ODE-Project Changing Coordinates

Section 6.1 Changing Coordinates

Objectives

To understand that a linear map $T$ converts solutions of $y^{'} = (T^{- 1} A T) y$ to solutions of $x^{'} = A x,$ and, conversely, the inverse of a linear map $T$ takes solutions of $x^{'} = A x$ to solutions of $y^{'} = (T^{- 1} A T) y .$
To understand that a change of coordinates converts the system $x^{'} = A x$ to one of the following special cases,

$(\begin{matrix} λ & 0 \\ 0 & μ \end{matrix}), (\begin{matrix} α & β \\ - β & α \end{matrix}), (\begin{matrix} λ & 0 \\ 0 & λ \end{matrix}), (\begin{matrix} λ & 1 \\ 0 & λ \end{matrix}) .$

In Chapter 5, we outlined procedures for solving systems of linear differential equations of the form

(\begin{matrix} d x / d t \\ d y / d t \end{matrix}) = (\begin{matrix} a & b \\ c & d \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = A (\begin{matrix} x \\ y \end{matrix})

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by determining the eigenvalues of

A .

In this section we will consider the following special cases for

A,

\begin{matrix} (6.1.1) & (\begin{matrix} λ & 0 \\ 0 & μ \end{matrix}), (\begin{matrix} α & β \\ - β & α \end{matrix}), (\begin{matrix} λ & 0 \\ 0 & λ \end{matrix}), (\begin{matrix} λ & 1 \\ 0 & λ \end{matrix}) . \end{matrix}

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Although it may seem that we have limited ourselves by attacking only a very small part of the problem of finding solutions for

x^{'} = A x,

we are actually very close to providing a complete classification of all solutions. We will now show that we can transform any

2 \times 2

system of first-order linear differential equations with constant coefficients into one of these special systems by using a change of coordinates.

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Subsection 6.1.1 Linear Maps

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First, we need to add a few things to our knowledge of matrices and linear algebra. A linear map or linear transformation on

R^{2}

is a function

T : R^{2} \to R^{2}

that is defined by a matrix. That is,

T (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} a & b \\ c & d \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) .

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When there is no confusion, we will think of the linear map

T : R^{2} \to R^{2}

and the matrix

(\begin{matrix} a & b \\ c & d \end{matrix})

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as interchangeable.

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We will say that

T : R^{2} \to R^{2}

is an invertible linear map if we can find a second linear map

S

such that

T \circ S = S \circ T = I,

where

I

is the identity transformation. In terms of matrices, this means that we can find a matrix

S

such that

T S = S T = I,

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where

I = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix})

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is the

2 \times 2

identity matrix. We write

T^{- 1}

for the inverse matrix of

T .

It is easy to check that the inverse of

T = (\begin{matrix} a & b \\ c & d \end{matrix})

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T^{- 1} = \frac{1}{det T} (\begin{matrix} d & - b \\ - c & a \end{matrix}) .

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Theorem 6.1.1.

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A linear map

T

is invertible if and only if

det T \neq 0 .

Proof.

det T = 0,

then there are infinitely many nonzero vectors

x

such that

T x = 0 .

Suppose that

T^{- 1}

exists and

x \neq 0

such that

T x = 0 .

Then

x = T^{- 1} T x = T^{- 1} 0 = 0,

which is a contradiction. On the other hand, we can certainly compute

T^{- 1},

at least in the

2 \times 2

case, if the determinant is nonzero.

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Subsection 6.1.2 Changing Coordinates

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In Subsection 5.1.2, we discussed what a basis was along with the coordinates with respect to a particular basis. The vectors

e_{1} = (1, 0)

and

e_{2} = (0, 1)

form a basis for

R^{2} .

Indeed, if

z = (- 5, - 4),

then we can write

z = - 5 e_{1} - 4 e_{2} .

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We say that the coordinates of

z

with respect to the basis

{e_{1}, e_{2}}

are

(- 5, - 4) .

Now consider the vectors

v_{1} = (2, 1)

and

v_{2} = (3, 2) .

Since

det (\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}) \neq 0,

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these vectors are linearly independent form a different basis for

R^{2} .

z = (- 5, - 4),

then we can write

z = 2 v_{1} - 3 v_{2} .

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The coordinates of

z

with respect to the basis

{v_{1}, v_{2}}

are

(2, - 3) .

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Suppose we wish to convert the coordinates with repect to one basis to a new set of coordinates with respect to a different basis; that is, we wish to do a change of coordinates. Observe that

\begin{aligned} v_{1} & = 2 e_{1} + e_{2} \\ v_{2} & = 3 e_{1} + 2 e_{2} . \end{aligned}

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It follows that

\begin{aligned} c_{1} v_{1} + c_{2} v_{2} & = c_{1} (2 e_{1} + e_{2}) + c_{2} (3 e_{1} + 2 e_{2}) \\ = (2 c_{1} + 3 c_{2}) e_{1} + (c_{2} + 2 c_{2}) e_{2} . \end{aligned}

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Thus, the coordinates of

c_{1} v_{1} + c_{2} v_{2}

with respect to the basis

{e_{1}, e_{2}}

can be determined by

(\begin{matrix} 2 c_{1} + 3 c_{2} \\ c_{2} + 2 c_{2} \end{matrix}) = (\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}) (\begin{matrix} c_{1} \\ c_{2} \end{matrix}) .

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If we let

T = (\begin{matrix} 2 & 3 \\ 1 & 2 \end{matrix}) and c = (\begin{matrix} c_{1} \\ c_{2} \end{matrix}),

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then the coordinates with respect to the basis

{e_{1}, e_{2}}

are given by

d = T c .

If we are given the coordinates with respect to the basis

{v_{1}, v_{2}}

for a vector, we simply need to multiply by the matrix

T .

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Now suppose that we wish to find the coordinates with respect to the basis

{v_{1}, v_{2}}

if we know that a vector

z = d_{1} e_{1} + d_{2} e_{2} .

Since

d = T c,

we need only multiply both sides of the equation by

T^{- 1}

to get

c = T^{- 1} d .

In our example,

T^{- 1} d = (\begin{matrix} 2 & - 3 \\ - 1 & 2 \end{matrix}) (\begin{matrix} d_{1} \\ d_{2} \end{matrix}) .

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In our particular example,

T^{- 1} d = (\begin{matrix} 2 & - 3 \\ - 1 & 2 \end{matrix}) (\begin{matrix} - 5 \\ - 4 \end{matrix}) = (\begin{matrix} 2 \\ - 3 \end{matrix}),

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which are the coordinates of

z

with respect to the basis

{v_{1}, v_{2}} .

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Subsection 6.1.3 Systems and Changing Coordinates

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The idea now is to use a change of coordinates to convert an arbitrary system

x^{'} = A x

into one of the special systems mentioned at the beginning of the section (6.1.1), solve the new system, and then convert our new solution back to a solution of the original system using another change of coordinates.

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Suppose that we consider a linear system

\begin{matrix} (6.1.2) & y^{'} = (T^{- 1} A T) y \end{matrix}

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where

T

is an invertible matrix. If

y (t)

is a solution of (6.1.2), we claim that

x (t) = T y (t)

solves the equation

x^{'} = A x .

Indeed,

\begin{aligned} x^{'} (t) & = (T y)^{'} (t) \\ = T y^{'} (t) \\ = T ((T^{- 1} A T) y (t)) \\ = A (T y (t)) \\ = A x (t) . \end{aligned}

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We can think of this in two ways.

A linear map $T$ converts solutions of $y^{'} = (T^{- 1} A T) y$ to solutions of $x^{'} = A x .$
The inverse of a linear map $T$ takes solutions of $x^{'} = A x$ to solutions of $y^{'} = (T^{- 1} A T) y .$

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W e are now in a position to solve our problem of finding solutions of an arbitrary linear system

(\begin{matrix} x^{'} \\ y^{'} \end{matrix}) = (\begin{matrix} a & b \\ c & d \end{matrix}) = (\begin{matrix} x \\ y \end{matrix}) .

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Subsection 6.1.4 Distinct Real Eigenvalues

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Consider the system

x^{'} = A x,

where

A

has two real, distinct eigenvalues

λ_{1}

and

λ_{2}

with eigenvectors

v_{1}

and

v_{2},

respectively. Let

T

be the matrix with columns

v_{1}

and

v_{2} .

e_{1} = (1, 0)

and

e_{2} = (0, 1),

then

T e_{i} = v_{i}

for

i = 1, 2 .

Consequently,

T^{- 1} v_{i} = e_{i}

for

i = 1, 2 .

Thus, we have

\begin{aligned} (T^{- 1} A T) e_{i} & = T^{- 1} A v_{i} \\ = T^{- 1} (λ_{i} v_{i}) \\ = λ_{i} T^{- 1} v_{i} \\ = λ_{i} e_{i} \end{aligned}

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for

i = 1, 2 .

Therefore, the matrix

T^{- 1} A T

is in canonical form,

T^{- 1} A T = (\begin{matrix} λ_{1} & 0 \\ 0 & λ_{2} \end{matrix}) .

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The eigenvalues of the matrix

T^{- 1} A T

are

λ_{1}

and

λ_{2}

with eigenvectors

(1, 0)

and

(0, 1),

respectively. Thus, the general solution of

y^{'} = (T^{- 1} A T) y

🔗

y (t) = α e^{λ_{1} t} (\begin{matrix} 1 \\ 0 \end{matrix}) + β e^{λ_{2} t} (\begin{matrix} 0 \\ 1 \end{matrix}) .

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Hence, the general solution of

x^{'} = A x

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\begin{aligned} T y (t) & = T (α e^{λ_{1} t} (\begin{array}{c} 1 \\ 0 \end{array}) + β e^{λ_{2} t} (\begin{array}{c} 0 \\ 1 \end{array})) \\ = α e^{λ_{1} t} T (\begin{array}{c} 1 \\ 0 \end{array}) + β e^{λ_{2} t} T (\begin{array}{c} 0 \\ 1 \end{array}) \\ = α e^{λ_{2} t} v_{1} + β e^{λ_{2} t} v_{2} . \end{aligned}

Example 6.1.2.

Suppose

d x / d t = A x,

where

A = (\begin{matrix} 1 & 2 \\ 4 & 3 \end{matrix}) .

The eigenvalues of

A

are

λ_{1} = 5

and

λ_{2} = - 1

and the associated eigenvectors are

(1, 2)

and

(1, - 1),

respectively. In this case, our matrix

T

(\begin{matrix} 1 & 1 \\ 2 & - 1 \end{matrix}) .

e_{1} = (1, 0)

and

e_{2} = (0, 1),

then

T e_{i} = v_{i}

for

i = 1, 2 .

Consequently,

T^{- 1} v_{i} = e_{i}

for

i = 1, 2,

where

T^{- 1} = (\begin{matrix} 1 / 3 & 1 / 3 \\ 2 / 3 & - 1 / 3 \end{matrix}) .

Thus,

T^{- 1} A T = (\begin{matrix} 1 / 3 & 1 / 3 \\ 2 / 3 & - 1 / 3 \end{matrix}) (\begin{matrix} 1 & 2 \\ 4 & 3 \end{matrix}) (\begin{matrix} 1 & 1 \\ 2 & - 1 \end{matrix}) = (\begin{matrix} 5 & 0 \\ 0 & - 1 \end{matrix}) .

The eigenvalues of the matrix

(\begin{matrix} 5 & 0 \\ 0 & - 1 \end{matrix})

are

λ_{1} = 5

and

λ_{2} = - 1

with eigenvectors

(1, 0)

and

(0, 1),

respectively. Thus, the general solution of

y^{'} = (T^{- 1} A T) y

y (t) = α e^{5 t} (\begin{matrix} 1 \\ 0 \end{matrix}) + β e^{- t} (\begin{matrix} 0 \\ 1 \end{matrix}) .

Hence, the general solution of

x^{'} = A x

\begin{aligned} T y (t) & = (\begin{array}{c} 1 & 1 \\ 2 & - 1 \end{array}) (α e^{5 t} (\begin{array}{c} 1 \\ 0 \end{array}) + β e^{- t} (\begin{array}{c} 0 \\ 1 \end{array})) \\ = α e^{5 t} (\begin{array}{c} 1 \\ 2 \end{array}) + β e^{- t} (\begin{array}{c} 1 \\ - 1 \end{array}) \end{aligned}

The linear map

T

converts the phase portrait of the system

y^{'} = (T^{- 1} A T) y

(Figure 6.1.3) to the phase portrait of the system

x^{'} = A x

(Figure 6.1.4).

described in detail following the image — Figure 6.1.3. Phase portrait for $y^{'} = (T^{- 1} A T) y$

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Activity 6.1.1. Distinct Real Eigenvalues and Transformation of Coordinates.

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Consider the system of linear differential equations

d x / d t = A x,

where

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

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(a)

🔗

Find the eigenvalues of

A .

You should find distinct real eigenvalues

λ

and

μ .

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(b)

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Find the general solution for

d x / d t = A x .

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(c)

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Construct the

2 \times 2

matrix

T = (v_{1}, v_{2})

and find

T^{- 1} .

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(d)

🔗

Calculate

T^{- 1} A T .

You should obtain the diagonal matrix

(\begin{matrix} λ & 0 \\ 0 & μ \end{matrix})

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with eigenvectors

e_{1} = (1, 0)

and

e_{2} = (0, 1) .

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(e)

🔗

The general solution of

y^{'} = (T^{- 1} A T) y

🔗

y (t) = α e^{λ t} (\begin{matrix} 1 \\ 0 \end{matrix}) + β e^{μ t} (\begin{matrix} 0 \\ 1 \end{matrix}) .

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Now calculate

T y

and compare this solution with the one that you obtained in Activity 5.2.1.

Of couurse, we have much quicker ways of solving a system

d x / d t = A x

with distinct real eigenvalues. The goal of this section is show that we have covered all possible cases for

2 \times 2

systems of linear differential equations and not to invent new methods of solution.

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Subsection 6.1.5 Complex Eigenvalues

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Suppose the matrix

A = (\begin{matrix} a & b \\ c & d \end{matrix})

🔗

in system

x^{'} = A x

has complex eigenvalues. In this case, the characteristic polynomial

p (λ) = λ^{2} - (a + d) λ + (a d - b c)

will have roots

λ = α + i β

and

\overset{―}{λ} = α - i β,

where

\begin{aligned} α & = \frac{a + d}{2} \\ β & = \frac{\sqrt{4 b c - (a - d)^{2}}}{2} . \end{aligned}

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The eigenvalues

λ

and

\overset{―}{λ}

are complex conjugates. Now, suppose that the eigenvalue

λ = α + i β

has an eigenvector of the form

v = v_{1} + i v_{2},

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where

v_{1}

and

v_{2}

are real vectors. Then

\overset{―}{v} = v_{1} - i v_{2}

is an eigenvector for

\overset{―}{λ},

since

A \overset{―}{v} = \overset{―}{A v} = \overset{―}{λ v} = \overset{―}{λ} \overset{―}{v} .

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Consequently, if

A

is a real matrix with complex eigenvalues, one of the eigenvalues determines the other.

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Proposition 6.1.5.

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λ = α + i β

is an eigenvalue of a real matrix

A

with

β \neq 0

and eigenvector the form

v_{1} + i v_{2},

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where

v_{1}

and

v_{2}

are real vectors, then the vectors

v_{1}

and

v_{2}

are linearly independent.

Proof.

v_{1}

and

v_{2}

are not linearly independent, then

v_{1} = c v_{2}

for some

c \in R .

On one hand, we have

A (v_{1} + i v_{2}) = A (c v_{2} + i v_{2}) = (c + i) A v_{2} .

However,

\begin{aligned} A (v_{1} + i v_{2}) & = (α + i β) (v_{1} + i v_{2}) \\ = (α + i β) (c + i) v_{2} \\ = (c + i) (α + i β) v_{2} \end{aligned}

In other words,

A v_{2} = (α + i β) v_{2} .

However, this is a contradiction since the left-side of the equation says that we have real eigenvector while the right-side of the equation is complex. Thus,

v_{1}

and

v_{2}

are linearly independent.

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Proposition 6.1.6.

🔗

Let

A

be a real matrix with eigenvalue

λ = α + i β,

where

β \neq 0 .

v_{1} + i v_{2},

🔗

is an eigenvector for

λ,

then there exists a matrix

T

such that

T^{- 1} A T = (\begin{matrix} α & β \\ - β & α \end{matrix}) .

Proof.

Since

v_{1} + i v_{2}

is an eigenvector associated to the eigenvalue

α + i β,

we have

A (v_{1} + i v_{2}) = (α + i β) (v_{1} + i v_{2}) .

Equating the real and imaginary parts, we find that

\begin{aligned} A v_{1} & = α v_{1} - β v_{2} \\ A v_{2} & = β v_{1} + α v_{2} . \end{aligned}

T

is the matrix with columns

v_{1}

and

v_{2},

then

\begin{aligned} T e_{1} & = v_{1} \\ T e_{2} & = v_{2} . \end{aligned}

Thus, we have

(T^{- 1} A T) e_{1} = T^{- 1} (α v_{1} - β v_{2}) = α e_{1} - β e_{2} .

Similarly,

(T^{- 1} A T) e_{2} = β e_{1} + α e_{2} .

Therefore, we can write the matrix

T^{- 1} A T

T^{- 1} A T = (\begin{matrix} α & β \\ - β & α \end{matrix}) .

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The system

y^{'} = (T^{- 1} A T) y

is in one of the canonical forms and has a phase portrait that is a spiral sink (

α < 0

), a center (

α = 0

), or a spiral source (

α > 0

). After a change of coordinates, the phase portrait of

x^{'} = A x

is equivalent to a sink, center, or source.

Example 6.1.7.

Suppose that we wish to find the solutions of the second order equation

2 x^{″} + 2 x^{'} + x = 0.

This particular equation might model a damped harmonic oscillator. If we rewrite this second-order equation as a first-order system, we have

\begin{aligned} x^{'} & = y \\ y^{'} & = - \frac{1}{2} x - y, \end{aligned}

or equivalently

x^{'} = A x,

where

A = (\begin{matrix} 0 & 1 \\ - 1 / 2 & - 1 \end{matrix}) .

The eigenvalues of

A

are

- \frac{1}{2} \pm i \frac{1}{2} .

The eigenvalue

λ = (1 + i) / 2

has an eigenvector

v = (\begin{matrix} 2 \\ - 1 + i \end{matrix}) = (\begin{matrix} 2 \\ - 1 \end{matrix}) + i (\begin{matrix} 0 \\ 1 \end{matrix}),

respectively. Therefore, we can take

T

to be the matrix

T = (\begin{matrix} 2 & 0 \\ - 1 & 1 \end{matrix}) .

Consequently,

T^{- 1} A T = (\begin{matrix} 1 / 2 & 0 \\ 1 / 2 & 1 \end{matrix}) (\begin{matrix} 0 & 1 \\ - 1 / 2 & - 1 \end{matrix}) (\begin{matrix} 2 & 0 \\ - 1 & 1 \end{matrix}) = (\begin{matrix} - 1 / 2 & 1 / 2 \\ - 1 / 2 & - 1 / 2 \end{matrix}),

which is in the canonical form

(\begin{matrix} α & β \\ - β & α \end{matrix}) .

The general solution to

y^{'} = (T^{- 1} A T) y

y (t) = c_{1} e^{- t / 2} (\begin{matrix} \cos (t / 2) \\ - \sin (t / 2) \end{matrix}) + c_{2} e^{- t / 2} (\begin{matrix} \sin (t / 2) \\ \cos (t / 2) \end{matrix}) .

The phase portrait of

y^{'} = (T^{- 1} A T) y

is given in Figure 6.1.8.

The general solution of

x^{'} = A x

\begin{aligned} T y (t) & = (\begin{array}{c} 2 & 0 \\ - 1 & 1 \end{array}) [c_{1} e^{- t / 2} (\begin{array}{c} \cos (t / 2) \\ - \sin (t / 2) \end{array}) + c_{2} e^{- t / 2} (\begin{array}{c} \sin (t / 2) \\ \cos (t / 2) \end{array})] \\ = c_{1} e^{- t / 2} (\begin{array}{c} 2 & 0 \\ - 1 & 1 \end{array}) (\begin{array}{c} \cos (t / 2) \\ - \sin (t / 2) \end{array}) + c_{2} e^{- t / 2} (\begin{array}{c} 2 & 0 \\ - 1 & 1 \end{array}) (\begin{array}{c} \sin (t / 2) \\ \cos (t / 2) \end{array}) \\ = c_{1} e^{- t / 2} (\begin{array}{c} 2 \cos (t / 2) \\ - \cos (t / 2) - \sin (t / 2) \end{array}) + c_{2} e^{- t / 2} (\begin{array}{c} 2 \sin (t / 2) \\ - \sin (t / 2) + \cos (t / 2) \end{array}) . \end{aligned}

The phase portrait for this system is given in Figure 6.1.9.

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Remark 6.1.10.

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Of course, we have a much more efficient way of solving the system

x^{'} = A x,

where

A = (\begin{matrix} 0 & 1 \\ - 1 / 2 & - 1 \end{matrix}) .

🔗

Since

A

has eigenvalue

λ = (- 1 + i) / 2

with an eigenvector

v = (2, - 1 + i),

we can apply Euler’s formula and write the solution as

\begin{aligned} x (t) & = e^{(- 1 + i) t / 2} v \\ = e^{- t / 2} e^{i t / 2} (\begin{array}{c} 2 \\ - 1 + i \end{array}) \\ = e^{- t / 2} (\cos (t / 2) + i \sin (t / 2)) (\begin{array}{c} 2 \\ - 1 + i \end{array}) \\ = e^{- t / 2} (\begin{array}{c} 2 \cos (t / 2) \\ - \cos (t / 2) - \sin (t / 2) \end{array}) + i e^{- t / 2} (\begin{array}{c} 2 \sin (t / 2) \\ - \sin (t / 2) + \cos (t / 2) \end{array}) . \end{aligned}

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Taking the real and the imaginary parts of the last expression, the general solution of

x^{'} = A x

x (t) = c_{1} e^{- t / 2} (\begin{matrix} 2 \cos (t / 2) \\ - \cos (t / 2) - \sin (t / 2) \end{matrix}) + c_{2} e^{- t / 2} (\begin{matrix} 2 \sin (t / 2) \\ - \sin (t / 2) + \cos (t / 2) \end{matrix}),

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which agrees with the solution that we found by transforming coordinates.

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Subsection 6.1.6 Repeated Eigenvalues

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Now suppose that

A

has a single real eigenvalue

λ .

Then the characteristic polynomial of

A

p (λ) = λ^{2} - (a + d) λ + (a d - b c),

then

A

has an eigenvalue

λ = (a + d) / 2 .

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Proposition 6.1.11.

🔗

A

has a single eigenvalue and a pair of linearly independent eigenvectors, then

A

must be of the form

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

Proof.

Suppose that

u

and

v

are linearly indeendent eigenvectors for

A,

and let

T

be the matrix whose first column is

u

and second column is

v .

That is,

T e_{1} = u

and

T e_{2} = v .

Since

u

and

v

are linearly independent,

det (T) \neq 0

and

T

is invertible. So, it must be the case that

A T = (A u, A v) = (λ u, λ v) = λ (u, v) = λ I T,

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

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In this case, the system is uncoupled and is easily solved. That is, we can solve each equation in the system

\begin{aligned} x^{'} & = λ x \\ y^{'} & = λ y \end{aligned}

🔗

separately to obtain the general solution

\begin{aligned} x & = c_{1} e^{λ t} \\ y^{'} & = c_{2} e^{λ t} . \end{aligned}

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Proposition 6.1.12.

🔗

Suppose that

A

has a single eigenvalue

λ .

v

is an eigenvector for

λ

and any other eigenvector for

λ

is a multiple of

v,

then there exists a matrix

T

such that

T^{- 1} A T = (\begin{matrix} λ & 1 \\ 0 & λ \end{matrix}) .

Proof.

w

is another vector in

R^{2}

such that

v

and

w

are linearly independent, then

A w

can be written as a linear combination of

v

and

w,

A w = α v + β w .

We can assume that

α \neq 0;

otherwise, we would have a second linearly independent eigenvector. We claim that

β = λ .

If this were not the case, then

\begin{aligned} A (w + (\frac{α}{β - λ}) v) & = A w + (\frac{α}{β - λ}) A v \\ = α v + β w + λ (\frac{α}{β - λ}) v \\ = β w + α (1 + \frac{λ}{β - λ}) v \\ = β w + α (\frac{β - λ + λ}{β - λ}) v \\ = β (w + (\frac{α}{β - λ}) v) \end{aligned}

and

β

would be an eigenvalue distinct from

λ .

Thus,

A w = α v + λ w .

If we will let

u = (1 / α) w,

then

A u = v + \frac{λ}{α} w = v + λ u .

We now define

T e_{1} = v

and

T e_{2} = u .

Since

\begin{aligned} A T & = A u + A v = v + λ u + λ v \\ T (\begin{array}{c} λ & 1 \\ 0 & λ \end{array}) & = T (λ e_{1}) + T e_{1} + T (λ e_{2}) = v + λ u + λ v, \end{aligned}

we have

T^{- 1} A T = (\begin{matrix} λ & 1 \\ 0 & λ \end{matrix}) .

Therefore,

x^{'} = A x

is in canonical form after a change of coordinates.

Example 6.1.13.

Consider the system

x^{'} = A x,

where

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

The characteristic polynomial of

A

λ^{2} - 6 λ + 9 = (λ - 3)^{2},

we have only a single eigenvalue

λ = 3

with eigenvector

v = (1, - 2) .

Any other eigenvector for

λ

is a multiple of

v .

If we choose

w = (1, 0),

then

v

and

w

are linearly independent. Furthermore,

A w = (\begin{matrix} 5 \\ - 4 \end{matrix}) = 2 (\begin{matrix} 1 \\ - 2 \end{matrix}) + λ (\begin{matrix} 1 \\ 0 \end{matrix}) = 2 (\begin{matrix} 1 \\ - 2 \end{matrix}) + 3 (\begin{matrix} 1 \\ 0 \end{matrix}) .

So we can let

u = (1 / 2) w = (1 / 2, 0) .

Therefore, the matrix that we seek is

T = (\begin{matrix} 1 & 1 / 2 \\ - 2 & 0 \end{matrix}),

and

T^{- 1} A T = (\begin{matrix} - 1 / 2 & 2 \\ 1 & 1 \end{matrix}) (\begin{matrix} 5 & 1 \\ - 4 & 1 \end{matrix}) (\begin{matrix} 1 & 1 / 2 \\ - 2 & 0 \end{matrix}) = (\begin{matrix} 3 & 1 \\ 0 & 3 \end{matrix}) .

From Section 5.3, we know that the general solution to the system

(\begin{matrix} d x / d t \\ d y / d t \end{matrix}) = (\begin{matrix} 3 & 1 \\ 0 & 3 \end{matrix}) (\begin{matrix} x \\ y \end{matrix})

y (t) = c_{1} e^{3 t} (\begin{matrix} 1 \\ 0 \end{matrix}) + c_{2} e^{3 t} (\begin{matrix} t \\ 1 \end{matrix}) .

Therefore, the general solution to

(\begin{matrix} d x / d t \\ d y / d t \end{matrix}) = (\begin{matrix} 5 & 1 \\ - 4 & 1 \end{matrix}) (\begin{matrix} x \\ y \end{matrix})

\begin{aligned} x (t) & = T y (t) \\ = c_{1} e^{3 t} T (\begin{array}{c} 1 \\ 0 \end{array}) + c_{2} e^{3 t} T (\begin{array}{c} t \\ 1 \end{array}) \\ = c_{1} e^{3 t} (\begin{array}{c} 1 \\ - 2 \end{array}) + c_{2} e^{3 t} (\begin{array}{c} 1 / 2 + t \\ - 2 t \end{array}) . \end{aligned}

This solution agrees with the solution that we found in Example 5.5.5.

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In practice, we find solutions to linear systems using the methods that we outlined in Sections 5.2–5.4. What we have demonstrated in this section is that those solutions are exactly the ones that we want.

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Subsection 6.1.7 Important Lessons

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A linear map $T$ is invertible if and only if $det T \neq 0 .$
A linear map $T$ converts solutions of $y^{'} = (T^{- 1} A T) y$ to solutions of $x^{'} = A x .$
The inverse of a linear map $T$ takes solutions of $x^{'} = A x$ to solutions of $y^{'} = (T^{- 1} A T) y$ .
A change of coordinates converts the system $x^{'} = A x$ to one of the following special cases,

$(\begin{matrix} λ & 0 \\ 0 & μ \end{matrix}), (\begin{matrix} α & β \\ - β & α \end{matrix}), (\begin{matrix} λ & 0 \\ 0 & λ \end{matrix}), (\begin{matrix} λ & 1 \\ 0 & λ \end{matrix}) .$