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

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96 Chapter 5 Higher Dimensional Linear Algebra

and where α, β, λ ∈ R with β = 0. The special cases where B

= (λ) or



αβ

−βα



are, of course, allowed.



We ﬁrst consider the case of R

.IfA has repeated eigenvalues in R

, then

all eigenvalues must be real. There are then two cases. Either there are two

distinct eigenvalues, one of which is repeated, or else all eigenvalues are the

same. The former case can be handled by a process similar to that described

in Chapter 3, so we restrict our attention here to the case where A has a single

eigenvalue λ of multiplicity 3.

Proposition. Suppose A is a 3 × 3 matrix for which λ is the only eigenvalue.

Then we may ﬁnd a change of coordinates T such that T

−1

AT assumes one of

the following three forms:

(i)

⎛

⎝

λ 00

0 λ 0

00λ

⎞

⎠

(ii)

⎛

⎝

λ 10

0 λ 0

00λ

⎞

⎠

(iii)

⎛

⎝

λ 10

0 λ 1

00λ

⎞

⎠

Proof: Let K be the kernel of A − λI . Any vector in K is an eigenvector of A.

There are then three subcases depending on whether the dimension of K is 1,

2, or 3.

If the dimension of K is 3, then (A − λI )V = 0 for any V ∈

. Hence

A = λI . This yields matrix (i).

Suppose the dimension of K is 2. Let R be the range of A − λI . Then R has

dimension 1 since dim K + dim R = 3, as we saw in the previous section. We

claim that R ⊂ K . If this is not the case, let V ∈ R be a nonzero vector. Since

(A −λI )V ∈ R and R is one dimensional, we must have (A −λI)V = μV for

some μ = 0. But then AV = (λ + μ)V , so we have found a new eigenvalue

λ + μ. This contradicts our assumption, so we must have R ⊂ K .

Now let V

∈ R be nonzero. Since V

∈ K , V

is an eigenvector and so

(A − λI)V

= 0. Since V

also lies in R, we may ﬁnd V

∈ R

− K with

(A − λI )V

= V

. Since K is two dimensional we may choose a second

vector V

∈ K such that V

and V

are linearly independent. Note that V

is also an eigenvector. If we now choose the change of coordinates TE

for j = 1, 2, 3, then it follows easily that T

−1

AT assumes the form of

case (ii).

Finally, suppose that K has dimension 1. Thus R has dimension 2. We claim

that, in this case, K ⊂ R. If this is not the case, then (A − λI )R = R and so

5.5 Repeated Eigenvalues 97

A − λI is invertible on R. Thus, if V ∈ R, there is a unique W ∈ R for which

(A − λI )W = V . In particular, we have

AV = A(A − λI )W

= (A

− λA)W

= (A − λI )(AW ).

This shows that, if V ∈ R, then so too is AV . Hence A also preserves the

subspace R. It then follows immediately that A must have an eigenvector in R,

but this then says that K ⊂ R and we have a contradiction.

Next we claim that (A − λI )R = K . To see this, note that (A − λI )R is

one dimensional, since K ⊂ R.If(A − λI )R = K , there is a nonzero vector

V ∈ K for which (A − λI )R ={tV } where t ∈

R. But then (A − λI)V = tV

for some t ∈

R, t = 0, and so AV = (t +λ)V yields another new eigenvalue.

Thus we must in fact have (A − λI )R = K .

Now let V

∈ K be an eigenvector for A. As above there exists V

∈ R such

that (A−λI )V

= V

. Since V

∈ R there exists V

such that (A−λI )V

= V

Note that (A −λI)

= V

. The V

are easily seen to be linearly independent.

Moreover, the linear map deﬁned by TE

= V

ﬁnally puts A into canonical

form (iii). This completes the proof.



Example. Suppose

A =

⎛

⎝

20−1

021

−1 −12

⎞

⎠

Expanding along the ﬁrst row, we ﬁnd

det(A − λI ) = (2 − λ)[(2 − λ)

+ 1]−(2 − λ) = (2 − λ)

so the only eigenvalue is 2. Solving (A −2I )V = 0 yields only one independent

eigenvector V

= (1, −1, 0), so we are in case (iii) of the proposition. We

compute

(A − 2I )

⎛

⎝

110

−1 −10

000

⎞

⎠

so that the vector V

= (1, 0, 0) solves (A − 2I )

= V

. We also have

(A − 2I )V

= V

= (0, 0, −1).

98 Chapter 5 Higher Dimensional Linear Algebra

As before, we let TE

= V

for j = 1, 2, 3, so that

T =

⎛

⎝

101

−100

0 −10

⎞

⎠

Then T

−1

AT assumes the canonical form

−1

AT =

⎛

⎝

210

021

002

⎞

⎠



Example. Now suppose

A =

⎛

⎝

110

−130

−112

⎞

⎠

Again expanding along the ﬁrst row, we ﬁnd

det(A − λI ) = (1 − λ)[(3 − λ)(2 − λ)]+(2 − λ) = (2 − λ)

so again the only eigenvalue is 2. This time, however, we have

A − 2I =

⎛

⎝

−110

⎞

⎠

so that we have two linearly independent eigenvectors (x, y, z) for which we

must have x = y while z is arbitrary. Note that (A −2I )

is the zero matrix, so

we may choose any vector that is not an eigenvector as V

, say, V

= (1, 0, 0).

Then (A − 2I )V

= V

= (−1, −1, −1) is an eigenvector. A second linearly

independent eigenvector is then V

= (0, 0, 1), for example. Deﬁning TE

as usual then yields the canonical form

−1

AT =

⎛

⎝

210

020

002

⎞

⎠



Now we turn to the 4 × 4 case. The case of all real eigenvalues is similar to

the 3 × 3 case (though a little more complicated algebraically) and is left as

5.5 Repeated Eigenvalues 99

an exercise. Thus we assume that A has repeated complex eigenvalues α ± iβ

with β = 0.

There are just two cases; either we can ﬁnd a pair of linearly independent

eigenvectors corresponding to α+iβ, or we can ﬁnd only one such eigenvector.

In the former case, let V

and V

be the independent eigenvectors. The V

and

are linearly independent eigenvectors for α −iβ. As before, choose the real

vectors

= (V

+ V

)/2

=−i(V

− V

)/2

= (V

+ V

)/2

=−i(V

− V

)/2.

If we set TE

= W

then changing coordinates via T puts A in canonical form

−1

AT =

⎛

⎜

⎝

αβ 00

−βα 00

00 αβ

00−βα

⎞

⎟

⎠

If we ﬁnd only one eigenvector V

for α + iβ, then we solve the system of

equations (A − (α + iβ)I)X = V

as in the case of repeated real eigenvalues.

The proof of the previous proposition shows that we can always ﬁnd a nonzero

solution V

of these equations. Then choose the W

as above and set TE

= W

Then T puts A into the canonical form

−1

AT =

⎛

⎜

⎝

αβ 10

−βα 01

00 αβ

00−βα

⎞

⎟

⎠

For example, we compute

−1

AT )E

= T

−1

= T

−1

A(V

+ V

)/2

= T

−1



+ (α + iβ)V

)/2 + (V

+ (α − iβ)V

)/2



= T

−1



+ V

)/2 + α



+ V



/2 + iβ



− V



= E

+ αE

− βE

100 Chapter 5 Higher Dimensional Linear Algebra

Example. Let

A =

⎛

⎜

⎝

1 −101

2 −110

00−12

00−11

⎞

⎟

⎠

The characteristic equation, after a little computation, is

(λ

+ 1)

= 0.

Hence A has eigenvalues ±i, each repeated twice.

Solving the system (A − iI )X = 0 yields one linearly independent complex

eigenvector V

= (1, 1 − i, 0, 0) associated to i. Then V

is an eigenvector

associated to the eigenvalue −i.

Next we solve the system (A −iI )X = V

to ﬁnd V

= (0, 0, 1 −i, 1). Then

solves the system (A − iI )X = V

. Finally, choose



+ V



/2 = Re V

=−i



− V



/2 = Im V



+ V



/2 = Re V

=−i



− V



/2 = Im V

and let TE

= W

for j = 1, ..., 4. We have

T =

⎛

⎜

⎝

1000

1 −10 0

001−1

0010

⎞

⎟

⎠

, T

−1

⎛

⎜

⎝

10 00

1 −100

00 01

00−11

⎞

⎟

⎠

and we ﬁnd the canonical form

−1

AT =

⎛

⎜

⎝

0110

−1001

0001

00−10

⎞

⎟

⎠



Example. Let

A =

⎛

⎜

⎝

2010

0201

0020

0 −102

⎞

⎟

⎠

5.6 Genericity 101

The characteristic equation for A is

(2 − λ)

((2 − λ)

+ 1) = 0

so the eigenvalues are 2 ± i and 2 (with multiplicity 2).

Solving the equations (A − (2 + i)I )X = 0 yields an eigenvector V =

(0, −i,0,1)for2+ i. Let W

= (0, 0, 0, 1) and W

= (0, −1, 0, 0) be the real

and imaginary parts of V .

Solving the equations (A −2I )X = 0 yields only one eigenvector associated

to 2, namely, W

= (1, 0, 0, 0). Then we solve (A − 2I)X = W

to ﬁnd

= (0, 0, 1, 0). Setting TE

= W

as usual puts A into the canonical form

−1

AT =

⎛

⎜

⎝

2100

−1200

0021

0002

⎞

⎟

⎠

as is easily checked.



5.6 Genericity

We have mentioned several times that “most” matrices have distinct eigenval-

ues. Our goal in this section is to make this precise.

Recall that a set

U ⊂ R

is open if whenever X ∈ U there is an open ball

about X contained in

U; that is, for some a>0 (depending on X) the open

ball about X of radius a,

{Y ∈



|Y − X | <a},

is contained in

U. Using geometrical language we say that if X belongs to an

open set

U, any point sufﬁciently near to X also belongs to U.

Another kind of subset of

is a dense set: U ⊂ R

is dense if there are

points in

U arbitrarily close to each point in R

. More precisely, if X ∈ R

then for every  > 0 there exists some Y ∈

U with |X −Y | < . Equivalently, U

is dense in R

if V ∩U is nonempty for every nonempty open set V ⊂ R

. For

example, the rational numbers form a dense subset of

R, as do the irrational

numbers. Similarly

{(x, y) ∈

|both x and y are rational}

is a dense subset of the plane.

102 Chapter 5 Higher Dimensional Linear Algebra

An interesting kind of subset of R

is a set that is both open and dense.

Such a set

U is characterized by the following properties: Every point in

the complement of

U can be approximated arbitrarily closely by points of

U (since U is dense); but no point in U can be approximated arbitrarily closely

by points in the complement (because

U is open).

Here is a simple example of an open and dense subset of

V ={(x, y) ∈ R

|xy = 1}.

This, of course, is the complement in

of the hyperbola deﬁned by xy = 1.

Suppose (x

, y

) ∈ V. Then x

= 1 and if |x −x

|, |y −y

|are small enough,

then xy = 1; this proves that

V is open. Given any (x

, y

) ∈ R

, we can ﬁnd

(x, y) as close as we like to (x

, y

) with xy = 1; this proves that V is dense.

An open and dense set is a very fat set, as the following proposition shows:

Proposition. Let

, ..., V

be open and dense subsets of R

. Then

V = V

∩ ...∩ V

is also open and dense.

Proof: It can be easily shown that the intersection of a ﬁnite number of open

sets is open, so

V is open. To prove that V is dense let U ⊂ R

be a nonempty

open set. Then

U ∩ V

is nonempty since V

is dense. Because U and V

are

open,

U ∩V

is also open. Since U ∩V

is open and nonempty, (U ∩V

) ∩V

is nonempty because V

is dense. Since V

is open, U ∩V

∩V

is open. Thus

(

U ∩ V

∩ V

) ∩ V

is nonempty, and so on. So U ∩ V is nonempty, which

proves that

V is dense in R

. 

We therefore think of a subset of R

as being large if this set contains an

open and dense subset. To make precise what we mean by “most” matrices, we

need to transfer the notion of an open and dense set to the set of all matrices.

Let L(

) denote the set of n ×n matrices, or, equivalently, the set of linear

maps of

. To discuss open and dense sets in L(R

), we need to have a notion

of how far apart two given matrices in L(

) are. But we can do this by simply

writing all of the entries of a matrix as one long vector (in a speciﬁed order)

and thereby thinking of L(

)asR

Theorem. The set

M of matrices in L(R

) that have n distinct eigenvalues is

open and dense in L(

Proof: We ﬁrst prove that

M is dense. Let A ∈ L(R

). Suppose that A has

some repeated eigenvalues. The proposition from the previous section states

5.6 Genericity 103

that we can ﬁnd a matrix T such that T

−1

AT assumes one of two forms. Either

we have a canonical form with blocks along the diagonal of the form

(i)

⎛

⎜

⎝

λ 1

⎞

⎟

⎠

or (ii)

⎛

⎜

⎝

⎞

⎟

⎠

where α, β, λ ∈

R with β = 0 and



αβ

−βα



, I





or else we have a pair of separate diagonal blocks (λ)orC

. Either case can be

handled as follows.

Choose distinct values λ

such that |λ −λ

|is as small as desired, and replace

block (i) above by

⎛

⎜

⎝

⎞

⎟

⎠

This new block now has distinct eigenvalues. In block (ii) we may similarly

replace each 2 × 2 block



αβ

−βα



with distinct α

’s. The new matrix thus has distinct eigenvalues α

±β. In this

fashion, we ﬁnd a new matrix B arbitrarily close to T

−1

AT with distinct eigen-

values. Then the matrix TBT

−1

also has distinct eigenvalues and, moreover,

this matrix is arbitrarily close to A. Indeed, the function F : L(

) → L(R

)

given by F(M) = TMT

−1

where T is a ﬁxed invertible matrix is a continuous

function on L(

) and hence takes matrices close to T

−1

AT to new matrices

close to A. This shows that

M is dense.

To prove that

M is open, consider the characteristic polynomial of a matrix

A ∈ L(

). If we vary the entries of A slightly, then the characteristic polyno-

mial’s coefﬁcients vary only slightly. Hence the roots of this polynomial in

104 Chapter 5 Higher Dimensional Linear Algebra

move only slightly as well. Thus, if we begin with a matrix whose eigenvalues

are distinct, nearby matrices have this property as well. This proves that

M is

open.



A property P of matrices is a generic property if the set of matrices having

property

P contains an open and dense set in L(R

). Thus a property is

generic if it is shared by some open and dense set of matrices (and perhaps

other matrices as well). Intuitively speaking, a generic property is one that

“almost all” matrices have. Thus, having all distinct eigenvalues is a generic

property of n × n matrices.

EXERCISES

1. Prove that the determinant of a 3 × 3 matrix can be computed by

expanding along any row or column.

2. Find the eigenvalues and eigenvectors of the following matrices:

(a)

⎛

⎝

001

010

100

⎞

⎠

(b)

⎛

⎝

001

020

300

⎞

⎠

(c)

⎛

⎝

111

⎞

⎠

(d)

⎛

⎝

002

020

−200

⎞

⎠

(e)

⎛

⎜

⎝

3001

0122

1 −2 −1 −4

−1003

⎞

⎟

⎠

3. Describe the regions in a, b, c–space where the matrix

⎛

⎝

00a

0 b 0

c 00

⎞

⎠

has real, complex, and repeated eigenvalues.

4. Describe the regions in a, b, c–space where the matrix

⎛

⎜

⎝

a 00a

0 ab0

0 ca0

a 00

⎞

⎟

⎠

has real, complex, and repeated eigenvalues.

Exercises 105

5. Put the following matrices in canonical form

(a)

⎛

⎝

001

010

100

⎞

⎠

(b)

⎛

⎝

101

010

001

⎞

⎠

(c)

⎛

⎝

010

−100

111

⎞

⎠

(d)

⎛

⎝

010

100

111

⎞

⎠

(e)

⎛

⎝

101

010

101

⎞

⎠

(f )

⎛

⎝

110

111

011

⎞

⎠

(g)

⎛

⎝

10−1

−11−1

00 1

⎞

⎠

(h)

⎛

⎜

⎝

1001

0110

0010

1000

⎞

⎟

⎠

6. Suppose that a 5 × 5 matrix has eigenvalues 2 and 1 ± i. List all possible

canonical forms for a matrix of this type.

7. Let L be the elementary matrix that interchanges the ith and jth rows of

a given matrix. That is, L has 1’s along the diagonal, with the exception

that 

= 

= 0 but 

= 

= 1. Prove that det L =−1.

8. Find a basis for both Ker T and Range T when T is the matrix

(a)





(b)

⎛

⎝

111

⎞

⎠

(c)

⎛

⎝

196

141

271

⎞

⎠

9. Suppose A isa4× 4 matrix that has a single real eigenvalue λ and only

one independent eigenvector. Prove that A may be put in canonical form

⎛

⎜

⎝

λ 100

0 λ 10

00λ 1

000λ

⎞

⎟

⎠

10. Suppose A isa4×4 matrix with a single real eigenvalue and two linearly

independent eigenvectors. Describe the possible canonical forms for A

and show that A may indeed be transformed into one of these canonical

forms. Describe explicitly the conditions under which A is transformed

into a particular form.

11. Show that if A and/or B are noninvertible matrices, then AB is also

noninvertible.

12. Suppose that

S is a subset of R

having the following properties:

(a) If X, Y ∈

S, then X + Y ∈ S;

(b) If X ∈

S and α ∈ R, then αX ∈ S.