Download Eigenvalues and Eigenvectors

Section 8.8 Eigenvalues and Eigenvectors
Matrix-vector multiplication can be thought of geometrically as
a linear transformation changing one vector into another
 4 2   −3   −4 

  = 

 5 1   4   −11
4x
5x
+
+
2y
y
=
=
0
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 4 2  x   0 

  =  
 5 1  y   6 
 4 2 2   0

  =  
 5 1   −4   6 
Ax = b
What vector x gets transformed
into the vector b ?
Ax = λx
 4 2   2   −2 
2

   =   = −1 
5
1
−
5
5
   
 −5 

 4 2   1  6 
1

  =   = 6 
1
1
6
5

   
1
eigenvector eigenvalue
What vector x gets transformed
into λx ( a scalar mult. of the orig. x ) ?
Section 8.8 Eigenvalues and Eigenvectors
Solve:
Ax = λ x
( An×n )
Ax − λ x = 0
A−λI )x = 0
(
↑ ↑
matrix
vector vector
Need ( A − λ I ) to not be invertible, because if it was
we would only have the trivial solution x = 0.
⇓
Set det ( A − λ I ) = 0
This leads to an equation in λ called the characteristic equation.
The roots of the characteristic equation are the eigenvalues λ .
For each eigenvalue λ , find its eigenvector by solving ( A − λ I ) x = 0
1
 −1 1 0 


Find the eigenvalues of  1 2 1  .
 0 3 −1 
det ( A − λ I ) = 0 ⇒


−1 − λ
1
0
1
2−λ
3
0
1
−1 − λ
= ( −1 − λ )
2−λ
1
3
−1 − λ
+ ( −1)1
1
1
0 −1 − λ
= ( −1 − λ ) ( 2 − λ )( −1 − λ ) − 3 − ( −1 − λ ) = 0
= ( −1 − λ ) ( −2 − λ + λ 2 ) − 3 − ( −1 − λ ) = 0
= ( −1 − λ )  λ 2 − λ − 5 − ( −1 − λ ) = 0
(
)
= ( −1 − λ ) ( λ 2 − λ − 5 ) − 1 = 0
= ( −1 − λ ) ( λ − λ − 6 ) = 0
2
= ( −1 − λ )( λ − 3)( λ + 2 ) = 0
λ1 = −1, λ2 = −2, λ3 = 3
There must be an easier way ☺
Shortcut to finding the characteristic equation
2× 2
λ 2 − ( trace ( A ) ) λ + det ( A) = 0
3× 3
sum of the
diagonal entries
λ 3 − ( trace ( A) ) λ 2 + ( C11 + C22 + C33 ) λ − det ( A) = 0
sum of the diagonal cofactors
The only problem now is that you have to factor a cubic Find one root, then use synthetic (or long) division to get the
quadratic factor, then factor that (if possible)
n × n triangular or diagonal matrix
The eigenvalues are on the diagonal!
2
 −1 1 0 


Find the eigenvalues of  1 2 1 
 0 3 −1 


trace ( A ) = −1 + 2 − 1
=0
2 1
C11 =
= −2 − 3 = −5
3 −1
det ( A ) = −1
C22 =
−1 0
=1
0 −1
C33 =
−1 1
= −2 − 1 = −3
1 2
⇒ C11 + C22 + C33 = −7
2 1
1 1
= −1( −2 − 3) − 1( −1) = 6
+ (1)( −1)
3 −1
0 −1
Possible roots:
Characteristic Equation:
λ − 0 ⋅ λ − 7 λ − 6 = 0 ⇒ λ − 7λ − 6 = 0
3
using the 3 × 3 shortcut
λ 3 − ( trace ( A) ) λ 2 + ( C11 + C22 + C33 ) λ − det ( A) = 0
2
3
( since the constant term is
±1, ± 2, ± 3, and ± 6
− 6)
Plug these in until you find all the roots.
λ = 2 ? ⇒ 8 − 14 − 6 ≠ 0 No
λ 3 − 7λ − 6 = 0
λ = −2? ⇒ −8 + 14 − 6 = 0 Yes
λ = 1? ⇒ 1 − 7 − 6 ≠ 0 No
λ = −1? ⇒ −1 + 7 − 6 = 0 Yes λ = 3? ⇒ 27 − 21 − 6 = 0 Yes
λ1 = −1, λ2 = −2, λ3 = 3
Long division:
λ2 − λ − 6
λ + 1 λ 3 + 0λ 2 − 7 λ − 6
Or you could find one root, then use synthetic (or long) division
to find the quadratic factor.
Synthetic division:
−1 1 0 −7 −6
↓ −1
1
6
1 −1 − 6
0
−(λ3 + λ2 )
− λ 2 − 7λ
− ( −λ 2 − λ )
⇒ λ − 7λ − 6 = ( λ + 1) ( λ − λ − 6 )
3
2
− 6λ − 6
λ 3 − 7λ − 6 = 0 ⇒ ( λ + 1)( λ + 2 )( λ − 3) = 0
− ( −6λ − 6 )
0
 −1 1 0 


Find the eigenvectors of  1 2 1 
 0 3 −1 


Eigenvalues:
λ1 = −1, λ2 = −2, λ3 = 3
For λ1 = −1
z is free, let z = 1
( A − λ I ) x = 0 ⇒ ( A + I ) v1 = 0
0 1 0

1 3 1
0 3 0

0  R1 ↔ R2

0
0 
|
|
|
1 3 1

0 1 0
0 3 0

|
|
|
0

0
0  −3R2 + R3
1 3 1

0 1 0
0 0 0

|
|
|
0 x + 3y + z = 0 ⇒ x = −z

0 ⇒ y = 0
0 
 −1
v1 =  0 
1
 
For λ2 = −2
z is free, let z = 3
( A − λ I ) x = 0 ⇒ ( A + 2 I ) v2 = 0
1 1 0

1 4 1
0 3 1

|
|
|
0

0  − R1 + R2
0 
1 1 0

0 3 1
0 3 1

|
|
|
0

0
0  − R2 + R3
1 1 0

0 3 1
0 0 0

|
|
|
1
0 x + y = 0 ⇒ x = −y ⇒ x = 3 z

0  3 y + z = 0 ⇒ y = −31 z
0 
1
v2 =  −1
3
 
For λ3 = 3
( A − λ I ) x = 0 ⇒ ( A − 3I ) v3 = 0
 −4 1 0 | 0  R1 ↔ R2


 1 −1 1 | 0 
 0 3 −4 | 0 


x − y + z = 0 ⇒ x = y − z ⇒ x = 13 z
 1 −1 1 | 0 
 1 −1 1 | 0 
 1 −1 1 | 0 



 4R + R 

4
2 0
−3 4 | 0 
 0 −3 4 | 0  −3 y + 4 z = 0 ⇒ y = 3 z
 −4 1 0 | 0  1

 0 3 −4 | 0 
 0 3 −4 | 0  R + R  0 0 0 | 0 




 2
3
1
 
z is free, let z = 3
v3 =  4 
 3
 
3
8.8 # 18
1 6 0


Find the eigenvalues and eigenvectors of  0 2 1  .
 0 1 2


det ( A − λ I ) = 0 ⇒
1− λ
6
0
0
2−λ
1
0
1
2−λ
= (1 − λ )
2−λ
1
1
2−λ
= (1 − λ ) ( 2 − λ )( 2 − λ ) − 1 = 0
(1 − λ )  4 − 4λ + λ 2 − 1 = 0
(1 − λ ) λ 2 − 4λ + 3 = 0 ⇒ (1 − λ )( λ − 1)( λ − 3) = 0
λ1 = λ2 = 1, λ3 = 3
For λ1 = λ2 = 1
For λ3 = 3
( A − I ) v1 = 0
0 6 0

0 1 1
0 1 1

0 1 1

0 6 0
0 0 0

|
|
|
|
|
|
0

0
0  − R3 + R2
0 6 0

0 1 1
0 0 0

0
 0


0  −6 R1 + R2  0
 0
0 

x is free, let x = 1
1
0
0
|
|
|
1
−6
0
( A − 3I ) v3 = 0
0  R1 ↔ R2

0
0 
|
|
|
 −2 6 0
 0 −1 1

 0 1 −1

0  y = −z ⇒ y = 0

0 ⇒ z =0
0 
1
0

0

|
|
|
−3
−1
0
1
|
|
0
0
|
0
0 
0 
0
0 
0 
−1
2
R1
1
0

0

−3
−1
0
1
|
|
1
−1
|
0
0 
0  R2 + R3
⇒ x = 3 y ⇒ x = 3z
⇒y=z
z is free, let z = 1
1
v1 =  0 
0
 
 3
 
v3 =  1 
1
 
8.8 # 14
7 0 
Find the eigenvalues and eigenvectors of 
.
 0 13 
7 0 
 0 13  is a diagonal matrix (its eigenvalues are on the diagonal)


λ1 = 7 and λ2 = 13
For λ1 = 7
For λ2 = 13
( A − 7 I ) v1 = 0
( A − 13I ) v2 = 0
0 0
0 6

1
v1 =  
 0
|
|
0  x is free,let x = 1
0  ⇒ y = 0
 −6
 0

0
|
0
|
0  ⇒x=0
0  y is free,let y = 1
 0
v2 =  
1
The eigenvectors of a diagonal matrix will be the columns of the identity matrix.
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