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Eigenvectors in
Complex Vector Spaces

9
...

Eigenvalues and eigenvectors are defined in the same way as before, except that
the scalars and the coordinates of the vectors are complex numbers
...
If for some A E e there exists a non-zero vector
z E e11 such that L(Z) AZ, then A is an eigenvalue of L and z is called an eigenvector
of L that corresponds to A
...
0, if
Let

�

=

Az= AZ
...
Since the Fundamental Theorem of Algebra guarantees that every n-th degree
polynomial has exactly n roots over C, the only way a matrix cannot be diagonalizable
over

C

is if it has an eigenvalue with geometric multiplicity less than its algebraic

multiplicity
...
However, a simple example is given to illustrate the theory
...
A

...


+

]


...

1·1zable
...
-2+6i -

(continued)

_

/I
...
-t2

- - 2i)tl+(3 - 3i) = 0
(1

Using the quadratic formula, we find that
±

(1
-2i)
/l =

[(1

2i)2 -4(1)(3 -3i)]112
2

-

Using the methods from Section 9
...
tl1 =1+i,
11
+
7
i
1
2
9i
A - = [ -1 - 4i -3+Si ] [ 0
[ ;l
=1+i [ ; 1 = -3i,
[
7
i
i
3
+
S
/
A /l2 =[ - 4t
...
] 0
2[ ; i]
...

= -3 [ ]
a

...
]
For

we get

�

/I
...
-t2

is

a

-1

Hence, the general solution is a
IS

l


...
Thus, an eigenvector corresponding to

-11

a

�

1

-20-i]

E C
...


Hence,

that

1

we get

_

l

/l1 = +i

the formula c

and

_

_

Complex Characteristic Roots of a Real Matrix and
a Real Canonical Form
Does diagonalizing a complex matrix tell us anything useful about diagonalizing a real
matrix? First, note that since the real numbers form a subset of the complex numbers,
we may regard a matrix

A

with real entries as being a matrix with complex entries;

all of the entries just happen to have zero imaginary part
...
-t, we speak of ,t as a complex eigenvalue

with a corresponding complex eigenvector
...


EXAMPLE2

Let A=

[; =n

Find its eigenvectors and diagonalize over IC
...
I =

[5

- A
...


]

2
so det(A-A
...
-4A
...
1 =2+3i
andA
...
1
...
1=2 + 3i,

] [

[

3- 3i
1
6
l
�
A-A
...
z1

...


1

For A
...
2 =

...
2 =2 - 3i is z2 =
If follows that A is diagonalized to

[

2

�

3i
2

� 3i


...


Observe in Example 2 that the eigenvalues of A were complex conjugates
...
1 2

...
Before proving this, we first extend the definition
of complex conjugates to matrices
...
We define the complex conjugate of A,

Suppose that A is an n x n matrix with real entries and that A
...
�mding eigenvector Z
...


Proof: Suppose that Az =A
...
Taking complex conjugates of both sides gives
Az =A
...


Now we note that the solution to Example 2 was not completely satisfying
...
However, in
Example 2, we have changed from a real matrix to a complex matrix
...
The rest of this section is concerned with the problem of
finding such a real matrix
...
Thus, we will just consider the case where A is an n x n real matrix
with at least one eigenvalue ;l = a + bi, where a, b E JR and b 'f
...

Since we are splitting the eigenvalue into real and imaginary parts, it makes sense
to also split the corresponding eigenvector z into real and imaginary parts:

=

z=

Thus, we have Az = ;lz, or

[Xl l
Xn
: +i

= x+iy'

x,y

E

JR
...
1)

Observe that equation (9
...
That is, if v E Span{x,y},
then Av E Span{x,y}
...


E

1lJ for all

Note that x and y must be linearly independent, for if x =
equation (9
...
0
...
) Moreover,
it can be shown that no vector in Span{x,Y} is a real eigenvector of A
...
) This discussion together with Problem D2 is summarized in Theorem 2
...
0 is an eigenvalue of an n x n real matrix A with cor­
responding eigenvector z=x+if
...

of
that is invariant under A and contains no real eigenvector of A
...
2
...
1) we get that the 13-matrix of the linear mapping L associated with A is
[L]!B =

[_: !]
...


Moreover, from our work in Section 4
...


Definition
Real Canonical Form

EXAMPLE3

[:

Let A be a 2 x 2 real matrix with eigenvalue A
...
0
...


Find a real canonical form of the matrix C of A =
coordinates matrix P such that p-I AP= C
...

IS
a

and find a change of

Solution: We have

1

det(A - A
...
2 +
_2
A
...
1 = 0 + i and A
...
1
...

] [1

b = 1 and hence a real canonical form of A is _
For A
...
ii=

[

2

1

-

i

-5
-2 - i

�

0

-2

0

-

i

]

so an eigenvector corresponding to A
...
= 2 + 3i and
Thus, a real canonical form of A is

EXERCISE 1

[ � ;J
...


:i' =

2 - 3i
...


[ �]

A matrix of the form �

b

can be rewritten as

va2 + b2

]

[?

c se - sine
sm e cose

where cose = a/Ya2 + b2 and sine= -b/Ya2 + b2
...

2
...


Thus,

is also a real canonical form of A
...
The complex eigenvector z is determined only up to multiplication by an arbitrary
non-zero complex number
...
For example, in Example 3, z =
) 2

(1

+i

is also an eigenvector corresponding to ti
...
p-IAP

EXERCISE 2

[ � i] [ \:�i]
[ � �]
=

would also

- [ _10 0
...


The Case of a 3

x

[-i �]

and find a change of

3 Matrix

If A is a 3 x 3 real matrix with one real eigenvalueµ with corresponding eigenvector
v and complex �igenvalues ,i = a + bi, b * and :i with corresponding eigenvectors
z

=

1 + iY and z, then we have the equations

Av= µv,

0

Ax=ax- by,

Then, the matrix of A with respect to the basis !B

Ay =bx+ ay

�

W
...


EXAMPLES
Find a real canonical form of the matrix C of A =
coordinates matrix P such that p-1AP = C
...


3, ,11 = 1 + 2i, and ,12 = 1 - 2i

we have

[�
-

A-µ/=

-4

-4

-2

-4

4

4

4

Thus, an eigenvector corresponding toµ is V=
For,11 = 1 + 2i, we have

[-2�

- 2i

A-,11/ =

= -(,1 - 3)(,12 - 2,1 +

-

6- 2i
-2
4

][
-

Ti
...

-

with the a's occurring on the diagonal in positions determined

by the position of x and y in the basis
...
In some cases, it is possible to find a basis of eigenvectors and diagonal­

ize the matrix over C
...


PROBLEMS 9
...
Also determine a real canonical form and
give a change of coordinates matrix P that brings

(b)

(c)

the matrix into this form
...
Also determine a real canonical form and
give a change of coordinates matrix P that brings
the matrix into this form
...

D2 Suppose that A is an n x n real matrix and that
tl a+ bi is a complex eigenvalue of A with b * 0
...

=

(a) Prove that x i- 0 and y i- 0
...
(Hint:
Suppose x
ky for some k, and use equation
(9
...
)
(c) Prove that Span{x,y} does not contain an
eigenvector of A corresponding to a real eigen­
value of A

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