complex eigenvalues

Announcements

Ï Today's material will not be a part of Test 2.

Ï Review for Test 2 tomorrow. Please come prepared with

questions.

Ï Exam 2 will be on Feb 25 Thurs in class.

Ï Please collect your graded exams on friday from Fisher 214.

Review of Complex Numbers

De�nition

An complex number is a number written in the form

z = a +bi

Here a and b are real numbers and i is a symbol that satis�es

i 2 =−1.

a is called the Real part of z and b is called the Imaginary part of Z

a = Re z,b = Im z

Review of Complex Numbers

1. If 2 complex numbers are equal, their real and imaginary parts

must be the same.

2. Any real number a is a special complex number (a +0i).

3. You can add 2 complex numbers as

(a +bi )+ (c +di ) = (a + c)+ (b +d)i

You can multiply two complex numbers as

(a +bi )(c +di ) = ac +adi +bci +bdi 2︸ ︷︷ ︸−bd

= (ac −bd)+ (ad +bc)i

Review of Complex Numbers

De�nition

If z = a +bi is a complex number, we de�ne the conjugate of z(denoted by z̄, read as "z bar") as

z̄ = a −bi

That is, we change the sign of the imaginary part of z.

Review of Complex Numbers

1.

zz̄ = (a +bi )(a −bi ) = a2 −abi +abi −b2 i 2︸︷︷︸−1

= a2 +b2

2. The absolute value or the modulus of z is the real number |z|given by

|z| =p

zz̄ =√

a2 +b2

3. If z 6= 0 we can de�ne the multiplicative inverse of z as

1

z= z−1 = z̄

|z|2

Back to Eigenvalues

1. The char equation of an n ×n matrix involves an nth degree

polynomial.

2. This equation has exactly n roots, if we include complex roots.

3. If the char equation of A has some complex roots, we get very

important information about A

4. Very important in problems involving vibrations and rotations

in space.

Complex Eigenvalues

1. Cn represents the set of all complex numbers

2. A complex scalar λ satis�es

det(A−λI) = 0

if and only if there is a NONZERO vector x in Cn such that

Ax= λx

3. λ is called a complex eigenvalue and x is a complex

eigenvector correspsonding to λ.

Example 1, section 5.5

Let the given matrix act on C2. Find the eigenvalues and a basis for

each eigenspace in C2. [1 −21 3

].

Solution: We have to look at the determinant of the matrix[1 −21 3

]−λ

[1 00 1

]=

[1−λ −2

1 3−λ

].

∣∣∣∣ 1−λ −21 3−λ

∣∣∣∣= (1−λ)(3−λ)+2.

Simplify this quantity.

=⇒ 3−4λ+λ2 +2 = λ2 −4λ+5

Example 1, section 5.5

Let the given matrix act on C2. Find the eigenvalues and a basis for

each eigenspace in C2. [1 −21 3

].

Solution: We have to look at the determinant of the matrix[1 −21 3

]−λ

[1 00 1

]=

[1−λ −2

1 3−λ

].

∣∣∣∣ 1−λ −21 3−λ

∣∣∣∣= (1−λ)(3−λ)+2.

Simplify this quantity.

=⇒ 3−4λ+λ2 +2 = λ2 −4λ+5

Example 1, section 5.5

Use the quadratic formula (factorization will not work here)

λ= 4±√

42 −4(1)(5)

2(1)= 4±p−4

2= 4±2i

2= 2± i

The 2 eigenvalues are λ1 = 2+ i and λ2 = 2− i (Observe that the

eigenvalues are conjugates of eachother or we have a conjugate

pair)

Example 1, section 5.5

Use the quadratic formula (factorization will not work here)

λ= 4±√

42 −4(1)(5)

2(1)= 4±p−4

2= 4±2i

2= 2± i

The 2 eigenvalues are λ1 = 2+ i and λ2 = 2− i (Observe that the

eigenvalues are conjugates of eachother or we have a conjugate

pair)

Example 1, section 5.5

We now have to �nd the eigenvector for each eigenvalue. Start

with λ1 = 2+ i . We want (A−λ1I)x= 0 to have nontrivial solution.

[1 −21 3

]−

[2+ i 0

0 2+ i

]=

[ −1− i −21 1− i

].

Whenever you deal with eigenvectors for a complex eigenvalue, we

do the following:

Use a convenient row to express x1 in terms x2 (or x2 in terms of

x1 if that is easier)

Example 1, section 5.5

We now have to �nd the eigenvector for each eigenvalue. Start

with λ1 = 2+ i . We want (A−λ1I)x= 0 to have nontrivial solution.[1 −21 3

]−

[2+ i 0

0 2+ i

]=

[ −1− i −21 1− i

].

Whenever you deal with eigenvectors for a complex eigenvalue, we

do the following:

Use a convenient row to express x1 in terms x2 (or x2 in terms of

x1 if that is easier)

Example 1, section 5.5

From row 2, we can write

x1 =−(1− i )x2

and so the solution will be[x1

x2

]=

[ −(1− i )x2

x2

]= x2

[ −1+ i1

]Pick x2 = 1 and so an eigenvector for λ1 will be[ −1+ i

1

]

Example 1, section 5.5

With complex eigenvalues, once we have an eigenvector for one

eigenvalue, an eigenvector for the second eigenvalue is found by

taking the conjugate of the �rst eigenvector.

That is, an eigenvector for λ2 = 2− i will be[ −1− i1

]Thus both eigenvalues and eigenvectors are conjugates.

Example 5, section 5.5

Let the given matrix act on C2. Find the eigenvalues and a basis for

each eigenspace in C2. [0 1−8 4

].

Solution: We have to look at the determinant of the matrix[0 1−8 4

]−λ

[1 00 1

]=

[0−λ 1−8 4−λ

].

∣∣∣∣ −λ 1−8 4−λ

∣∣∣∣= (−λ)(4−λ)+8.

Simplify this quantity.

=⇒−4λ+λ2 +8 = λ2 −4λ+8

Example 5, section 5.5

Let the given matrix act on C2. Find the eigenvalues and a basis for

each eigenspace in C2. [0 1−8 4

].

Solution: We have to look at the determinant of the matrix[0 1−8 4

]−λ

[1 00 1

]=

[0−λ 1−8 4−λ

].

∣∣∣∣ −λ 1−8 4−λ

∣∣∣∣= (−λ)(4−λ)+8.

Simplify this quantity.

=⇒−4λ+λ2 +8 = λ2 −4λ+8

Example 5, section 5.5

Use the quadratic formula (factorization will not work here)

λ= 4±√

42 −4(1)(8)

2(1)= 4±p−16

2= 4±4i

2= 2±2i

The 2 eigenvalues are λ1 = 2+2i and λ2 = 2−2i (Observe that the

eigenvalues are again conjugates of eachother or we have a

conjugate pair)

Example 5, section 5.5

Use the quadratic formula (factorization will not work here)

λ= 4±√

42 −4(1)(8)

2(1)= 4±p−16

2= 4±4i

2= 2±2i

The 2 eigenvalues are λ1 = 2+2i and λ2 = 2−2i (Observe that the

eigenvalues are again conjugates of eachother or we have a

conjugate pair)

Example 5, section 5.5

We now have to �nd the eigenvector for each eigenvalue. Start

with λ1 = 2+2i . We want (A−λ1I)x= 0 to have nontrivial solution.

[0 1−8 4

]−

[2+2i 0

0 2+2i

]=

[ −2−2i 1−8 2−2i

].

Whenever you deal with eigenvectors for a complex eigenvalue, we

do the following:

Use a convenient row to express x1 in terms x2 (or x2 in terms of

x1 if that is easier)

Example 5, section 5.5

We now have to �nd the eigenvector for each eigenvalue. Start

with λ1 = 2+2i . We want (A−λ1I)x= 0 to have nontrivial solution.[0 1−8 4

]−

[2+2i 0

0 2+2i

]=

[ −2−2i 1−8 2−2i

].

Whenever you deal with eigenvectors for a complex eigenvalue, we

do the following:

Use a convenient row to express x1 in terms x2 (or x2 in terms of

x1 if that is easier)

Example 5, section 5.5

From row 1, we can write

(2+2i )x1 = x2

and so the solution will be[x1

x2

]=

[x1

(2+2i )x1

]= x1

[1

2+2i

]Pick x2 = 1 and so an eigenvector for λ1 will be[

12+2i

]

Example 5, section 5.5

With complex eigenvalues, once we have an eigenvector for one

eigenvalue, an eigenvector for the second eigenvalue is found by

taking the conjugate of the �rst eigenvector.

That is, an eigenvector for λ2 = 2−2i will be[1

2−2i

]Thus both eigenvalues and eigenvectors are conjugates.

Understanding Complex Eigenvalues

Consider the following matrix where a and b are real numbers and

both a and b are never zero.

C =[

a −bb a

].

1. The eigenvalues of A are a +bi and a −bi

2. We can write C as follows for better understanding

C =[

r 00 r

][cosθ −sinθsinθ cosθ

].

where r = |λ| =p

a2 +b2 and θ is the angle between the

positive X-axis and the line joining (0,0) and (a,b).

3. The angle θ is called the argument of λ= a +bi . From basic

trigonometry, we can see that a = r cosθ and b = r sinθ.

Understanding Complex Eigenvalues

Consider the following matrix where a and b are real numbers and

both a and b are never zero.

C =[

a −bb a

].

1. The eigenvalues of A are a +bi and a −bi

2. We can write C as follows for better understanding

C =[

r 00 r

][cosθ −sinθsinθ cosθ

].

where r = |λ| =p

a2 +b2 and θ is the angle between the

positive X-axis and the line joining (0,0) and (a,b).

3. The angle θ is called the argument of λ= a +bi . From basic

trigonometry, we can see that a = r cosθ and b = r sinθ.

What does a complex eigenvalue mean?y

x0

xRotation by θ

θ

Scaling by |λ|Ax

What does a complex eigenvalue mean?y

x0

xRotation by θ

θ

Scaling by |λ|Ax

What does a complex eigenvalue mean?y

x0

xRotation by θ

θ

Scaling by |λ|Ax

Example 7 section 5.5

List the eigenvalues of A. Give the angle of rotation θ and the scale

factor r if multiplying this matrix e�ectively rotates and scales a

given vector.

A =[ p

3 −11

p3

].

Here a =p3 and b = 1. The eigenvalues are thus

λ=p3± i

The scale factor is the modulus of λ which is

r = |λ| =√

(p

3)2 +12 =p3+1 = 2

To �nd θ, use the fact that cosθ= ar =

p3

2 . Thus θ= π6 .

Use of Eigenvalues in Long Term Behavior

We can explain lots of dynamical systems that "evolve" through

time by the equation

xt+1 = Axt

where t denotes time with proper units.

Thus if x0 is the initial vector of any quantity (for ex. population)

then the population at t=1 is

x1 = Ax0,

the population at t = 2 is

x2 = Ax1 = A2x0

and so on.

Use of Eigenvalues in Long Term Behavior

In general, if A is an n ×n matrix, it will have n eigenvalues. Call

them

λ1,λ2 . . . ,λn

Let the corresponding linearly independent eigenvectors be

v1,v2 . . . ,vn

These vectors form a basis for Rn . We can write the initial vector

x0 as

x0 = c1v1 + c2v2 + . . .+ cnvn

x1 = Ax0 = c1Av1 + c2Av2 + . . .+ cn Avn

From the de�nition of eigenvalue, we know that Av1 = λv1,

Av2 = λv2 etc. Substitute these and we get

Use of Eigenvalues in Long Term Behavior

x1 = Ax0 = c1λ1v1 + c2λ2v2 + . . .+ cnλnvn

Based on this,

x2 = Ax1 = c1λ21v1 + c2λ

22v2 + . . .+ cnλ

2nvn

x3 = Ax2 = c1λ31v1 + c2λ

32v2 + . . .+ cnλ

3nvn

...

xt = Axt−1 = c1λt1v1 + c2λ

t2v2 + . . .+ cnλ

tnvn

Based on this, we can see what happens as t →∞

Simple Predator-Prey Model

Let O stand for owl and R for rats (in thousands). Studies show

that the population of these species in an ecosystem evolve

according to the following model (t is in months) and[Ot+1

Rt+1

]=

[0.5 0.4

−0.104 1.1

]︸ ︷︷ ︸

A

[Ot

Rt

].

Meaning of the numbers in A.

1. With no rats for food only 50 percent of owls will survive each

month(the 0.5 term)

2. With no owls as predators, the rat population goes up by 10

percent every month (the 1.1 term)

3. The -0.104 represents the decrease in rat population as a

result of being hunted by owls

4. The 0.4 represents the increase in owl population if there are

lots of rats around.

Simple Predator-Prey Model

How does this system behave in the long run? Can you predict the

rat and owl population eventually?

The eigenvalues of A are λ1 = 1.02 and λ2 = 0.58. (omitting details

here, you know how to do it). The corresponding eigenvectors are

v1 =[

1013

]and v2 =

[51

]Based on the time evolution equation we saw above, we have

xt = c1λt1v1 + c2λ

t2v2 = c1(1.02)t

[1013

]+ c2(0.58)t

[51

]Since 0.58 < 1, higher exponents of 0.58 will be smaller and as

t →∞, (0.58)t → 0.

Simple Predator-Prey Model

After a very long period of time,

xt ≈ c1λt1v1 = c1(1.02)t

[1013

]So if t increases, we have a better approximation. Eventually,

xt+1 ≈ c1λt+11 v1 = c1(1.02)t+1

[1013

]= (1.02)xt

Both owls and rats grow by 2 percent each month. The proportion

of owls to rats stays the same. For every 13 thousand rats, there

are 10 owls.