4.14 Complex Eigenvalues
4.14.1 Matrices with Complex Eigenvalues
The version of the QR algorithm discussed in the previous sections won't approximate complex eigenvalues.
The reason is that complex eigenvalues of matrices with real entries occur in conjugate pairs. Complex
numbers which are complex conjugates of each other have the same absolute value. The result is that the
versions of power iteration that we have discussed so far won't separate them if one only uses real shifts.
The key to approximating the complex eigenvalues is to allow shifts that are complex. We look at this in
the next section. However, before this we discuss properties of complex eigenvalues of matrices with real
entries in this section. We begin with an example.
Example 1. Find the eigenvalues and eigenvectors of A =
A - I =
 3 2 . For the eigenvalues one has
-8 3
 3- 2 
 -8 3-
0 = det( A - I ) =
 3 -  2  = (3 - )2 + 16
 -8 3-
( - 3)2 = - 16
 - 3 =  - 16 =  4i
So the eigenvalues are
 = 3  4i
So
1 = 3 + 4i
and
2 = 3 - 4i
If z = x + yi is a complex number with real and imaginary parts x and y then
(1)
_
z =
x – yi = complex conjugate of z
= the number obtained by reversing the sign of the imaginary part of z
= reflection of z across the real axis.
_____
For example, 3 - 4i = 3 + 4i. The complex number z = x + yi can be identified with the point in the plane
x
with coordinates  y . With this correspondence, the real numbers lie on the x axis and the pure imaginary
_
x
numbers yi lie on the y axis. The complex conjugage z = x – yi of z = x + yi is identified with  - y  which
x
is the reflection of  y  across the real axis.
4.14.1 - 1
Note that the two eigenvalues in this example are complex conjugates of each other. This is a general
feature of complex eigenvalues of matrices that have real entries, i.e. they occur in complex conjugate pairs.
One reason is that the characteristic equation det( A - I ) = 0 is a polynomial equation in  with real
coefficients and for such equations the roots occur in complex conjugate pairs. We shall see another reason
below.
For the eigenvectors for 1 = 3 + 4i one has
A - 1I = A – (3 + 4i)I =
 - 4i 2 
 - 8 - 4i 
x
So an eigenvector v =  y  satisfies
 0  = (A - I)v =  - 4i 2   x  =  - 4ix + 2y 
0
 - 8 - 4i   y 
 - 8x - 4iy 
So
- 4ix + 2y = 0
- 8x - 4iy = 0
8x
2ix
= - 2 = 2ix. So the two equations
4i
i
are multiples of each other. So an eigenvector v for 1 = 3 + 4i has the form
The first equation implies y = 2ix. The second equation implies y = -
v =
x =  x  = x 1
y
 2ix 
 2i 
1
So any multiple of the vector v1 =  2i  is an eigenvector for 1 = 3 + 4i.
For 2 = 3 - 4i all the previous computation that we did for 1 = 3 + 4i remain the same except we replace i
1
by – i. So it is not hard to see that any multiple of the vector  - 2i  is an eigenvector for 2 = 3 - 4i.
This example illustrates a general features of the eigenvectors for complex eigenvalues, namely the
eigenvector for complex conjugate eigenvalues have complex conjugate components. It was not hard to
see why this was true in the above example, and the same argument can be used in general. However, there
is a slightly different argument that is useful in other similar situations. The argument involves the
algebraic properties of taking complex conjugates. Some of these are
(2)
_____
_
_
z+w = z + w
(3)
__
__
zw = z w
(4)
__
_
zn = ( z )n
_____
_ _
z-w = z - w
_____
_ _
z/w = z/w
4.14.1 - 2
 zz21 
The operation of taking complex conjugates can be extended to vectors and matrices. If v =  .  is a
 zn 
 __z 
a
_



vector with complex components then its complex conjugate is v = z . If A =  a

 __. 

a
 
__
1
2

_
matrix with complex components then its complex conjugate is A = 


zn
__
a11
__
a21

__
am1
21
a12  a1n
a22  a2n
m1
am2  amn
11

 is a


.

__
a 
__
__
a12  a1n
__
__
a22  a2n
__
am2 
mn1
_
_
2 - 3i
2 + 3i
2 - 3i 7 + i
2 + 3i
Example 2. If v =  5 + 4i  then v =  5 - 4i . If A =  5 + 4i 6 - 8i  then A =  5 - 4i
7-i 
6 + 8i .
The algebraic properties (2) - (4) of complex conjugates for numbers extends to complex conjugates of
vectors and matrices, e.g. if  is a complex number, u and v are vectors and A and B are matrices then
_____
_
_
A+B = A + B
_____
_
_
u+v = u + v
__
(5)
__
u =  u
__
__
A =  A
__
__
Au = A u
__
__
AB = A B
The following proposition shows that complex eigenvalues of matrices with real entries occur in conjugate
pairs.
Proposition 1. Suppose A is a matrix with real entries and  is an eigenvalue of A with eigenvector v.
_
_
_
Then  is also an eigenvalue of A and v is an eigenvector for .
__ __
__ _
Proof. One has Av = v. Taking complex conjugates of both sides gives Av = v. Using (3) gives A v = 
_
v . which proves the proposition. //
x
Example 3. Consider the mapping that takes a point v =  y  and rotates it by an angle  = /4 to the new
r
point w =  s . Thent w = Rv where R = R/4 is the matrix for a rotation by /4. In general the matrix R
cos  - sin  
 sin  cos  . In the case  = /4 one has
1 1 -1
sin(/4) = cos(/4) = 1/ 2 . So R =
. The eigenvalues of R can not be real since no non-zero
2 1 1
for a rotation by  is given by R = 
4.14.1 - 3
vector v is mapped on to the line through itself when it is rotated by /4. To find the eigenvalues of R we
proceed as usual.
R - I =



1
-
2
1
2
0 = det( R - I ) =


1
-

2
-



1
2


1
-

2
1
-
2
1
2
1
2
-
2
=
 1 - + 1
2
 2

So
2
 1 - = -1
2
 2

-

1
2
= 
1
2
So the eigenvalues are
1 =
1
1
+
i
2
2
For the eigenvectors for 1 =
2 = -
and
1
1
i
2
2
1
1
+
i one has
2
2
A - 1I = A – (
1
1
1
+
i)I =
2
2
2
-i -1
 1 -i
x
So an eigenvector v =  y  satisfies
 0  = (A - I)v =
0
1
2
 - i - 1   x  = 1  - ix - y 
 1 -iy
2  x - iy 
So
- ix -
y = 0
x - iy = 0
If one multiplies the first equation by i one obtains the second. So any solution to the first equation is also
a solution to the second. So it suffices to solve the first equation whose solution is y = - iy. So an
1
1
x
x
1
eigenvector v for 1 =
+
i has the form v =  y  =  - ix  = x  - i . So any multiple of the vector
2
2
1
1
1
1
v1 =  - i  is an eigenvector for 1 =
+
i. Taking complex conjugates one sees that v2 =  i  is an
2
2
1
1
eigenvector for 2 =
i.
2
2
4.14.1 - 4
- sin  
are 1 = cos  + (sin ) i and
cos  
1
1
2 = cos  - (sin ) i and the corresponding eigenvectors are v1 =  - i  and v2 =  i .
Problem 1. Show that the eigenvalues of R = 
cos 
 sin 
0 -2
Problem 2. Show that the eigenvalues of A =  1 2  are 1 = 1 + i and 2 = 1 - i and the corresponding
2
2
eigenvectors are v1 =  - 1 - i  and v2 =  - 1 + i .
What happens when we use power iteration with a matrix with complex eigenvalues.
Example 4. Let A =
 3 2  and x =  1 . Find xk as k grows and check for convergence.
-8 3
0
c1
First we must write x as a linear combination x = c1v1 + c2v2 of the eigenvectors. One has  c  = V-1x
2
1
1
1 1 -i
where V =  2i - 2i  is the matrix whose columns are the eigenvectors. One has V-1 = 4  1 i , so
 c1  = 1  1 - i   1  = 1  1 . So x = 1 v1 + 1 v2. So
4
4
 c2  4  1 i   0  4  1 
(6)
k
k
1
1
1
1 (3 + 4i) + (3 - 4i)
Ak 0  = 4 (3 + 4i)k v1 + 4 (3 - 4i)k v2 = 4  2i[(3 + 4i)k - (3 - 4i)k] 
In order to see the behavior of Akx it is convenient to use the polar form of a complex number. Given a
complex number z = x + yi we can express it as
(7)
z = r (cos  + i sin  )
where
r = |z| =
x2 + y2 =
_
z z = absolute value of z
 = arg z = angle  y  makes with the positive x axis
x
Note that if (7) holds then
_
z = r (cos  - i sin  )
Example 5. | 3 + 4i | = 5 and arg(3 + 4i) =  where  = tan-1
4
 0.927. So
3
3 + 4i = 5(cos  + i sin  ). Similarly 3 + 4i = 5(cos  - i sin  ).
4.14.1 - 5
One reason the polar form of a complex number is useful is because it allows us to easily compute powers
of complex numbers. If z = r (cos  + i sin  ) then it is not hard to show
zk = rk (cos(k) + i sin(k) ).
_
(z )k = rk (cos(k) - i sin(k) ).
_
zk + (z )k = 2rkcos(k)
_
zk - (z )k = 2irksin(k)
Example 6. (3 + 4i)k = 5k(cos(k) + i sin(k) ) and (3 - 4i)k = 5k(cos(k) - i sin(k) ) where  = tan-1
 0.927.
Using these relations, (6) becomes
1
5k  cos(k) 
Ak 0  = =
2  - 2isin(k) 
where  = tan-1
4
 0.927. Also
3
k
xk =
Ax
|| Akx ||
=
5k  cos(k) 
2  - 2isin(k) 
5k  cos(k) 
||
||
2  - 2isin(k) 
=
1
1 + sin2(k)
 cos(k) 
 - 2isin(k) 
As one can see xk does not approach a limit.
So the basic QR algorithm is not going to work with matrices with complex eigenvalues. The
modifications necessary to handle complex eigenvalues are discussed in the next section.
While we are on the subject of absolute values of complex numbers, it will be convenient to define
_
sign(z) = z / z
Example 7. sign(3 + 4i) =
3 4
+ i.
5 5
4.14.1 - 6
4
3