Hirsch M., Smale S., Devaney R., Differential Equations, Dynamical Systems, and an Introduction to Chaos

Подождите немного. Документ загружается.

86 Chapter 5 Higher Dimensional Linear Algebra

These equations reduce to

x − 3z = 0

y − 2z = 0.

Hence V

= (3, 2, 1) is an eigenvector associated to λ = 2. In similar fashion

we ﬁnd that (1, 0, 0) is an eigenvector associated to λ = 1, while (0, 1, 2) is an

eigenvector associated to λ =−1. Then we set

T =

⎛

⎝

310

201

102

⎞

⎠

A simple calculation shows that

AT = T

⎛

⎝

20 0

01 0

00−1

⎞

⎠

Since det T =−3, T is invertible and we have

−1

AT =

⎛

⎝

20 0

01 0

00−1

⎞

⎠



5.3 Complex Eigenvalues

Now we treat the case where A has nonreal (complex) eigenvalues. Suppose

α + iβ is an eigenvalue of A with β = 0. Since the characteristic equation for

A has real coefﬁcients, it follows that if α + iβ is an eigenvalue, then so is its

complex conjugate

α + iβ = α − iβ.

Another way to see this is the following. Let V be an eigenvector associated

to α + iβ. Then the equation

AV = (α + iβ)V

shows that V is a vector with complex entries. We write

V =

⎛

⎜

⎝

+ iy

⎞

⎟

⎠

5.3 Complex Eigenvalues 87

Let V denote the complex conjugate of V :

V =

⎛

⎜

⎝

− iy

⎞

⎟

⎠

Then we have

V = AV = (α + iβ)V = (α − iβ)V ,

which shows that

V is an eigenvector associated to the eigenvalue α − iβ.

Notice that we have (temporarily) stepped out of the “real” world of

and

into the world

of complex vectors. This is not really a problem, since all of

the previous linear algebraic results hold equally well for complex vectors.

Now suppose that A isa2n × 2n matrix with distinct nonreal eigenvalues

± iβ

for j = 1, ..., n. Let V

and V

denote the associated eigenvectors.

Then, just as in the previous proposition, this collection of eigenvectors is

linearly independent. That is, if we have



j=1

+ d

) = 0

where the c

and d

are now complex numbers, then we must have c

= d

= 0

for each j.

Now we change coordinates to put A into canonical form. Let

2j−1



+ V



−i



− V



Note that W

2j−1

and W

are both real vectors. Indeed, W

2j−1

is just the real

part of V

while W

is its imaginary part. So working with the W

brings us

back home to

Proposition. The vectors W

, ..., W

are linearly independent.

Proof: Suppose not. Then we can ﬁnd real numbers c

, d

for j = 1, ..., n

such that



j=1



2j−1

+ d



= 0

88 Chapter 5 Higher Dimensional Linear Algebra

but not all of the c

and d

are zero. But then we have



j=1



+ V

) − id

− V

)



= 0

from which we ﬁnd



j=1



− id

+ (c

+ id



= 0.

Since the V

and V

’s are linearly independent, we must have c

± id

= 0,

from which we conclude c

= d

= 0 for all j. This contradiction establishes

the result.



Note that we have

2j−1

(AV

+ AV

)



(α + iβ)V

+ (α − iβ)V



+ V

) +

iβ

− V

)

= αW

2j−1

− βW

Similarly, we compute

= βW

2j−1

+ αW

Now consider the linear map T for which TE

= W

for j = 1, ...,2n.

That is, the matrix associated to T has columns W

, ..., W

. Note that this

matrix has real entries. Since the W

are linearly independent, it follows from

Section 5.2 that T is invertible. Now consider the matrix T

−1

AT . We have

−1

AT )E

2j−1

= T

−1

2j−1

= T

−1

(αW

2j−1

− βW

)

= αE

2j−1

− βE

and similarly

−1

AT )E

= βE

2j−1

+ αW

5.4 Bases and Subspaces 89

Therefore the matrix associated to T

−1

AT is

−1

AT =

⎛

⎜

⎝

⎞

⎟

⎠

where each D

isa2× 2 matrix of the form



−β



This is our canonical form for matrices with distinct nonreal eigenvalues.

Combining the results of this and the previous section, we have:

Theorem. Suppose that the n ×n matrix A has distinct eigenvalues. Then there

exists a linear map T so that

−1

AT =

⎛

⎜

⎝



⎞

⎟

⎠

where the D

are 2 × 2 matrices in the form



−β





5.4 Bases and Subspaces

To deal with the case of a matrix with repeated eigenvalues, we need some

further algebraic concepts. Recall that the collection of all linear combinations

of a given ﬁnite set of vectors is called a subspace of

. More precisely, given

, ..., V

∈ R

, the set

S ={α

+···+α

|α

∈ R}

is a subspace of

. In this case we say that S is spanned by V

, ..., V

90 Chapter 5 Higher Dimensional Linear Algebra

Definition

Let

S be a subspace of R

. A collection of vectors V

, ..., V

is a

basis of

S if the V

are linearly independent and span S.

Note that a subspace always has a basis, for if S is spanned by V

, ..., V

,we

can always throw away certain of the V

in order to reach a linearly independent

subset of these vectors that spans

S. More precisely, if the V

are not linearly

independent, then we may ﬁnd one of these vectors, say, V

, for which

= β

+···+β

k−1

Hence we can write any vector in

S as a linear combination of the V

, ..., V

k−1

alone; the vector V

is extraneous. Continuing in this fashion, we eventually

reach a linearly independent subset of the V

that spans S.

More important for our purposes is:

Proposition. Every basis of a subspace

S ⊂ R

has the same number of

elements.

Proof: We ﬁrst observe that the system of k linear equations in k+ unknowns

given by

+···+a

1 k+

k+

= 0

+···+a

kk+

k+

= 0

always has a nonzero solution if  > 0. Indeed, using row reduction, we may

ﬁrst solve for one unknown in terms of the others, and then we may eliminate

this unknown to obtain a system of k − 1 equations in k +  − 1 unknowns.

Thus we are ﬁnished by induction (the ﬁrst case, k = 1, being obvious).

Now suppose that V

, ..., V

is a basis for the subspace S. Suppose that

, ..., W

k+

is also a basis of S, with  > 0. Then each W

is a linear

combination of the V

, so we have constants a

such that



i=1

, for j = 1, ..., k + .

5.4 Bases and Subspaces 91

By the previous observation, the system of k equations

k+



j=1

= 0, for i = 1, ..., k

has a nonzero solution (c

, ..., c

k+

). Then

k+



j=1

k+



j=1





i=1





i=1



k+



j=1



= 0

so that the W

are linearly dependent. This contradiction completes the

proof.



As a consequence of this result, we may deﬁne the dimension of a subspace

S as the number of vectors that form any basis for S. In particular, R

a subspace of itself, and its dimension is clearly n. Furthermore, any other

subspace of

must have dimension less than n, for otherwise we would have

a collection of more than n vectors in

that are linearly independent. This

cannot happen by the previous proposition. The set consisting of only the 0

vector is also a subspace, and we deﬁne its dimension to be zero. We write

dim

S for the dimension of the subspace S.

Example. A straight line through the origin in

forms a one-dimensional

subspace of

, since any vector on this line may be written uniquely as tV

where V ∈

is a ﬁxed nonzero vector lying on the line and t ∈ R is arbitrary.

Clearly, the single vector V forms a basis for this subspace.



Example. The plane P in R

deﬁned by

x + y + z = 0

is a two-dimensional subspace of

. The vectors (1, 0, −1) and (0, 1, −1) both

lie in

P and are linearly independent. If W ∈ P, we may write

W =

⎛

⎝

−y − x

⎞

⎠

= x

⎛

⎝

−1

⎞

⎠

+ y

⎛

⎝

−1

⎞

⎠

so these vectors also span

P. 

92 Chapter 5 Higher Dimensional Linear Algebra

As in the planar case, we say that a function T : R

→ R

is linear if

T (X) = AX for some n × n matrix A. T is called a linear map or linear

transformation. Using the properties of matrices discussed in Section 5.1, we

have

T (αX + βY ) = αT (X) + βT (Y )

for any α, β ∈

R and X, Y ∈ R

. We say that the linear map T is invertible if

the matrix A associated to T has an inverse.

For the study of linear systems of differential equations, the most important

types of subspaces are the kernels and ranges of linear maps. We deﬁne the

kernel of T , denoted Ker T , to be the set of vectors mapped to 0 by T . The

range of T consists of all vectors W for which there exists a vector V for which

TV = W . This, of course, is a familiar concept from calculus. The difference

here is that the range of T is always a subspace of

Example. Consider the linear map

T (X) =

⎛

⎝

010

001

000

⎞

⎠

If X = (x, y, z), then

T (X) =

⎛

⎝

⎞

⎠

Hence Ker T consists of all vectors of the form (α, 0, 0) while Range T is the

set of vectors of the form (β, γ , 0), where α, β, γ ∈

R. Both sets are clearly

subspaces.



Example. Let

T (X) = AX =

⎛

⎝

123

456

789

⎞

⎠

For Ker T , we seek vectors X that satisfy AX = 0. Using row reduction, we

ﬁnd that the reduced row echelon form of A is the matrix

⎛

⎝

10−1

01 2

00 0

⎞

⎠

5.4 Bases and Subspaces 93

Hence the solutions X = (x, y, z)ofAX = 0 satisfy x = z, y =−2z.

Therefore any vector in Ker T is of the form (z, −2z, z), so Ker T has dimen-

sion one. For Range T , note that the columns of A are vectors in Range T ,

since they are the images of (1, 0, 0), (0, 1, 0), and (0, 0, 1), respectively. These

vectors are not linearly independent since

−1

⎛

⎝

⎞

⎠

+ 2

⎛

⎝

⎞

⎠

⎛

⎝

⎞

⎠

However, (1, 4, 7) and (2, 5, 8) are linearly independent, so these two vectors

give a basis of Range T .



Proposition. Let T : R

→ R

be a linear map. Then Ker T and Range T

are both subspaces of

. Moreover,

dim Ker T + dim Range T = n.

Proof: First suppose that Ker T ={0}. Let E

, ..., E

be the standard basis

. Then we claim that TE

, ..., TE

are linearly independent. If this is not

the case, then we may ﬁnd α

, ..., α

, not all 0, such that



j=1

= 0.

But then we have

⎛

⎝



j=1

⎞

⎠

= 0,

which implies that



∈ Ker T , so that



= 0. Hence each α

= 0,

which is a contradiction. Thus the vectors TE

are linearly independent. But

then, given V ∈

, we may write

V =



j=1

94 Chapter 5 Higher Dimensional Linear Algebra

for some β

, ..., β

. Hence

V = T

⎛

⎝



j=1

⎞

⎠

which shows that Range T =

. Hence both Ker T and Range T are

subspaces of

and we have dim Ker T = 0 and dim Range T = n.

If Ker T ={0}, we may ﬁnd a nonzero vector V

∈ Ker T . Clearly,

T (αV

) = 0 for any α ∈ R, so all vectors of the form αV

lie in Ker T .

If Ker T contains additional vectors, choose one and call it V

. Then Ker T

contains all linear combinations of V

and V

, since

T (α

+ α

) = α

+ α

= 0.

Continuing in this fashion we obtain a set of linearly independent vectors that

span Ker T , thus showing that Ker T is a subspace. Note that this process must

end, since every collection of more than n vectors in

is linearly dependent.

A similar argument works to show that Range T is a subspace.

Now suppose that V

, ..., V

form a basis of Ker T where 0 <k<n

(the case where k = n being obvious). Choose vectors W

k+1

, ..., W

so that

, ..., V

, W

k+1

, ..., W

form a basis of R

. Let Z

= TW

for each j. Then

the vectors Z

are linearly independent, for if we had

k+1

+···+α

= 0,

then we would also have

T (α

k+1

+···+α

) = 0.

This implies that

k+1

+···+α

∈ Ker T .

But this is impossible, since we cannot write any W

(and hence any linear

combination of the W

) as a linear combination of the V

. This proves that the

sum of the dimensions of Ker T and Range T is n.



We remark that it is easy to ﬁnd a set of vectors that spans Range T ; simply

take the set of vectors that comprise the columns of the matrix associated

to T . This works since the ith column vector of this matrix is the image of the

standard basis vector E

under T . In particular, if these column vectors are

5.5 Repeated Eigenvalues 95

linearly independent, then Ker T ={0} and there is a unique solution to the

equation T (X) = V for every V ∈

. Hence we have:

Corollary. If T :

→ R

is a linear map with dim Ker T = 0, then T is

invertible.



5.5 Repeated Eigenvalues

In this section we describe the canonical forms that arise when a matrix has

repeated eigenvalues. Rather than spending an inordinate amount of time

developing the general theory in this case, we will give the details only for

3 × 3 and 4 × 4 matrices with repeated eigenvalues. More general cases are

relegated to the exercises. We justify this omission in the next section where

we show that the “typical” matrix has distinct eigenvalues and hence can be

handled as in the previous section. (If you happen to meet a random matrix

while walking down the street, the chances are very good that this matrix will

have distinct eigenvalues!) The most general result regarding matrices with

repeated eigenvalues is given by:

Proposition. Let A be an n ×n matrix. Then there is a change of coordinates

T for which

−1

AT =

⎛

⎜

⎝

⎞

⎟

⎠

where each of the B

’s is a square matrix (and all other entries are zero) of one of

the following forms:

(i)

⎛

⎜

⎝

λ 1

⎞

⎟

⎠

(ii)

⎛

⎜

⎝

⎞

⎟

⎠

where



αβ

−βα



, I



