Download Complex numbers A complex number is a number written in the form

Complex numbers
A complex number is a number written in the form
a + bi
where :
a and b are ordinary “real” numbers,
i is the “imaginary” number satisfying i2 = −1.
Let z = a + bi. Then :
a is called the real part of z and is represented by Re z
b is called the imaginary part of z and is represented by Im z.
Example
• z = 3 − 5i is a complex number. We have Re z = 3 and Im z = −5.
• z = −1 + i is a complex number. We have Re z = −1 and Im z = 1.
• −i is a complex number. We have Re z = 0 and Im z = −1.
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Addition of complex numbers
How do we add complex numbers?
(a + bi) + (c + di) = (a + c) + (b + d)i
Example
• Suppose we want to add 3 − i and 1 + 2i.
(3 − i) + (1 + 2i) = (3 + 1) + (−1 + 2)i = 4 + i
• Suppose we want to add −1 + i and 2 + 3i.
(−1 + i) + (2 + 3i) = (−1 + 2) + (1 + 3)i = 1 + 4i
• Suppose we want to add i − 2 and 1 − 2i.
We first note that i − 2 = −2 + i. Then :
(−2 + i) + (1 − 2i) = (−2 + 1) + (1 − 2)i = −1 − i
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• We can subtract complex numbers similarly :
Subtraction of complex numbers
(a + bi) − (c + di) = (a − c) + (b − d)i
Example.
(3 − 5i) − (2 − 4i) = (3 − 2) + (−5 − (−4))i = 1 − i
Complex conjugation
For a complex number z = a + bi, the complex conjugate of z is defined as
z = a − bi
Example.
3 − 2i = 3 + 2i
1
2
+
√
3i =
1
2
−
√
3i
3 / 26
Multiplication of complex numbers
How do we multiply complex numbers?
(a + bi)(c + di) = ac + adi + bic + bidi = ac + adi + bci + bdi2 = (ac − bd) + (ad + bc)i
Example
(2 − 3i)(−1 + 2i) = 2(−1) + 2(2)i + (−3)(−1)i + (−3)(2)i2 = −2 + 4i + 3i + 6 = 4 + 7i
(i − 1)(3 − i) = 3i + i(−i) + (−1)(3) + (−1)(−i) = 3i − i2 − 3 + i = −2 + 4i
i4 = i · i · i · i = (−1) · i · i = (−1)(−1) = 1
4 / 26
Absolute value
Consider a complex number z = a + bi. The absolute value (or modulus or
norm) of z is defined by
√
|z| = a2 + b2
Note that zz = (a + bi)(a − bi) = a2 + b2 . Therefore we have
√
|z| = zz
• The absolute value of a complex number is always a nonnegative real
number.
Example.p
√
√
|3 − 4i| = 32 + (−4)2 = 9 + 16 = 25 = 5.
| − 1 − 2i| =
p
√
(−1)2 + (−2)2 = 5
Note that :
|ai| =
For instance | − 5i| =
p
(−5)2 =
√
a2 = |a|
√
25 = 5.
5 / 26
Basic properties of complex numbers
Let z, z1 , and z2 be complex numbers. Then :
1
z = z if and only if z is a real number, i.e., Im z = 0.
2
z1 + z2 = z1 + z2
3
z1 z2 = z1 z2
4
zz = |z|2 > 0.
5
6
|z1 z2 | = |z1 ||z2 |
Triangle Inequality : |z1 + z2 | 6 |z1 | + |z2 |
6 / 26
Division of complex numbers
Example
Simplify
1 + 3i
.
−1 + 5i
Solution.
(1 + 3i)(−1 − 5i)
−1 − 5i − 3i − 15i2
14 − 8i
14 − 8i
1 + 3i
=
=
=
=
−1 + 5i (−1 + 5i)(−1 − 5i)
1 + 5i − 5i − 25i2
1 − (−25)
26
14
8
1
(14 − 8i) =
− i
=
26
26 26
Division of complex numbers
In general, if z = a + bi and w = c + di are complex numbers and w , 0, then
is obtained as follows :
z
w
1
zw
z
=
=
(zw)
w ww |w|2
7 / 26
Example
3−i
−1 + i
=
=
(3 − i)(−1 − i)
−3 − 3i + (−i)(−1) + (−i)2
=
(−1 + i)(−1 − i)
1 + i − i − i2
−3 − 3i + i − 1 −4 − 2i
=
= −2 − i
1+i−i+1
2
In particular, using the above method one can see that if z = a + bi is a
complex number then
1
a
b
= 2
−
i
z
a + b2 a2 + b2
Example. Simplify
Solution.
1
.
−2 + i
−2
1
2 1
−
i = − − i.
((−2)2 + 12 ) (−2)2 + 12
5 5
You can check that
2 1
(−2 + i)(− − i) = 1
5 5
8 / 26
Eigenvalues and Eigenvectors
Problem. Consider the matrix A =
"
#
−1
. What does multiplication by A
3
0
2
do to vectors in R2 ?
x 7→ Ax
"
0
2
"
0
2
"
0
2
"
0
2
Ax
4
3
x
2
Ax
1
Ax
x x
Ax
0
-1
-2
x
#" # " #
1
−1 2
=
1
3 −1
#" # " #
2
−1 3
=
0
3 −2
#" # " #
−2
−1 −1
=
4
2
3
#" # " #
1
−1 1
=
−1
3 −1
-3
-4
-4
-3
-2
-1
0
1
2
3
4
9 / 26
Eigenvalues and Eigenvectors
Let A be an n × n matrix.
We are interested in directions which remain invariant by the matrix.
This means we want to find vectors x in Rn such that Ax = λx for some
scalar λ.
Let A be an n × n matrix.
If a nonzero vector x in Rn satisfies Ax = λx for some scalar λ, then :
The scalar λ is called an eigenvalue of A.
The vector x is called an eigenvector of A corresponding to λ.
Example
#
"
" #
0 −1
−1
corresponding to λ = 2:
is an eigenvector of
2 3
2
" #
#" # " #
"
−1
−2
0 −1 −1
=2
=
2
4
2
2 3
Remark
The eigenvector must be nonzero, but the eigenvalue can be zero or nonzero.
10 / 26
Let A be an n × n matrix.
If a nonzero vector x satisfies Ax = λx for some scalar λ, then :
The scalar λ is called an eigenvalue of A.
The vector x is called an eigenvector of A corresponding to λ.
Example
#
"
" #
1 3
1
?
an eigevector of
• Is
−2 2
−2
" #
" #
#" # " #
"
1
−5
−5
1 3 1
.
=λ
and there is no λ such that
=
No! Because
−2
−6
−6
−2 2 −2
Example

 
1 1
 1 

 
• The vector −1 is an eigenvector of 3 −1

 
0 1
2

 
   
0   1  0
1 1
 1 

 
   
because 3 −1 −2 −1 = 0 = 0 −1

 
  
1 
0 1
2
0
2
2

0 
−2 corresponding to λ = 0
1 
2
11 / 26
Let A be an n × n matrix.
If a nonzero vector x in Rn satisfies Ax = λx for some scalar λ, then :
The scalar λ is called an eigenvalue of A.
The vector x is called an eigenvector of A corresponding to λ.
Example
" # " # "√ #
"
1 −2 √3
2
,
,
The vectors
are eigenvectors of
1 −2
1
3
" #
#" # " #
"
1
3
2 1 1
=3
=
1
3
1 2 1
" #
#" # " #
"
−2
−6
2 1 −2
=3
=
−2
−6
1 2 −2
#"√ # " √ #
"
"√ #
2 1 √3
3 √3
3
=
=3 √
1 2
3
3
3 3
#
1
corresponding to λ = 3:
2
If x is an eigenvector of A corresponding to λ, then for every scalar c , 0, the
vector cx is also an eigenvector of A corresponding to λ.
Reason : A(cx) = c(Ax) = cλx = λ(cx).
12 / 26
Examples
Example
Is −2 an eigenvalue of A =
Solution.
Ax = (−2)x"
1
where I =
0
=⇒
#
0
.
1
"
1
3
#
3
?
1
Ax − (−2)x = 0
"
1
A + 2I =
3
=⇒
# "
2
3
+
0
1
Ax + 2x = 0
# "
3
0
=
3
2
=⇒
(A + 2 I)x = 0
#
3
3
" #
"
x1
3
x=
⇒ we should solve the homogeneous linear system
x2
3
#" # " #
0
3 x1
=
0
3 x2
#
"
#
"
R1 → 31 R1
R2 → R2 − 3R1
1 1 0
1
3 0
−−−−−−−−−−−−→
−−−−−−−−−−−−−−−−→
3 0
3 3 0
0
n
=⇒
x1 + x2 = 0
x1 : basic, x2 : free.
" # "
#
" #
x
−x2
−1
x1 + x2 = 0 =⇒ x1 = −x2
⇒ x= 1 =
= x2
x2
x2
1
1
0
"
3
3
0
0
#
13 / 26
Examples
Example
Is −2 an eigenvalue of A =
"
1
3
#
3
?
1
#
" # "
" #
x1
−x2
−1
Recall: Ax = (−2)x =⇒ (A + 2I)x = 0 =⇒ x =
=
= x2
x2
x2
1
It follows that −2 is an eigenvalue of A.
"
" #
1
−1
, with c , 0, is an eigenvector of
Any vector of the form c
3
1
corresponding to λ = −2.
" #
#" # " #
"
−1
2
1 3 −1
= (−2)
=
One can check that
1
−2
3 1 1
#
3
1
14 / 26
Examples
Example
Is 4 an eigenvalue of A =
"
#
3
?
1
1
3
Solution.
Ax = 4x
Ax − 4x = 0
=⇒
=⇒
(A − 4 I)x = 0 where I =
"
#
0
.
1
1
0
#
3
−3
#" # " #
" #
"
0
x1
−3 3 x1
=
x=
⇒ we should solve the homogeneous system
0
3 −3 x2
x2
"
1
3
−−−−−−−−−−→
"
A − 4I =
"
−3
3
=⇒
3
−3
0
0
#
n
x1 − x2 = 0
x1 − x2 = 0
=⇒
R1 → − 13 R1
# "
−4
3
+
0
1
1
3
−1
−3
0
0
# "
−3
0
=
3
−4
#
R2 →R2 −3R1
−−−−−−−−→
"
1
0
−1
0
0
0
#
x1 : basic, x2 : free.
" # " #
" #
x
x
1
x1 = x2
⇒ x = 1 = 2 = x2
x2
x2
1
15 / 26
Examples
Example
Is 4 an eigenvalue of A =
"
1
3
#
3
?
1
" # " #
" #
x1
x2
1
Recall: Ax = 4x =⇒ (A − 4 I)x = 0 =⇒ x =
=
= x2
x2
x2
1
It follows that 4 is an eigenvlue of A.
"
" #
1
1
Any vector of the form c , with c , 0, is an eigenvector of
3
1
corresponding to λ = 4.
" #
#" # " #
"
1
4
1 3 1
=4
=
One can check that
1
4
3 1 1
#
3
1
16 / 26
Examples
Example
Is 2 an eigenvalue of A =
Ax = 2x
A − 2I =
"
−1
2
"
1
2
Ax − 2x = 0 ⇒
#
# "
"
−1 3
2 0
3
=
−
2 3
0 2
5
#
"
R2 →R2 +2R1
−1 3
−−−−−−−−→
0 9
#
"
R1 →−R1
−1 0 0
−−−−−−→
0 1 0
=⇒
"
1
2
3
3
R1 →R1 −3R2
−−−−−−−−→
0
0
"
#
3
?
5
#
(A − 2I)x = 0.
0
0
1
0
#
0
1
R2 → 91 R2
−−−−−−→
0
0
"
−1
0
3
1
0
0
#
#
There are no free variables.
Therefore the homogeneous system has a unique solution, which is the trivial
solution x = 0.
#
"
1 3
It follows that 2 is NOT an eigenvalue of
2 5
17 / 26
Theorem
Let A be an n × n matrix.
• A scalar λ is an eigenvalue of A if and only if the homogeneous linear
system
(A − λI)x = 0
has nontrivial solutions.
• Here “I” denotes the n × n identity matrix.
Eigenspaces
Let A be an n × n matrix.
If λ is an eigenvalue of A, the set of vectors x in Rn satisfying
Ax = λx
forms a subspace of Rn . It is called the eigenspace of A corresponding to λ.
• Indeed the eigenspace corresponding to λ is identical to Nul(A − λI).
Ax = λx ⇔ (A − λI)x = 0
18 / 26
Example


1
1
 1
−2 −2 −1
Is −1 an eigenvalue of A = 
? If so, then determine a basis for


2
1
0
the eigenspace corresponding to −1.
Solution.
Ax = −x ⇒ Ax + x = 0 ⇒ (A + I)x = 0.

 
 
1
1  1 0 0  2
 1
−2 −2 −1 0 1 0 −2
A + I = 
 + 
 = 

 
 
2
1
0
0 0 1
2

 2
 −2


2
1
−1
1
1
−1
1
0
0
0
 R2 → R2 + R1 
 R3 → R3 − R1  2


 −−−−−−−−−−−−−−−−→  0


0
1
−1
1
1
0
0
 
x1 
 
x = x2 
 
x3

1 
−1

1
1
0
0
0
0
0


 R1 → 1 R1  1
2


 −−−−−−−→  0


0
1
2
0
0
1
2
0
0
0
0
0
The system has nontrivial solutions =⇒ λ = −1 is an eigenvalue of A.
   1

 1
 1
x1  − 2 x2 − 12 x3 
− 2 
− 2 
x  

 
 
1
1
1
1
x1 + 2 x2 + 2 x3 = 0 ⇒ x1 = − 2 x2 − 2 x3 ⇒  2  = 
x2
 = x2  1  + x3  0 
  

 
 
x3
x3
0
1
  1  1 

− 
−



  2   2  

is a basis for the eigenspace corresponding to λ = −1.
 1  ,  0  









 0
1 





19 / 26
Example

4 −1

Is 3 an eigenvalue of A = 1 4

2 −1
eigenspace corresponding to 3.

0 
−1? If so, then determine a basis for the
5 
2
Solution.
Ax = 3x ⇒ Ax − 3x = 0 ⇒ (A − 3I)x = 0.

 
 
0  3 0 0 1 −1
4 −1
1




1
4
−1 − 0 3 0 = 1
A − 3I = 

 
5  0
0 3
2 −1
2 −1
2

 1
 1


2
−1
1
−1
0
−1
− 12
 R2 → R2 − R1 
0  R
 1
3 → R3 − 2R1
0  −−−−−−−−−−→  0


0
0
−1
2
1
0
−1
− 12

0 
−1 

− 12
 
x1 
 
x = x2 
 
x3
R → 1R
2
2
 R1 → R2 1 + R2 
0  R
 1
3 → R3 − R2 

0  −−−−−−−−−−→  0


0
0
The system has nontrivial solutions =⇒ λ = 3 is an eigenvalue of A.
1
  1 

 2 
1

x1   21 x3 
 
x1 = 2 x3




⇒ x2  =  2 x3  = x3  12 

x = 1 x


 


3
2
2
x3
x3
1
 1 

 



  21  






 is a basis for the eigenspace corresponding to λ = 3.
2

  1  

0
1
0
− 12
− 12
0

0 

0 

0
20 / 26
Recall : Triangular matrices
An n × n matrix is called upper triangular if every entry below the
diagonal is zero.

−1
 0


 0
0
3
3
0
0
−1
2
−2
0

3

0 

5 

−3
An n × n matrix is called lower triangular if every entry above the
diagonal is zero.

 1

−2

 3
 4

0
0
2
0
1
−3
0
0
−4
5
4
0
0
0
2
5

0 

0 

0 
0 

−1
A matrix is called triangular if it is either upper triangular or lower
triangular.
21 / 26
Eigenvalues of a triangular matrix
Theorem
The eigenvalues of a triangular matrix are the diagonal entries of the matrix.

 1
−2

Example. The eigenvalues of A =  3
 4

0
We disregard repetitions.
0
2
0
1
−3
0
0
0
5
4
0
0
0
2
5
0

0 

0  are -1, 0, 1, 2.
0 
−1

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Eigenvalues of powers of a matrix
Theorem
Let A be a square matrix whose eigenvalues are λ1 , ..., λk . Then the
eigenvalues of Ar are λr1 , ..., λrk .
Key idea : If Ax = λx then :
A2 x = A(Ax) = A(λx) = λ(Ax) = λ(λx) = λ2 x
Similarly A3 x = A(A(Ax)) = λ3 x, etc.


0
0 0
0
 1

−1 2 0 0 0 
 2 −1 0 0 0 
3
4
Example If A = 
, what are the eigenvalues of A and A ?
 4
1
5 2
0 

0
Solution.
5
4
5
−1
Since A is triangular, the eigenvalues of A are its diagonal
entries, i.e., -1, 0, 1, 2.
The eigenvalues of A3 are (−1)3 , 03 , 13 , 23 ; i.e., -1, 0, 1, 8.
The eigenvalues of A4 are (−1)4 , 04 , 14 , 24 ; i.e., 0, 1, 16.
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Example

 i
0

Let A = 
0
0
−1 + i
−i
0
0
1 + 2i
4+i
−1 − i
0

4i − 3

−5 
.
2 

−1
• What are the eigenvalues of A?
Solution. The eigenvalues of A are i, −i, −1 − i, −1.
• What are the eigenvalues of A2 ?
Solution. The eigenvalues of A2 are i2 = −1, (−i)2 = −1, (−1 − i)2 = 2i,
(−1)2 = 1.
• What are the eigenvalues of A−1 ?
Solution. The eigenvalues of A−1 are
1
= −1.
−1
1
1
1
1 1
= −i,
= i,
= − + i, and
i
−i
−1 − i
2 2
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Summary
Let A be an n × n matrix.
If a nonzero vector x in Rn satisfies Ax = λx for some scalar λ, then :
The scalar λ is called an eigenvalue of A.
The vector x is called an eigenvector of A corresponding to λ.
Theorem
Let A be an n × n matrix.
• A scalar λ is an eigenvalue of A if and only if the homogeneous linear
system
(A − λI)x = 0
has nontrivial solutions.
• Here “I” denotes the n × n identity matrix.
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Summary
Eigenspaces
Let A be an n × n matrix.
If λ is an eigenvalue of A, the set of vectors x in Rn satisfying
Ax = λx
forms a subspace of Rn . It is called the eigenspace of A corresponding to λ.
• Indeed the eigenspace corresponding to λ is identical to Nul(A − λI).
Ax = λx ⇔ (A − λI)x = 0
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