ODE-ProjectODE-Project The Matrix Exponential

Section 3.9 The Matrix Exponential

Objectives

To understand that $x (t) = e^{t A} x_{0}$ is the solution to the initial value problem

$\begin{aligned} x^{'} & = A x \\ x (0) & = x_{0}, \end{aligned}$

where $e^{t A}$ is the matrix exponential.
To understand that if $λ$ be an eigenvalue of an $n \times n$ matrix A and $v$ is an eigenvector for $λ,$ then $e^{t A} v = e^{λ t} v .$
To understand and be able to apply the properties of the matrix exponential to $n \times n$ matrices $A$ and $B .$
To understand and be able to use the matrix exponential to solve a linear system

$\begin{aligned} x^{'} & = A x \\ x (0) & = x_{0} . \end{aligned}$

Consider the linear system

\begin{aligned} x^{'} & = - 17 x - 46 y - 7 z \\ y^{'} & = 8 x + 21 y + 3 z \\ z^{'} & = - 7 x - 15 y - z \end{aligned}

The matrix associated with this system is

🔗

A = (\begin{matrix} - 17 & - 46 & - 7 \\ 8 & 21 & 3 \\ - 7 & - 15 & - 1 \end{matrix}) .

The characteristic polynomial of

A

🔗

p (λ) = det (A - λ I) = λ^{3} - 3 λ^{2} + 3 λ - 1 = (λ - 1)^{3}

hence, there is only a single eigenvalue

λ = 1 .

Moreover, we can only find a single linearly independent eigenvector

(1, - 1, 4) .

Thus,

🔗

x (t) = c e^{- t} (\begin{matrix} 1 \\ - 1 \\ 4 \end{matrix})

is a solution to our system. However, this is not the general solution to the system. We can only solve initial value problems where the initial condition lies on the line through the origin containing the vector

(1, - 1, 4) .

To construct a general solution to our system, we will need two other linearly independent solutions. One way of doing this is with the matrix exponential.

🔗

Subsection 3.9.1 The Exponential of a Matrix

Our goal is to construct a solution to the initial value problem

🔗

\begin{aligned} x^{'} & = A x \\ x (0) & = x_{0}, \end{aligned}

where

A

is an

n \times n

matrix. Recalling that the solution to the initial value problem

🔗

\begin{aligned} x^{'} & = k x \\ x (0) & = x_{0} \end{aligned}

x (t) = x_{0} e^{k t},

we might guess that a solution to the initial value problem

🔗

\begin{aligned} x^{'} & = A x \\ x (0) & = x_{0} \end{aligned}

has the form

🔗

x (t) = e^{t A} x_{0}

if we can make sense of the expression

e^{t A} .

🔗

We will define the exponential of

A

using the power series for

e^{t} .

Thus,

🔗

e^{A} = I + A + \frac{1}{2!} A^{2} + \frac{1}{3!} A^{3} + \dots = \sum_{k = 0}^{\infty} \frac{1}{k!} A^{k},

where

A

is an

n \times n

matrix, where

A^{0} = I

by convention. Each term makes sense in our definition since each is an expression of

n \times n

matrices; however, there are some issues surrounding the convergence of the power series. A series, even a series whose individual terms are matrix expressions, converges if and only if its partial sums,

🔗

S_{N} = \sum_{k = 0}^{N} \frac{1}{k!} A^{k}

converge. Although we shall not provide a proof, the matrix exponential

e^{A}

converges for all

A .

🔗

Example 3.9.1.

Let us compute the exponential of

A = (\begin{matrix} s & 0 \\ 0 & t \end{matrix}) .

Actually, this is quite easy,

\begin{aligned} e^{A} & = I + A + \frac{1}{2!} A^{2} + \frac{1}{3!} A^{3} + \dots \\ = (\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}) + (\begin{array}{c} s & 0 \\ 0 & t \end{array}) + \frac{1}{2!} (\begin{array}{c} s^{2} & 0 \\ 0 & t^{2} \end{array}) + \frac{1}{3!} (\begin{array}{c} s^{3} & 0 \\ 0 & t^{3} \end{array}) + \dots \\ = (\begin{array}{c} 1 + s + s^{2} / 2! + s^{3} / 3! + \dots & 0 \\ 0 & 1 + t + t^{2} / 2! + t^{3} / 3! + \dots \end{array}) \\ = (\begin{array}{c} e^{s} & 0 \\ 0 & e^{t} \end{array}) . \end{aligned}

Theorem 3.9.2.

A

is an

n \times n

matrix, then

🔗

\frac{d}{d t} e^{t A} = A e^{t A} .

🔗

Proof.

We simply need to differentiate

e^{t A} = I + t A + \frac{t^{2}}{2!} A^{2} + \frac{t^{3}}{3!} A^{3} + \dots

term by term.

Since we are differentiating an infinite series, we still need to show that differentiating term by term is something that can be done. We will, however, leave the details to a course in advanced calculus.

However,

\begin{aligned} \frac{d}{d t} e^{t A} & = \frac{d}{d t} (I + t A + \frac{t^{2}}{2!} A^{2} + \frac{t^{3}}{3!} A^{3} + \dots) \\ = A + t A^{2} + \frac{t^{2}}{2!} A^{3} + \dots \\ = A (I + t A + \frac{t^{2}}{2!} A^{2} + \frac{t^{3}}{3!} + \dots) \\ = A e^{t A} . \end{aligned}

Corollary 3.9.3.

Let

A

be an

n \times n

matrix. Then

x (t) = e^{t A} x_{0}

is the solution to the initial value problem

🔗

\begin{aligned} x^{'} & = A x \\ x (0) & = x_{0} \end{aligned}

🔗

Proof.

The corollary follows immediately from Theorem 3.9.2. If

x (t) = e^{t A} x_{0},

then

x^{'} (t) = \frac{d}{d t} (e^{t A} x_{0}) = \frac{d}{d t} (e^{t A}) x_{0} = A e^{t A} x_{0} = A x (t) .

Thus, solving linear systems is simply a matter of computing matrix exponentials. The problem is that matrix exponentials may not be so easy to compute.

🔗

Example 3.9.4.

The matrix

A = (\begin{matrix} 4 & 4 \\ - 9 & - 8 \end{matrix})

has repeated eigenvalues

λ = - 2 .

If we try to compute

e^{t A},

then

\begin{aligned} e^{t A} & = I + t A + \frac{t^{2}}{2!} A^{2} + \frac{t^{3}}{3!} A^{3} + \dots \\ = (\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}) + t (\begin{array}{c} 4 & 4 \\ - 9 & - 8 \end{array}) + \frac{t^{2}}{2!} {(\begin{array}{c} 4 & 4 \\ - 9 & - 8 \end{array})}^{2} + \frac{t^{3}}{3!} {(\begin{array}{c} 4 & 4 \\ - 9 & - 8 \end{array})}^{3} + \dots \\ = (\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}) + t (\begin{array}{c} 4 & 4 \\ - 9 & - 8 \end{array}) + \frac{t^{2}}{2!} (\begin{array}{c} - 20 & - 16 \\ - 36 & 28 \end{array}) + \frac{t^{3}}{3!} (\begin{array}{c} 64 & 48 \\ - 108 & - 80 \end{array}) + \dots \\ = (\begin{array}{c} 1 + t - 20 t^{2} / 2! + 64 t^{3} / 3! + \dots & 4 t - 16 t^{2} / 2! + 48 t^{3} / 3! + \dots \\ - 9 t - 36 t^{2} / 2! - 108 t^{3} / 3! + \dots & 1 - 8 t + 28 t^{2} / 2! - 80 t^{3} / 3! + \dots \end{array}) . \end{aligned}

It is not at all clear that this series converges to a matrix whose entries can be expressed in terms of elementary functions.

Now let us see how we can use the matrix exponential to solve a linear system as well as invent a more direct way to compute the matrix exponential.

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Theorem 3.9.5.

Let

λ

be an eigenvalue of an

n \times n

matrix

A .

v

is an eigenvector for

λ,

then

e^{t A} v = e^{λ t} v .

🔗

Proof.

Since

v

is an eigenvalue for

λ,

we know that

A v = λ v .

Using mathematical induction, we can show that

A^{n}

has eigenvalue

λ^{n}

with associated eigenvector

v .

Indeed,

A^{n} v = A (A^{n - 1} v) = A (λ^{n - 1} v) = λ^{n - 1} A v = λ^{n} v .

Hence,

\begin{aligned} e^{t A} v & = (I + t A + \frac{t^{2}}{2!} A^{2} + \frac{t^{3}}{3!} A^{3} + \dots) v \\ = v + t A v + \frac{t^{2}}{2!} A^{2} v + \frac{t^{3}}{3!} A^{3} v + \dots \\ = v + t λ v + \frac{t^{2}}{2!} λ^{2} v + \frac{t^{3}}{3!} λ^{3} v + \dots \\ = (I + t λ + \frac{t^{2}}{2!} λ^{2} + \frac{t^{3}}{3!} λ^{3} + \dots) v \\ = e^{λ t} v . \end{aligned}

The matrix exponential shares several properties with the exponential function

e^{x}

that we studied in calculus.

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Theorem 3.9.6.

Let

A

and

B

n \times n

matrices. Then

🔗

$A e^{A} = e^{A} A;$
if $A B = B A,$ then $e^{A} e^{B} = e^{A + B};$
$e^{A}$ has inverse $e^{- A} .$

🔗

Proof.

To prove (1), we can simply expand both sides of the equality in a power series,

\begin{aligned} A e^{A} & = A (I + A + \frac{1}{2!} A^{2} + \frac{1}{3!} A^{3} + \dots) \\ = A + A^{2} + \frac{1}{2!} A^{3} + \frac{1}{3!} A^{4} + \dots \\ = (I + A + \frac{1}{2!} A^{2} + \frac{1}{3!} A^{3} + \dots) A \\ = e^{A} A . \end{aligned}

Proving (2) is a only bit more complicated if we notice that the binomial expansion holds for matrices,

(A + B)^{n} = \sum_{k = 0}^{n} (\begin{matrix} n \\ k \end{matrix}) A^{k} B^{n - k},

where

(\begin{matrix} n \\ k \end{matrix}) = \frac{n!}{k! (n - k)!},

providing

A B = B A .

\begin{aligned} e^{A + B} & = \sum_{n = 0}^{\infty} \frac{1}{n!} (A + B)^{n} \\ = \sum_{n = 0}^{\infty} \frac{1}{n!} (\sum_{k = 0}^{n} (\begin{array}{c} n \\ k \end{array}) A^{k} B^{n - k}) \\ = \sum_{n = 0}^{\infty} (\sum_{k = 0}^{n} \frac{1}{k! (n - k)!} A^{k} B^{n - k}) \\ = (\sum_{n = 0}^{\infty} \frac{1}{n!} A^{n}) (\sum_{n = 0}^{\infty} \frac{1}{n!} B^{n}) \\ = e^{A} e^{B} . \end{aligned}

Simply expand each series out to see that this is true. Part (3) follows directly from Part (2), since

A

and

- A

commute.

Example 3.9.7.

Now let us compute

e^{t A}

once again for

A = (\begin{matrix} 4 & 4 \\ - 9 & - 8 \end{matrix}) .

First notice that

A = λ I + (A - λ I) .

Since the identity matrix

I

commutes with every matrix, we know that

e^{t A} = e^{t (λ I + (A - λ I))} = e^{t λ I} e^{t (A - λ I)} .

We also know that

e^{λ t I} = e^{λ t} I

by Example 3.9.1. Thus,

e^{t A} = e^{t (λ I + (A - λ I))} = e^{t λ I} e^{t (A - λ I)} = e^{λ t} e^{t (A - λ I)} .

The matrix

A

has repeated eigenvalue

λ = - 2 .

Consequently,

A - λ I = A + 2 I = (\begin{matrix} 6 & 4 \\ - 9 & - 6 \end{matrix}),

and

(A - λ I)^{2}

is the zero matrix. Thus,

\begin{aligned} e^{t A} & = e^{λ t} e^{t (A - λ I)} \\ = e^{λ t} (I + t (A - λ I) + \frac{t^{2}}{2!} (A - λ I)^{2} + \frac{t^{2}}{3!} (A - λ I)^{3} + \dots) \\ = e^{- 2 t} (I + t (A + 2 I) + \frac{t^{2}}{2!} (A + 2 I)^{2} + \frac{t^{2}}{3!} (A + 2 I)^{3} + \dots) \\ = e^{- 2 t} (I + t (A + 2 I)) \\ = e^{- 2 t} (\begin{array}{c} 1 + 6 t & 4 t \\ - 9 t & - 6 t \end{array}) . \end{aligned}

Our example suggests at the following proposition. We leave the proof of this proposition as an exercise.

🔗

Proposition 3.9.8.

A

is an

n \times n

matrix with a single eigenvalue

λ,

then there exists a nonnegative integer

k < n

such that

🔗

e^{t A} = e^{λ t} (I + t (A - λ I) + \frac{t^{2}}{2!} (A - λ I)^{2} + \dots + \frac{t^{k}}{k!} (A - λ I)^{k}) .

🔗

Example 3.9.9.

We are now ready to return to our original system

x^{'} = A x,

where

A = (\begin{matrix} - 17 & - 46 & - 7 \\ 8 & 21 & 3 \\ - 7 & - 15 & - 1 \end{matrix}) .

This matrix has a single eigenvalue

λ = 1 .

It is easy to show that the only nonzero powers of

A - λ I = A - I

are

\begin{aligned} (A + I) & = (\begin{array}{c} - 18 & - 46 & - 7 \\ 8 & 20 & 3 \\ - 7 & - 15 & - 2 \end{array}) \\ (A + I)^{2} & = (\begin{array}{c} 5 & 13 & 2 \\ - 5 & - 13 & - 2 \\ 20 & 52 & 8 \end{array}) . \end{aligned}

Therefore,

\begin{aligned} e^{t A} & = e^{t} (I + t (A + I) + \frac{t^{2}}{2!} (A + I)^{2}) \\ = e^{t} (\begin{array}{c} 1 - 18 t + 5 t^{2} / 2 & - 46 t + 13 t^{2} / 2 & - 7 t + t^{2} \\ 8 t - 5 t^{2} / 2 & 1 + 20 t - 13 t^{2} / 2 & 3 t - t^{2} \\ - 7 t + 10 t^{2} & - 15 t + 26 t^{2} & 1 - 2 t + 4 t^{2} \end{array}) . \end{aligned}

Now, to compute three linearly independent solutions for

x^{'} = A x,

we simply compute

e^{t A} v

for three linearly independent vectors. We will use the standard basis vectors

e_{1} = (\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}), e_{2} = (\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}), e_{3} = (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}) .

Thus, the general solution to our system is

\begin{aligned} x (t) & = c_{1} e^{- t} (\begin{array}{c} 1 - 18 t + 5 t^{2} / 2 \\ 8 t - 5 t^{2} / 2 \\ - 7 t + 10 t^{2} \end{array}) + c_{2} e^{- t} (\begin{array}{c} - 46 t + 13 t^{2} / 2 \\ 1 + 20 t - 13 t^{2} / 2 \\ - 15 t + 26 t^{2} \end{array}) \\ + c_{3} e^{- t} (\begin{array}{c} - 7 t + t^{2} \\ 3 t - t^{2} \\ 1 - 2 t + 4 t^{2} \end{array}) . \end{aligned}

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Subsection 3.9.2 Generalized Eigenvalues

Example 3.9.10.

Consider the system

x^{'} = A x,

where

A = (\begin{matrix} 7 & - 1 & 2 \\ 18 & - 2 & 6 \\ - 9 & 2 & - 1 \end{matrix}) .

The characteristic polynomial of

A

det (A - λ I) = λ^{3} - 4 λ^{2} + 5 λ - 2 = (λ - 1)^{2} (λ - 2) .

The eigenvalue

λ_{1} = 1

has eigenvector

v_{1} = (2, 6, - 3)

and the eigenvalue

λ_{2} = 2

has eigenvector

v_{2} = (1, 3, - 1) .

Thus, we can find two linearly independent solutions in this case

x_{1} (t) = e^{t} (\begin{matrix} 2 \\ 6 \\ - 3 \end{matrix}) and x_{2} (t) = e^{2 t} (\begin{matrix} 1 \\ 3 \\ - 1 \end{matrix}) .

Since

λ_{1} = 1

has multiplicity two and we can find only one linearly independent eigenvector, it is not possible to apply Proposition 3.9.8 in this case.

If we consider the exponential

\begin{aligned} e^{t A} v & = e^{λ_{1} t} e^{t (A - λ_{1} I)} v \\ = e^{λ_{1} t} (v + t (A - λ_{1} I) v + \frac{t^{2}}{2!} (A - λ_{1} I)^{2} v + \frac{t^{2}}{3!} (A - λ_{1} I)^{3} v + \dots) \end{aligned}

where

v, v_{1}

and

v_{2}

are linearly independent, our goal is to choose

v

for which the series truncates. That is, we must look for vectors

v

such that

(A - λ_{1} I)^{k} v = (A - I)^{k} v = 0 .

k = 1,

then

(A - λ_{1} I) v = (A - I) v = 0,

which means that

v

is an eigenvector. Thus,

v

must be a multiple of

v_{1} = (2, 6, - 3)

in this case. Since we already know that the eigenspace associated with this eigenvector has dimension one and is generated by

v_{1},

we must consider higher powers.

Since

(A - λ I) = (A - I) = (\begin{matrix} 6 & - 1 & 2 \\ 18 & - 3 & 6 \\ - 9 & 2 & - 2 \end{matrix}),

we have

(A - λ I)^{2} = (A - I)^{2} = (\begin{matrix} 0 & 1 & 2 \\ 0 & 3 & 6 \\ 0 & - 1 & - 2 \end{matrix}) .

The nullspace of this matrix has dimension two. Certainly,

v_{1} = (2, 6, - 3)

is in the nullspace of

(A - I)^{2},

since it is the nullspace of

A - I .

We wish to find a vector that is not a multiple of the vector

v_{1}

that is also in the nullspace of

(A - I)^{2} .

The vector

v = (0, 2, - 1)

will do. Now our series truncates,

\begin{aligned} x_{3} (t) & = e^{t A} v \\ = e^{λ_{1} t} e^{t (A - λ_{1} I)} v \\ = e^{λ_{1} t} (v + t (A - λ_{1} I) v) \\ = e^{λ_{1} t} [(\begin{array}{c} 0 \\ 2 \\ - 1 \end{array}) + t (A - I) (\begin{array}{c} 0 \\ 2 \\ - 1 \end{array})] \\ = e^{t} (\begin{array}{c} - 4 t \\ 2 - 12 t \\ - 1 + 6 t \end{array}) \end{aligned}

We now have a general solution for our system,

x (t) = c_{1} e^{t} (\begin{matrix} 2 \\ 6 \\ - 3 \end{matrix}) + c_{2} e^{2 t} (\begin{matrix} 1 \\ 3 \\ - 1 \end{matrix}) + c_{3} e^{t} (\begin{matrix} - 4 t \\ 2 - 12 t \\ - 1 + 6 t \end{matrix}) .

λ

is an eigenvalue of

A

and

(A - λ I)^{p} v = 0

for some

p \geq 1,

then

v

is called a generalized eigenvector of

A .

When eigenvalues have algebraic multiplicity greater than one, we can compute extra solutions by looking for vectors in the nullspace of

(A - λ I)^{p}

for

p > 1 .

The following theorem tells us that this is always possible. We leave the proof of the theorem as an exercise in linear algebra.

🔗

Theorem 3.9.11.

Suppose that

λ

is an eigenvalue of

A

with multiplicity

q .

Then there exists an integer

p \leq q

such that the dimension of the nullspace of

(A - λ I)^{p}

q .

🔗

We now have a procedure for finding

q

linearly independent solutions corresponding to an eigenvalue

λ

of multiplicity

q .

🔗

Find the smallest integer $p$ such that the nullspace of $(A - λ I)^{p}$ has dimension $q .$
Find a basis ${v_{1}, \dots, v_{q}}$ for the nullspace of $(A - λ I)^{p} .$
For each $v_{j}$ ( $j = 1, 2, \dots, q$ ), we have the solution

$\begin{aligned} x_{j} (t) & = e^{t A} v_{j} \\ = e^{λ t} (v_{j} + t (A - λ I) v_{j} + \dots + \frac{t^{p - 1}}{(p - 1)!} (A - λ I)^{p - 1} v_{j} .) . \end{aligned}$

This procedure works for complex eigenvalues as well as real. If

λ = α + i β

has eigenvector

z = x + i y,

then set

x = Re z

and

y = Im z .

🔗

Activity 3.9.1. Solving Systems Using the Matrix Exponential.

Consider the system

d x / d t = A x,

where

🔗

A = (\begin{matrix} - 12 & 26 & - 1 \\ - 8 & 17 & 0 \\ 3 & - 5 & 4 \end{matrix})

🔗

(a)

Find the eigenvalue,

λ

A

(there is only one).

🔗

(b)

Find eigenvectors,

v

for the eigenvalue

λ .

🔗

(c)

Compute

A - λ I, (A - λ I)^{2}, \dots .

🔗

(d)

Compute

e^{t A}

as in Example 3.9.9.

🔗

(e)

Find the general solution to

d x / d t = A x .

🔗

Subsection 3.9.3 Important Lessons

If $A$ is an $n \times n$ matrix, we define the exponential of $A$ to be

$e^{A} = I + A + \frac{1}{2!} A^{2} + \frac{1}{3!} A^{3} + \dots = \sum_{k = 0}^{\infty} \frac{1}{k!} A^{k} .$
If $A$ is an $n \times n$ matrix, then

$\frac{d}{d t} e^{t A} = A e^{t A} .$
Let $A$ be an $n \times n$ matrix. Then $x (t) = e^{t A} x_{0}$ is the solution to the initial value problem

$\begin{aligned} x^{'} & = A x \\ x (0) & = x_{0} \end{aligned}$
Let $λ$ be an eigenvalue of an $n \times n$ matrix A. If $v$ is an eigenvector for $λ,$ then $e^{t A} v = e^{λ t} v .$
Let $A$ and $B$ be $n \times n$ matrices. Then
- $A e^{A} = e^{A} A;$
- if $A B = B A,$ then $e^{A} e^{B} = e^{A + B};$
- $e^{A}$ is nonsingular with inverse $e^{- A} .$
If $A$ is an $n \times n$ matrix with a single eigenvalue $λ,$ then there exists nonnegative integer $k < n$ such that

$e^{t A} = e^{λ t} (I + t (A - λ I) + \frac{t^{2}}{2!} (A - λ I)^{2} + \dots + \frac{t^{k}}{k!} (A - λ I)^{k}) .$
Suppose that $λ$ is an eigenvalue of $A$ with multiplicity $q .$ Then there exists an integer $p \leq q$ such that the dimension of the nullspace of $(A - λ I)^{p}$ is $q .$
The procedure for finding $q$ linearly independent solutions corresponding to an eigenvalue $λ$ of multiplicity $q$ is the following.
- Find the smallest integer $p$ such that the nullspace of $(A - λ I)^{p}$ has dimension $q .$
- Find a basis ${v_{1}, \dots, v_{q}}$ for the nullspace of $(A - λ I)^{p} .$
- For each $v_{j}$ ( $j = 1, 2, \dots, q$ ), we have the solution
  
  $\begin{aligned} x_{j} (t) & = e^{t A} v_{j} \\ = e^{λ t} (v_{j} + t (A - λ I) v_{j} + \dots + \frac{t^{p - 1}}{(p - 1)!} (A - λ I)^{p - 1} v_{j} .) . \end{aligned}$
The procedure works for complex eigenvalues as well as real. If $λ = α + i β$ has eigenvector $z,$ then set $x = Re z$ and $y = Im z .$

🔗

Reading Questions 3.9.4 Reading Questions

1.

What is the exponential of a matrix

A ?

🔗

2.

Explain what a generalized eigenvector for a matrix

A

is.

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Exercises 3.9.5 Exercises

Finding Solutions Using the Matrix Exponential.

For each of the linear systems

d x / d t = A x

in Exercise Group 3.9.5.1–8

🔗

Find the eigenvalues of $A .$ You may find Sage useful.
Find the eigenvectors for each eigenvalue of $A .$
Compute $A - λ I, (A - λ I)^{2}, \dots .$
Compute $e^{t A}$
Find the general solution to $d x / d t = A x .$

🔗

1.

A = (\begin{matrix} 3 & 1 \\ - 4 & - 1 \end{matrix})

🔗

2.

A = (\begin{matrix} - 1 & 1 \\ - 4 & - 5 \end{matrix})

🔗

3.

A = (\begin{matrix} 7 & 1 \\ - 4 & 3 \end{matrix})

🔗

4.

A = (\begin{matrix} - 2 & 1 \\ - 4 & - 6 \end{matrix})

🔗

5.

A = (\begin{matrix} - 5 & 3 & - 2 \\ - 10 & 6 & - 3 \\ 2 & - 1 & 2 \end{matrix})

🔗

6.

A = (\begin{matrix} - 8 & 3 & - 2 \\ - 10 & 3 & - 3 \\ 2 & - 1 & - 1 \end{matrix})

🔗

7.

A = (\begin{matrix} - 3 & 3 & - 2 \\ - 10 & 8 & - 3 \\ 2 & 4 & 4 \end{matrix})

🔗

8.

A = (\begin{matrix} - 10 & 3 & - 2 \\ - 10 & 1 & - 3 \\ 2 & - 1 & - 3 \end{matrix})

🔗

Solving Initial-value Problems.

Solve each of initial-value problems in Exercise Group 3.9.5.9–16

🔗

9.

\begin{aligned} x^{'} & = 3 x + y \\ y^{'} & = - 4 x - y \\ x (0) & = - 1 \\ y (0) & = 1 \end{aligned}

🔗

10.

\begin{aligned} x^{'} & = - x + y \\ y^{'} & = - 4 x - 5 y \\ x (0) & = 2 \\ y (0) & = - 1 \end{aligned}

🔗

11.

\begin{aligned} x^{'} & = 7 x + y \\ y^{'} & = - 4 x + 3 y \\ x (0) & = 1 \\ y (0) & = 1 \end{aligned}

🔗

12.

\begin{aligned} x^{'} & = - 2 x + y \\ y^{'} & = - 4 x - 6 y \\ x (0) & = 1 \\ y (0) & = 2 \end{aligned}

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13.

\begin{aligned} x^{'} & = - 5 x + 3 y - 2 z \\ y^{'} & = - 10 x + 6 y - 3 z \\ z^{'} & = 2 x - y + 2 z \\ x (0) & = - 1 \\ y (0) & = - 2 \\ z (0) & = - 3 \end{aligned}

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14.

\begin{aligned} x^{'} & = - 8 x + 3 y - 2 z \\ y^{'} & = - 10 x + 3 y - 3 z \\ z^{'} & = 2 x - y - z \\ x (0) & = - 2 \\ y (0) & = 2 \\ z (0) & = 0 \end{aligned}

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15.

\begin{aligned} x^{'} & = - 3 x + 3 y - 2 z \\ y^{'} & = - 10 x + 8 y - 3 z \\ z^{'} & = 2 x + 4 y + 4 z \\ x (0) & = 1 \\ y (0) & = 1 \\ z (0) & = 3 \end{aligned}

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16.

\begin{aligned} x^{'} & = - 10 x + 3 y - 2 z \\ y^{'} & = - 10 x + y - 3 z \\ z^{'} & = 2 x - y - 3 z \\ x (0) & = - 1 \\ y (0) & = 0 \\ z (0) & = 2 \end{aligned}

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17.

Let

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A = (\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}) and B = (\begin{matrix} 0 & 0 \\ - 1 & 0 \end{matrix}) .

Verify that

e^{A t} e^{B t} \neq e^{(A + B) t} .

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18.

Prove Proposition 3.9.8.

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19.

Suppose that we are given the system

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\begin{matrix} (3.9.1) & \frac{d x}{d t} = A x (t) + f (t), x (t_{0}) = c, \end{matrix}

where

A

is a matrix of constants. For example, the system

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\begin{aligned} \frac{d x_{1}}{d t} & = x_{2} \\ \frac{d x_{2}}{d t} & = 8 x_{1} - 2 x_{2} + e^{t} \\ x_{1} (0) & = 1 \\ x_{2} (0) & = - 4 \end{aligned}

can be written in the form

d x / d t = A x (t) + f (t),

x (t_{0}) = c,

where

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A = (\begin{matrix} 0 & 1 \\ 8 & - 2 \end{matrix}), f (t) = (\begin{matrix} 0 \\ e^{t} \end{matrix}), c = (\begin{matrix} 1 \\ - 4 \end{matrix}) .

Show that

$x (t) = e^{A (t - t_{0})} c + e^{A t} \int_{t_{0}}^{t} e^{- A s} f (s) d s$

is a solution to (3.9.1).
Use Item 3.9.5.19.a to solve the system

$\begin{aligned} \frac{d x_{1}}{d t} & = x_{2} \\ \frac{d x_{2}}{d t} & = 8 x_{1} - 2 x_{2} + e^{t} \\ x_{1} (0) & = 1 \\ x_{2} (0) & = - 4. \end{aligned}$

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