is, that

c = Φ (0) - 1 x 0 .

$c = Φ {(0)}^{- 1} x_{0} .$ (6)

When we substitute (6) in Eq. (4), we get the conclusion of the following theorem.

Section 7.3 tells us how to find a fundamental matrix for the system

x' = A x

$x^{′} = Ax$ (9)

with constant n×n $n \times n$ coefficient matrix A, at least in the case where A has a complete set of n linearly independent eigenvectors v1, v2, …, vn $v_{1}, v_{2}, \dots, v_{n}$ associated with the (not necessarily distinct) eigenvalues λ1, λ2, …,λn, $λ_{1}, λ_{2}, \dots, λ_{n},$ respectively. In this event the corresponding solution vectors of Eq. (9) are given by

x i (t) = v i e λ i t

$x_{i} (t) = v_{i} e^{λ_{i} t}$

for i=1, 2, …, $i = 1, 2, \dots,$ n. Therefore, the n×n $n \times n$ matrix

Φ (t) = ⎡ ⎣ ⎢ | v 1 e λ 1 t | | v 2 e λ 2 t | \dots | v n e λ n t | ⎤ ⎦ ⎥

$Φ (t) = [\begin{matrix} | & | & | \\ v_{1} e^{λ_{1} t} & v_{2} e^{λ_{2} t} & \dots & v_{n} e^{λ_{n} t} \\ | & | & | \end{matrix}]$ (10)

having the solutions x1, x2, …, xn $x_{1}, x_{2}, \dots, x_{n}$ as column vectors is a fundamental matrix for the system x'=Ax. $x^{′} = Ax .$

In order to apply Eq. (8), we must be able to compute the inverse matrix Φ(0)−1. $Φ {(0)}^{- 1} .$ The inverse of the nonsingular 2×2 $2 \times 2$ matrix

A = [a c b d]

$A = [\begin{matrix} a & b \\ c & d \end{matrix}]$

A - 1 = 1 Δ [d - c - b a],

$A^{- 1} = \frac{1}{Δ} [\begin{array}{r} d & - b \\ - c & a \end{array}],$ (11)

where Δ=det(A)=ad−bc≠0. $Δ = \det (A) = a d - b c \neq 0.$ The inverse of the nonsingular 3×3 $3 \times 3$ matrix A=[aij] $A = [a_{i j}]$ is given by

A - 1 = 1 Δ ⎡ ⎣ ⎢ + A 11 - A 21 + A 31 - A 12 + A 22 - A 32 + A 13 - A 23 + A 33 ⎤ ⎦ ⎥ T,

$A^{- 1} = \frac{1}{Δ} {[\begin{array}{l} + A_{11} & - A_{12} & + A_{13} \\ - A_{21} & + A_{22} & - A_{23} \\ + A_{31} & - A_{32} & + A_{33} \end{array}]}^{T},$ (12)

where Δ=det(A)≠0 $Δ = \det (A) \neq 0$ and Aij $A_{i j}$ denotes the determinant of the 2×2 $2 \times 2$ submatrix of A obtained by deleting the i th row and j th column of A. (Do not overlook the symbol T for transpose in Eq. (12).) The formula in (12) is also valid upon generalization to n×n $n \times n$ matrices, but in practice inverses of larger matrices are usually computed instead by row reduction methods (see any linear algebra text) or by using a calculator or computer algebra system.

Example 1

Find a fundamental matrix for the system

x' y' = = 4 x 3 x + - 2 y, y,

$\begin{array}{rcrcr} x^{′} & = & 4 x & + & 2 y, \\ y^{′} & = & 3 x & - & y, \end{array}$ (13)

and then use it to find the solution of (13) that satisfies the initial conditions x(0)=1 $x (0) = 1$ , y(0)=−1 $y (0) = - 1$ .

Solution

The linearly independent solutions

x 1 (t) = [e - 2 t - 3 e - 2 t] and x 2 (t) = [2 e 5 t e 5 t]

$x_{1} (t) = [\begin{array}{c} \begin{array}{r} e^{- 2 t} \\ - 3 e^{- 2 t} \end{array} \end{array}] and x_{2} (t) = [\begin{array}{r} 2 e^{5 t} \\ e^{5 t} \end{array}]$

found in Example 1 of Section 7.3 yield the fundamental matrix

Φ (t) = [e - 2 t - 3 e - 2 t 2 e 5 t e 5 t] .

$Φ (t) = [\begin{array}{r} e^{- 2 t} & 2 e^{5 t} \\ - 3 e^{- 2 t} & e^{5 t} \end{array}] .$ (14)

Then

Φ (0) = [1 - 3 21],

$Φ (0) = [\begin{array}{r} 1 & 2 \\ - 3 & 1 \end{array}],$

and the formula in (11) gives the inverse matrix

Φ (0) - 1 = 1 7 [13 - 2 1] .

$Φ {(0)}^{- 1} = \frac{1}{7} [\begin{array}{r} 1 & - 2 \\ 3 & 1 \end{array}] .$ (15)

Hence the formula in (8) gives the solution

x (t) = [e - 2 t - 3 e - 2 t 2 e 5 t e 5 t] (1 7) [13 - 2 1] [1 - 1] = (1 7) [e - 2 t - 3 e - 2 t 2 e 5 t e 5 t] [32],

$x (t) = [\begin{array}{r} e^{- 2 t} & 2 e^{5 t} \\ - 3 e^{- 2 t} & e^{5 t} \end{array}] (\frac{1}{7}) [\begin{array}{r} 1 & - 2 \\ 3 & 1 \end{array}] [\begin{array}{r} 1 \\ - 1 \end{array}] = (\frac{1}{7}) [\begin{array}{r} e^{- 2 t} & 2 e^{5 t} \\ - 3 e^{- 2 t} & e^{5 t} \end{array}] [\begin{array}{l} 3 \\ 2 \end{array}],$

and so

x (t) = 1 7 [3 e - 2 t + 4 e 5 t - 9 e - 2 t + 2 e 5 t] .

$x (t) = \frac{1}{7} [\begin{array}{r} 3 e^{- 2 t} + 4 e^{5 t} \\ - 9 e^{- 2 t} + 2 e^{5 t} \end{array}] .$

Thus the solution of the original initial value problem is given by

x (t) = 3 7 e - 2 t + 4 7 e 5 t, y (t) = - 9 7 e - 2 t + 2 7 e 5 t .

$x (t) = \frac{3}{7} e^{- 2 t} + \frac{4}{7} e^{5 t}, y (t) = - \frac{9}{7} e^{- 2 t} + \frac{2}{7} e^{5 t} .$

Remark

An advantage of the fundamental matrix approach is this: Once we know the fundamental matrix Φ(t) $Φ (t)$ and the inverse matrix Φ(0)−1, $Φ {(0)}^{- 1},$ we can calculate rapidly by matrix multiplication the solutions corresponding to different initial conditions. For example, suppose that we seek the solution of the system in (13) satisfying the new initial conditions x(0)=77, y(0)=49 $x (0) = 77, y (0) = 49$ . Then substitution of (14) and (15) in (8) gives the new particular solution

x (t) = = 1 7 [e - 2 t - 3 e - 2 t 2 e 5 t e 5 t] [13 - 2 1] [7749] 1 7 [e - 2 t - 3 e - 2 t 2 e 5 t e 5 t] [- 21 280] = [- 3 e - 2 t + 80 e 5 t 9 e - 2 t + 40 e 5 t] .

$\begin{array}{rcl} x (t) & = & \frac{1}{7} [\begin{array}{r} e^{- 2 t} & 2 e^{5 t} \\ - 3 e^{- 2 t} & e^{5 t} \end{array}] [\begin{array}{r} 1 & - 2 \\ 3 & 1 \end{array}] [\begin{array}{r} 77 \\ 49 \end{array}] \\ = & \frac{1}{7} [\begin{array}{r} e^{- 2 t} & 2 e^{5 t} \\ - 3 e^{- 2 t} & e^{5 t} \end{array}] [\begin{array}{r} - 21 \\ 280 \end{array}] = [\begin{array}{r} - 3 e^{- 2 t} + 80 e^{5 t} \\ 9 e^{- 2 t} + 40 e^{5 t} \end{array}] . \end{array}$

Exponential Matrices

We now discuss the possibility of constructing a fundamental matrix for the constant-coefficient linear system x'=Ax $x^{′} = Ax$ directly from the coefficient matrix A-that is, without first applying the methods of earlier sections to find a linearly independent set of solution vectors.

We have seen that exponential functions play a central role in the solution of linear differential equations and systems, ranging from the scalar equation x'=kx $x ′ = k x$ with solution x(t)=x0ekt $x (t) = x_{0} e^{k t}$ to the vector solution x(t)=veλt $x (t) = v e^{λ t}$ of the linear system x'=Ax $x^{′} = Ax$ whose coefficient matrix A has eigenvalue λ $λ$ with associated eigenvector v. We now define exponentials of matrices in such a way that

X (t) = e A t

$X (t) = e^{A t}$

is a matrix solution of the matrix differential equation

X' = A X

$X^{′} = A X$

with n×n $n \times n$ coefficient matrix A—in analogy with the fact that the ordinary exponential function x(t)=eat $x (t) = e^{a t}$ is a scalar solution of the first-order differential equation x'=ax $x ′ = a x$ .

The exponential ez $e^{z}$ of the complex number z may be defined (as in Section 5.3) by means of the exponential series

e z = 1 + z + z 2 2 ! + z 3 3 ! + \dots + z n n ! + \dots .

$e^{z} = 1 + z + \frac{z^{2}}{2!} + \frac{z^{3}}{3!} + \dots + \frac{z^{n}}{n!} + \dots .$ (16)

Similarly, if A is an n×n $n \times n$ matrix, then the exponential matrix eA $e^{A}$ is the n×n $n \times n$ matrix defined by the series

e A = I + A + A 2 2 ! + \dots + A n n ! + \dots,

$e^{A} = I + A + \frac{A^{2}}{2!} + \dots + \frac{A^{n}}{n!} + \dots,$ (17)

where I is the identity matrix. The meaning of the infinite series on the right in (17) is given by

\sum n = 0 \infty A n n ! = lim k \to \infty (\sum n = 0 k A n n !),

$\sum_{n = 0}^{\infty} \frac{A^{n}}{n!} = \lim_{k \to \infty} (\sum_{n = 0}^{k} \frac{A^{n}}{n!}),$ (18)

where $A^{0} = I, A^{2} = AA, A^{3} = A A^{2},$ and so on; inductively, $A^{n + 1} = A A^{n}$ if $n ≧ 0.$ It can be shown that the limit in (18) exists for every $n \times n$ square matrix A. That is, the exponential matrix $e^{A}$ is defined (by Eq. (17)) for every square matrix A.

Example 2

Consider the $2 \times 2$ diagonal matrix

$A = [\begin{array}{r} a & 0 \\ 0 & b \end{array}] .$

Then it is apparent that

$A^{n} = [\begin{array}{c} a^{n} & 0 \\ 0 & b^{n} \end{array}]$

for each integer $n ≧ 1.$ It therefore follows that

$\begin{array}{rcl} e^{A} & = & I + A + \frac{A^{2}}{2!} + \dots \\ = & [\begin{array}{r} 1 & 0 \\ 0 & 1 \end{array}] + [\begin{array}{r} a & 0 \\ 0 & b \end{array}] + [\begin{array}{c} a^{2} / 2! & 0 \\ 0 & b^{2} / 2! \end{array}] + \dots \\ = & [\begin{array}{c} 1 + a + a^{2} / 2! + \dots & 0 \\ 0 & 1 + b + b^{2} / 2! + \dots \end{array}] . \end{array}$

Thus

$e^{A} = [\begin{array}{c} e^{a} & 0 \\ 0 & e^{b} \end{array}],$

so the exponential of the diagonal $2 \times 2$ matrix A is obtained simply by exponentiating each diagonal element of A.

The $n \times n$ analog of the $2 \times 2$ result in Example 2 is established in the same way. The exponential of the $n \times n$ diagonal matrix

$D = [\begin{array}{c} a_{1} & 0 & \dots & 0 \\ 0 & a_{2} & \dots & 0 \\ ⋮ & ⋮ & ⋮ \\ 0 & 0 & \dots & a_{n} \end{array}]$ (19)

is the $n \times n$ diagonal matrix

$e^{D} = [\begin{array}{c} e^{a_{1}} & 0 & \dots & 0 \\ 0 & e^{a_{2}} & \dots & 0 \\ ⋮ & ⋮ & ⋮ \\ 0 & 0 & \dots & e^{a_{n}} \end{array}],$ (20)

obtained by exponentiating each diagonal element of D.

The exponential matrix $e^{A}$ satisfies most of the exponential relations that are familiar in the case of scalar exponents. For instance, if 0 is the $n \times n$ zero matrix, then Eq. (17) yields

$e^{0} = I,$ (21)

the $n \times n$ identity matrix. In Problem 31 we ask you to show that a useful law of exponents holds for $n \times n$ matrices that commute:

$If A B = B A, then e^{A + B} = e^{A} e^{B} .$ (22)

In Problem 32 we ask you to conclude that

${(e^{A})}^{- 1} = e^{- A} .$ (23)

In particular, the matrix $e^{A}$ is nonsingular for every $n \times n$ matrix A (reminiscent of the fact that $e^{z} \neq 0$ for all z). It follows from elementary linear algebra that the column vectors of $e^{A}$ are always linearly independent.

If t is a scalar variable, then substitution of At for A in Eq. (17) gives

$e^{A t} = I + A t + A^{2} \frac{t^{2}}{2!} + \dots + A^{n} \frac{t^{n}}{n!} + \dots .$ (24)

(Of course, At is obtained simply by multiplying each element of A by t.)

Example 3

If,

$A = [\begin{array}{r} 0 & 3 & 4 \\ 0 & 0 & 6 \\ 0 & 0 & 0 \end{array}],$

then

$A^{2} = [\begin{array}{r} 0 & 0 & 18 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}] and A^{3} = [\begin{array}{r} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}],$

so $A^{n} = 0$ for $n ≧ 3.$ It therefore follows from Eq. (24) that

$\begin{array}{rcl} e^{A t} & = & I + A t + \frac{1}{2} A^{2} t^{2} \\ = & [\begin{array}{r} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] + [\begin{array}{r} 0 & 3 & 4 \\ 0 & 0 & 6 \\ 0 & 0 & 0 \end{array}] t + \frac{1}{2} [\begin{array}{r} 0 & 0 & 18 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}] t^{2}; \end{array}$

that is,

$e^{A t} = [\begin{array}{c} 1 & 3 t & 4 t + 9 t^{2} \\ 0 & 1 & 6 t \\ 0 & 0 & 1 \end{array}] .$

Remark

If $A^{n} = 0$ for some positive integer n, then the exponential series in (24) terminates after a finite number of terms, so the exponential matrix $e^{A}$ (or $e^{A t}$ ) is readily calculated as in Example 3. Such a matrix—with a vanishing power—is said to be nilpotent.

Example 4

$A = [\begin{array}{r} 2 & 3 & 4 \\ 0 & 2 & 6 \\ 0 & 0 & 2 \end{array}],$

then

$A = [\begin{array}{r} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}] + [\begin{array}{r} 0 & 3 & 4 \\ 0 & 0 & 6 \\ 0 & 0 & 0 \end{array}] = D + B,$

where $D = 2 I$ is a diagonal matrix and B is the nilpotent matrix of Example 3. Therefore, (20) and (22) give

$e^{A t} = e^{(D + B) t} = e^{D t} e^{B t} = [\begin{array}{c} e^{2 t} & 0 & 0 \\ 0 & e^{2 t} & 0 \\ 0 & 0 & e^{2 t} \end{array}] [\begin{array}{c} 1 & 3 t & 4 t + 9 t^{2} \\ 0 & 1 & 6 t \\ 0 & 0 & 1 \end{array}];$

thus

$e^{A t} = [\begin{array}{c} e^{2 t} & 3 t e^{2 t} & (4 t + 9 t^{2}) e^{2 t} \\ 0 & e^{2 t} & 6 t e^{2 t} \\ 0 & 0 & e^{2 t} \end{array}] .$

Matrix Exponential Solutions

It happens that term-by-term differentiation of the series in (24) is valid, with the result

$\frac{d}{d t} (e^{A t}) = A + A^{2} t + A^{3} \frac{t^{2}}{2!} + \dots = A (I + A t + A^{2} \frac{t^{2}}{2!} + \dots);$

that is,

$\frac{d}{d t} (e^{A t}) = A e^{A t},$ (25)

in analogy to the formula $D_{t} (e^{k t}) = k e^{k t}$ from elementary calculus. Thus the matrix-valued function

$X (t) = e^{A t}$

satisfies the matrix differential equation

$X ′ = AX .$

Because the matrix $e^{A t}$ is nonsingular, it follows that the matrix exponential $e^{A t}$ is a fundamental matrix for the linear system $x ′ = Ax$ . In particular, it is the fundamental matrix X(t) such that $X (0) = I$ . Therefore, Theorem 1 implies the following result.

Thus the solution of homogeneous linear systems reduces to the task of computing exponential matrices. Conversely, if we already know a fundamental matrix $Φ (t)$ for the linear system $x ′ = Ax$ , then the facts that $e^{A t} = Φ (t) C$ (by Eq. (4′)) and $e^{A \cdot 0} = e^{0} = I$ (the identity matrix) yield

$e^{A t} = Φ (t) Φ {(0)}^{- 1} .$ (28)

So we can find the matrix exponential $e^{A t}$ by solving the linear system $x ′ = Ax$ .

Example 5

In Example 1 we found that the system $x ′ = Ax$ with

$A = [\begin{array}{r} 4 & 2 \\ 3 & - 1 \end{array}]$

has fundamental matrix

$Φ (t) = [\begin{array}{r} e^{- 2 t} & 2 e^{5 t} \\ - 3 e^{- 2 t} & e^{5 t} \end{array}] with Φ {(0)}^{- 1} = \frac{1}{7} [\begin{array}{r} 1 & - 2 \\ 3 & 1 \end{array}] .$

Hence Eq. (28) gives

$\begin{array}{rcl} e^{A t} & = & \frac{1}{7} [\begin{array}{r} e^{- 2 t} & 2 e^{5 t} \\ - 3 e^{- 2 t} & e^{5 t} \end{array}] [\begin{array}{r} 1 & - 2 \\ 3 & 1 \end{array}] \\ = & \frac{1}{7} [\begin{array}{c} e^{- 2 t} + 6 e^{5 t} & - 2 e^{- 2 t} + 2 e^{5 t} \\ - 3 e^{- 2 t} + 3 e^{5 t} & 6 e^{- 2 t} + e^{5 t} \end{array}] . \end{array}$

Example 6

Use an exponential matrix to solve the initial value problem

$x ′ = [\begin{array}{l} 2 & 3 & 4 \\ 0 & 2 & 6 \\ 0 & 0 & 2 \end{array}] x, x (0) = [\begin{array}{l} 19 \\ 29 \\ 39 \end{array}] .$ (29)

Solution

The coefficient matrix A in (29) evidently has characteristic equation ${(2 - λ)}^{3} = 0$ and thus the triple eigenvalue $λ = 2, 2, 2$ . It is easy to see that the eigenvector equation

$(A - 2 I) v = [\begin{array}{l} 0 & 3 & 4 \\ 0 & 0 & 6 \\ 0 & 0 & 0 \end{array}] [\begin{array}{l} a \\ b \\ c \end{array}] = [\begin{array}{l} 0 \\ 0 \\ 0 \end{array}]$

has (to within a constant multiple) the single solution $v = {[\begin{array}{l} 1 & 0 & 0 \end{array}]}^{T} .$ Thus there is only a single eigenvector associated with the eigenvalue $λ = 2$ , and so we do not yet have the three linearly independent solutions needed for a fundamental matrix. But we note that A is the same matrix whose matrix exponential

$e^{A t} = [\begin{array}{c} e^{2 t} & 3 t e^{2 t} & (4 t + 9 t^{2}) e^{2 t} \\ 0 & e^{2 t} & 6 t e^{2 t} \\ 0 & 0 & e^{2 t} \end{array}]$

was calculated in Example 4. Hence, using Theorem 2, the solution of the initial value problem in (29) is given by

$\begin{array}{rcl} x (t) = e^{A t} x (0) & = & [\begin{array}{c} e^{2 t} & 3 t e^{2 t} & (4 t + 9 t^{2}) e^{2 t} \\ 0 & e^{2 t} & 6 t e^{2 t} \\ 0 & 0 & e^{2 t} \end{array}] [\begin{array}{c} 19 \\ 29 \\ 39 \end{array}] \\ = & [\begin{array}{c} (19 + 243 t + 351 t^{2}) e^{2 t} \\ (29 + 234 t) e^{2 t} \\ 39 e^{2 t} \end{array}] . \end{array}$

Remark

The same particular solution x(t) as in Example 6 could be found using the generalized eigenvector method of Section 7.6. One would start by finding the chain of generalized eigenvectors

$v_{1} = [\begin{array}{c} 18 \\ 0 \\ 0 \end{array}], v_{2} = [\begin{array}{c} 4 \\ 6 \\ 0 \end{array}], v_{3} = [\begin{array}{c} 0 \\ 0 \\ 1 \end{array}]$

corresponding to the triple eigenvalue $λ = 2$ of the matrix A. Then one would—using Eqs. (27) in Section 7.6—assemble the linearly independent solutions

$x_{1} (t) = v_{1} e^{2 t}, x_{2} (t) = (v_{1} t + v_{2}) e^{2 t}, x_{3} (t) = (\frac{1}{2} v_{1} t^{2} + v_{2} t + v_{3}) e^{2 t}$

of the differential equation $x ′ = Ax$ in (29). The final step would be to determine values of the coefficients $c_{1}, c_{2}, c_{3}$ so that the particular solution $x (t) = c_{1} x_{1} (t) + c_{2} x_{2} (t) + c_{3} x_{3} (t)$ satisfies the initial condition in (29). At this point it should be apparent that—especially if the matrix exponential $e^{A t}$ is readily available (for instance, from a computer algebra system)—the method illustrated in Example 6 can well be more “computationally routine” than the generalized eigenvector method.

General Matrix Exponentials

The relatively simple calculation of $e^{A t}$ carried out in Example 4 (and used in Example 6) was based on the observation that if

$A = [\begin{array}{l} 2 & 3 & 4 \\ 0 & 2 & 6 \\ 0 & 0 & 2 \end{array}],$

then $A - 2 I$ is nilpotent:

${(A - 2 I)}^{3} = {[\begin{array}{l} 0 & 3 & 4 \\ 0 & 0 & 6 \\ 0 & 0 & 0 \end{array}]}^{3} = [\begin{array}{l} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}] = 0 .$ (30)

A similar result holds for any $3 \times 3$ matrix A having a triple eigenvalue r, in which case its characteristic equation reduces to ${(λ - r)}^{3} = 0$ . For such a matrix, an explicit computation similar to that in Eq. (30) will show that

${(A - r I)}^{3} = 0 .$ (31)

(This particular result is a special case of the Cayley-Hamilton theorem of Section 6.3, according to which every matrix satisfies its own characteristic equation.) Thus the matrix $A - r I$ is nilpotent, and it follows that

$e^{A t} = e^{(r I + A - r I) t} = e^{r I t} \cdot e^{(A - r I) t} = e^{r t} I \cdot [I + (A - r I) t + \frac{1}{2} {(A - r I)}^{2} t^{2}],$ (32)

the exponential series here terminating because of Eq. (31). In this way, we can rather easily calculate the matrix exponential $e^{A t}$ for any square matrix having only a single eigenvalue.

The calculation in Eq. (32) motivates a method of calculating $e^{A t}$ for any $n \times n$ matrix A whatsoever. As we saw in Section 7.6, A has n linearly independent generalized eigenvectors $u_{1}, u_{2}, \dots,$ $u_{n}$ . Each generalized eigenvector u is associated with an eigenvalue $λ$ of A and has a rank $r ≧ 1$ such that

${(A - λ I)}^{r} u = 0 but {(A - λ I)}^{r - 1} u \neq 0 .$ (33)

(If $r = 1$ , then u is an ordinary eigenvector such that $A u = λ u$ .)

Even if we do not yet know $e^{A t}$ explicitly, we can consider the function $x (t) = e^{A t} u$ , which is a linear combination of the column vectors of $e^{A t}$ and is therefore a solution of the linear system $x ′ = Ax$ with $x (0) = u$ . Indeed, we can calculate x explicitly in terms of A, u, $λ$ , and r:

$\begin{array}{rcl} x (t) & = & e^{A t} u = e^{(λ I + A - λ I) t} u = e^{λ I t} e^{(A - λ I) t} u \\ = & e^{λ t} I [I + (A - λ I) t + \dots + {(A - λ I)}^{r - 1} \frac{t^{r - 1}}{(r - 1)!} + \dots] u, \end{array}$

$\begin{array}{rcl} x (t) & = & e^{λ t} [u + (A - λ I) u t + {(A - λ I)}^{2} u \frac{t^{2}}{2!} + \dots \\ + {(A - λ I)}^{r - 1} u \frac{t^{r - 1}}{(r - 1)!}], \end{array}$ (34)

using (33) and the fact that $e^{λ I t} = e^{λ t} I$ .

If the linearly independent solutions $x_{1} (t), x_{2} (t), \dots, x_{n} (t)$ of $x ′ = Ax$ are calculated using (34) with the linearly independent generalized eigenvectors $u_{1}, u_{2}, \dots,$ $u_{n}$ , then the $n \times n$ matrix

$Φ (t) = [\begin{array}{l} x_{1} (t) & x_{2} (t) & \dots & x_{n} (t) \end{array}]$ (35)

is a fundamental matrix for the system $x ′ = Ax$ . Finally, the specific fundamental matrix $X (t) = Φ (t) Φ {(0)}^{- 1}$ satisfies the initial condition $X (0) = I$ , and thus is the desired matrix exponential $e^{A t}$ . We have therefore outlined a proof of the following theorem.

Example 7

Find $e^{A t}$ if

$A = [\begin{array}{l} 3 & 4 & 5 \\ 0 & 5 & 4 \\ 0 & 0 & 3 \end{array}] .$ (37)

Solution

Theorem 3 would apply even if the matrix A were not upper triangular. But because A is upper triangular, this fact enables us to see quickly that its characteristic equation is

$(5 - λ) {(3 - λ)}^{2} = 0.$

Thus A has the distinct eigenvalue $λ_{1} = 5$ and the repeated eigenvalue $λ_{2} = 3$ .

Case 1: $λ_{1} = 5$ . The eigenvector equation $(A - λ I) u = 0$ for $u = {[\begin{array}{l} a & b & c \end{array}]}^{T}$ is

$(A - 5 I) u = [\begin{array}{r} - 2 & 4 & 5 \\ 0 & 0 & 4 \\ 0 & 0 & - 2 \end{array}] [\begin{array}{l} a \\ b \\ c \end{array}] = [\begin{array}{l} 0 \\ 0 \\ 0 \end{array}] .$

The last two scalar equations $4 c = 0$ and $- 2 c = 0$ give $c = 0$ . Then the first equation $- 2 a + 4 b = 0$ is satisfied by $a = 2$ and $b = 1$ . Thus the eigenvalue $λ_{1} = 5$ has the (ordinary) eigenvector $u_{1} = {[\begin{array}{l} 2 & 1 & 0 \end{array}]}^{T} .$ The corresponding solution of the system $x ′ = Ax$ is

$x_{1} (t) = e^{5 t} u_{1} = e^{5 t} {[\begin{array}{l} 2 & 1 & 0 \end{array}]}^{T} .$ (38)

Case 2: $λ_{2} = 3 .$ The eigenvector equation $(A - λ I) u = 0$ for $u = {[\begin{array}{l} a & b & c \end{array}]}^{T}$ is

$(A - 3 I) u = [\begin{array}{l} 0 & 4 & 5 \\ 0 & 2 & 4 \\ 0 & 0 & 0 \end{array}] [\begin{array}{l} a \\ b \\ c \end{array}] = [\begin{array}{l} 0 \\ 0 \\ 0 \end{array}] .$

The first two equations $4 b + 5 c = 0$ and $2 b + 4 c = 0$ imply that $b = c = 0$ , but leave a arbitrary. Thus the eigenvalue $λ_{2} = 3$ has the single (ordinary) eigenvector $u_{2} = {[\begin{array}{l} 1 & 0 & 0 \end{array}]}^{T} .$ The corresponding solution of the system $x ′ = Ax$ is

$x_{2} (t) = e^{3 t} u_{2} = e^{3 t} {[\begin{array}{l} 1 & 0 & 0 \end{array}]}^{T} .$ (39)

To look for a generalized eigenvector of rank $r = 2$ in Eq. (33), we consider the equation

${(A - 3 I)}^{2} u = [\begin{array}{llr} 0 & 8 & 16 \\ 0 & 4 & 8 \\ 0 & 0 & 0 \end{array}] [\begin{array}{r} a \\ b \\ c \end{array}] = [\begin{array}{r} 0 \\ 0 \\ 0 \end{array}] .$

The first two equations $8 b + 16 c = 0$ and $4 b + 8 c = 0$ are satisfied by $b = 2$ and $c = - 1$ , but leave a arbitrary. With $a = 0$ we get the generalized eigenvector $u_{3} = {[\begin{array}{r} 0 & 2 & - 1 \end{array}]}^{T}$ of rank $r = 2$ associated with the eigenvalue $λ = 3$ . Because ${(A - 3 I)}^{2} u = 0$ , Eq. (34) yields the third solution

$\begin{aligned} x_{3} (t) & = & e^{3 t} [u_{3} + (A - 3 I) u_{3} t] \\ = & e^{3 t} ([\begin{array}{r} 0 \\ 2 \\ - 1 \end{array}] + [\begin{array}{r} 0 & 4 & 5 \\ 0 & 2 & 4 \\ 0 & 0 & 0 \end{array}] [\begin{array}{r} 0 \\ 2 \\ - 1 \end{array}] t) = e^{3 t} [\begin{array}{r} 3 t \\ 2 \\ - 1 \end{array}] . \end{aligned}$ (40)

With the solutions listed in Eqs. (39) and (40), the fundamental matrix

$Φ (t) = [\begin{array}{l} x_{1} (t) & x_{2} (t) & x_{3} (t) \end{array}]$

defined by Eq. (35) is

$Φ (t) = [\begin{array}{r} 2 e^{5 t} & e^{3 t} & 3 t e^{3 t} \\ e^{5 t} & 0 & 2 e^{3 t} \\ 0 & 0 & - e^{3 t} \end{array}] with Φ {(0)}^{- 1} = [\begin{array}{r} 0 & 1 & 2 \\ 1 & - 2 & - 4 \\ 0 & 0 & - 1 \end{array}] .$

Hence Theorem 3 finally yields

$\begin{array}{rcl} e^{A t} & = & Φ (t) Φ {(0)}^{- 1} \\ = & [\begin{array}{l} 2 e^{3 t} & e^{3 t} & 3 t e^{3 t} \\ e^{5 t} & 0 & 2 e^{3 t} \\ 0 & 0 & - e^{3 t} \end{array}] [\begin{array}{r} 0 & 1 & 2 \\ 1 & - 2 & - 4 \\ 0 & 0 & - 1 \end{array}] \\ = & [\begin{array}{r} e^{3 t} & 2 e^{5 t} - 2 e^{3 t} & 4 e^{5 t} - (4 + 3 t) e^{3 t} \\ 0 & e^{5 t} & 2 e^{5 t} - 2 e^{3 t} \\ 0 & 0 & e^{3 t} \end{array}] . \end{array}$

Remark

As in Example 7, Theorem 3 suffices for the computation of $e^{A t}$ provided that a basis consisting of generalized eigenvectors of A can be found. Alternatively, a computer algebra system can be used as indicated in the project material for this section.

8.1 Problems

Find a fundamental matrix of each of the systems in Problems 1 through 8, then apply Eq. (8) to find a solution satisfying the given initial conditions.

$x ′ = [\begin{array}{r} 2 & 1 \\ 1 & 2 \end{array}] x, x (0) = [\begin{array}{r} 3 \\ - 2 \end{array}]$
$x ′ = [\begin{array}{r} 2 & - 1 \\ - 4 & 2 \end{array}] x, x (0) = [\begin{array}{r} 2 \\ - 1 \end{array}]$
$x ′ = [\begin{array}{r} 2 & - 5 \\ 4 & - 2 \end{array}] x, x (0) = [\begin{array}{r} 0 \\ 1 \end{array}]$
$x ′ = [\begin{array}{r} 3 & - 1 \\ 1 & 1 \end{array}] x, x (0) = [\begin{array}{r} 1 \\ 0 \end{array}]$
$x ′ = [\begin{array}{r} - 3 & - 2 \\ 9 & 3 \end{array}] x, x (0) = [\begin{array}{r} 1 \\ - 1 \end{array}]$
$x ′ = [\begin{array}{r} 7 & - 5 \\ 4 & 3 \end{array}] x, x (0) = [\begin{array}{r} 2 \\ 0 \end{array}]$
$x ′ = [\begin{array}{r} 5 & 0 & - 6 \\ 2 & - 1 & - 2 \\ 4 & - 2 & - 4 \end{array}] x, x (0) = [\begin{array}{r} 2 \\ 1 \\ 0 \end{array}]$
$x ′ = [\begin{array}{r} 3 & 2 & 2 \\ - 5 & - 4 & - 2 \\ 5 & 5 & 3 \end{array}] x, x (0) = [\begin{array}{r} 1 \\ 0 \\ - 1 \end{array}]$

Compute the matrix exponential $e^{A t}$ for each system $x ′ = Ax$ given in Problems 9 through 20.

$x_{1}^{′} = 5 x_{1} - 4 x_{2}, x_{2}^{′} = 2 x_{1} - x_{2}$
$x_{1}^{′} = 6 x_{1} - 6 x_{2}, x_{2}^{′} = 4 x_{1} - 4 x_{2}$
$x_{1}^{′} = 5 x_{1} - 3 x_{2}, x_{2}^{′} = 2 x_{1}$
$x_{1}^{′} = 5 x_{1} - 4 x_{2}, x_{2}^{′} = 3 x_{1} - 2 x_{2}$
$x_{1}^{′} = 9 x_{1} - 8 x_{2}, x_{2}^{′} = 6 x_{1} - 5 x_{2}$
$x_{1}^{′} = 10 x_{1} - 6 x_{2}, x_{2}^{′} = 12 x_{1} - 7 x_{2}$
$x_{1}^{′} = 6 x_{1} - 10 x_{2}, x_{2}^{′} = 2 x_{1} - 3 x_{2}$
$x_{1}^{′} = 11 x_{1} - 15 x_{2}, x_{2}^{′} = 6 x_{1} - 8 x_{2}$
$x_{1}^{′} = 3 x_{1} + x_{2}, x_{2}^{′} = x_{1} + 3 x_{2}$
$x_{1}^{′} = 4 x_{1} + 2 x_{2}, x_{2}^{′} = 2 x_{1} + 4 x_{2}$
$x_{1}^{′} = 9 x_{1} + 2 x_{2}, x_{2}^{′} = 2 x_{1} + 6 x_{2}$
$x_{1}^{′} = 13 x_{1} + 4 x_{2}, x_{2}^{′} = 4 x_{1} + 7 x_{2}$

In Problems 21 through 24, show that the matrix A is nilpotent and then use this fact to find (as in Example 3) the matrix exponential $e^{A t}$ .

$A = [\begin{array}{r} 1 & - 1 \\ 1 & - 1 \end{array}]$
$A = [\begin{array}{r} 6 & 4 \\ - 9 & - 6 \end{array}]$
$A = [\begin{array}{r} 1 & - 1 & - 1 \\ 1 & - 1 & 1 \\ 0 & 0 & 0 \end{array}]$
$A = [\begin{array}{r} 3 & 0 & - 3 \\ 5 & 0 & 7 \\ 3 & 0 & - 3 \end{array}]$

Each coefficient matrix A in Problems 25 through 30 is the sum of a nilpotent matrix and a multiple of the identity matrix. Use this fact (as in Example 6) to solve the given initial value problem.

$x ′ = [\begin{array}{r} 2 & 5 \\ 0 & 2 \end{array}] x, x (0) = [\begin{array}{r} 4 \\ 7 \end{array}]$
$x ′ = [\begin{array}{r} 7 & 0 \\ 11 & 7 \end{array}] x, x (0) = [\begin{array}{r} 5 \\ - 10 \end{array}]$
$x ′ = [\begin{array}{r} 1 & 2 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}] x, x (0) = [\begin{array}{r} 4 \\ 5 \\ 6 \end{array}]$
$x ′ = [\begin{array}{r} 5 & 0 & 0 \\ 10 & 5 & 0 \\ 20 & 30 & 5 \end{array}] x, x (0) = [\begin{array}{r} 40 \\ 50 \\ 60 \end{array}]$
$x ′ = [\begin{array}{r} 1 & 2 & 3 & 4 \\ 0 & 1 & 6 & 3 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & 0 & 1 \end{array}] x, x (0) = [\begin{array}{r} 1 \\ 1 \\ 1 \\ 1 \end{array}]$
$x ′ = [\begin{array}{r} 3 & 0 & 0 & 0 \\ 6 & 3 & 0 & 0 \\ 9 & 6 & 3 & 0 \\ 12 & 9 & 6 & 3 \end{array}] x, x (0) = [\begin{array}{r} 1 \\ 1 \\ 1 \\ 1 \end{array}]$
Suppose that the $n \times n$ matrices A and B commute; that is, that $A B = B A$ . Prove that $e^{A + B} = e^{A} e^{B}$ . (Suggestion: Group the terms in the product of the two series on the right-hand side to obtain the series on the left.)
Deduce from the result of Problem 31 that, for every square matrix A, the matrix $e^{A}$ is nonsingular with ${(e^{A})}^{- 1} = e^{- A}$ .
Suppose that

$A = [\begin{array}{r} 0 & 1 \\ 1 & 0 \end{array}] .$

Show that $A^{2 n} = I$ and that $A^{2 n + 1} = A$ if n is a positive integer. Conclude that

$e^{A t} = I \cosh t + A \sinh t,$

and apply this fact to find a general solution of $x ′ = Ax$ . Verify that it is equivalent to the general solution found by the eigenvalue method.
Suppose that

$A = [\begin{array}{r} 0 & 2 \\ - 2 & 0 \end{array}] .$

Show that $e^{A t} = I \cos 2 t + \frac{1}{2} A \sin 2 t$ . Apply this fact to find a general solution of $x ′ = Ax$ , and verify that it is equivalent to the solution found by the eigenvalue method.

Apply Theorem 3 to calculate the matrix exponential $e^{A t}$ for each of the matrices in Problems 35 through 40.

$A = [\begin{array}{r} 3 & 4 \\ 0 & 3 \end{array}]$
$A = [\begin{array}{r} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{array}]$
$A = [\begin{array}{r} 2 & 3 & 4 \\ 0 & 1 & 3 \\ 0 & 0 & 1 \end{array}]$
$A = [\begin{array}{r} 5 & 20 & 30 \\ 0 & 10 & 20 \\ 0 & 0 & 5 \end{array}]$
$A = [\begin{array}{r} 1 & 3 & 3 & 3 \\ 0 & 1 & 3 & 3 \\ 0 & 0 & 2 & 3 \\ 0 & 0 & 0 & 2 \end{array}]$
$A = [\begin{array}{r} 2 & 4 & 4 & 4 \\ 0 & 2 & 4 & 4 \\ 0 & 0 & 2 & 4 \\ 0 & 0 & 0 & 3 \end{array}]$

8.1 Application Automated Matrix Exponential Solutions

If A is an $n \times n$ matrix, then a computer algebra system can be used first to calculate the fundamental matrix $e^{A t}$ for the system

$X^{′} = Ax,$ (1)

then to calculate the matrix product $x (t) = e^{A t} x_{0}$ to obtain a solution satisfying the initial condition $x (0) = x_{0}$ . For instance, suppose that we want to solve the initial value problem

$\begin{array}{rcl} x_{1}^{′} & = & 13 x_{1} + 4 x_{2}, \\ x_{2}^{′} & = & 4 x_{1} + 7 x_{2}; \\ x_{1} (0) & = & 11, x_{2} (0) = 23. \end{array}$

After the matrices

$A = [\begin{array}{r} 13 & 4 \\ 4 & 7 \end{array}], x_{0} = [\begin{array}{r} 11 \\ 23 \end{array}]$ (2)

have been entered, either the Maple command


with(linalg): exponential(A∗t)

the Mathematica command


MatrixExp[A t]

or the Matlab command


syms t, expm(A∗t)

yields the matrix exponential

$exp At = \frac{1}{5} [\begin{array}{c} e^{5 t} + 4 e^{15 t} & - 2 e^{5 t} + 2 e^{15 t} \\ - 2 e^{5 t} + 2 e^{15 t} & 4 e^{5 t} + e^{15 t} \end{array}] .$

Then either the Maple product multiply(expAt,x0), the Mathematica product expAt.x0, or the Matlab product expAt∗x0 gives the solution vector

$x = [\begin{array}{r} - 7 e^{5 t} + 18 e^{15 t} \\ 14 e^{5 t} + 9 e^{15 t} \end{array}] .$

Obviously this, finally, is the way to do it!

Matrix exponentials also allow for convenient interactive exploration of the system (1). For example, a version of the Mathematica commands


A = {{13, 4}, {4, 7}};
field = VectorPlot[A.{x, y}, {x, −25, 25},
 {y, −25, 25}];
Manipulate[
 curves = ParametricPlot[
 MatrixExp[A t, #]&/@pt, {t, −1, 1},
 PlotRange −> 25];
 Show[curves, field],
 {{pt, {{11, 23}, {20, −10}, {−20, −10},
 {−20, 10}}}, Locator}]

was used to generate Fig. 8.1.1, which shows four solution curves of the system (1) with the matrix A chosen as in Eq. (2). The initial conditions for each solution curve-one of which initially passes through the point (11, 23) of our initial value problem, while the other three pass through the points $(20, - 10), (- 20, - 10)$ , and (20, 10)—are specified by a “locator point” which can be freely dragged to any desired position in the phase plane, with the corresponding solution curve being instantly redrawn.

FIGURE 8.1.1.

Interactive display of the linear system (1). As the “locator points” are dragged to different positions, the solution curves are immediately redrawn, illustrating the behavior of the system.

Experimenting with such interactive displays can shed considerable light on the behavior of linear systems. For example, notice the straight line solution in Fig. 8.1.1; if you could drag the corresponding locator point around the phase plane, what other straight line solution could you find?How could you have predicted this by examining the matrix A?

For a three-dimensional example, solve the initial value problem

$\begin{array}{c} \begin{array}{rcrcrcr} x_{1}^{′} & = & - 149 x_{1} & - & 50 x_{2} & - & 154 x_{3}, \\ x_{2}^{′} & = & 537 x_{1} & + & 180 x_{2} & + & 546 x_{3}, \\ x_{3}^{′} & = & - 27 x_{1} & - & 9 x_{2} & - & 25 x_{3}; \end{array} \\ x_{1} (0) = 17, x_{2} (0) = 43, x_{3} (0) = 79. \end{array}$

And here’s a four-dimensional problem:

$\begin{array}{c} \begin{array}{rcrcrcrcr} x_{1}^{′} & = & 4 x_{1} & + & x_{2} & + & x_{3} & + & 7 x_{4}, \\ x_{2}^{′} & = & x_{1} & + & 4 x_{2} & + & 10 x_{3} & + & x_{4}, \\ x_{3}^{′} & = & x_{1} & + & 10 x_{2} & + & 4 x_{3} & + & x_{4}, \\ x_{4}^{′} & = & 7 x_{1} & + & x_{2} & + & x_{3} & + & 4 x_{4}; \end{array} \\ x_{1} (0) = 15, x_{2} (0) = 35, x_{3} (0) = 55, x_{4} (0) = 75. \end{array}$

If at this point you’re having too much fun with matrix exponentials to stop, make up some problems of your own. For instance, choose any homogeneous linear system appearing in this chapter and experiment with different initial conditions. The exotic $5 \times 5$ matrix A of the Section 7.6 application may suggest some interesting possibilities.

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Table of Contents for 8 Matrix Exponential Methods

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Table of Contents for
8 Matrix Exponential Methods