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Chapter 6 Jordan Forms of Matrices
6-1 Jordan Forms and Dot Diagrams
Generalized eigenvector: X  V, X≠0. If  λ fulfills (A-λI)PX=0 for some positive integer P.
Generalized eigenspace: Kλ={x  V: (A-λI)P(x)=0 for some P}.
 3 1  2
Eg. For A=  1 0
5  , find the generalized eigenspace of A.

 1  1 4 
(Sol.) f ( )  det( A  I )  (  3)(  2) 2  0 , 1  3, 2  2
For 1 , K 1  E1  N ( A  3I ) ,
  1

1  2  x 
x   0
 







( A  3I )  y    1  3 5    y   0  K 1  E 1  t  2  : t  C 
1

 z   1  1 1   z 
 

 1

1  2  x  0
x 
1
x   1
  













For 2 , ( A  2 I ) y   1  2 5  y  0   y     3  E 2    3,   C  ,
  
    
   1

 z 
  1
 z   1  1 2   z  0
  

  1   1

1  1  x  0
x   2
    









( A  2 I )  y    4  2 2    y   0  K 2    3  v  2  :  , v  C 
   1  0 

 z   2  1 1   z  0
    

2
 K 2  E 2 in this case,
 1  1   1 
3 0 0 
      
   2 ,  3,  2   and J  0 2 1  is a Jordan form.
  1    1  0  
0 0 2
      
0  2  3
Eg. For A= 1 3
3  , find the generalized eigenspace of A.

0 0
1 
(Sol.) det( A  I )  (  1) 2 (  2)  0 , 1  1, 2  2
For  2 , K 2
   1

  

 E2  N ( A  2 I )  u 1 : u  C 
 
 0

  

For 1 , K1  N (( A  I ) 2 ) , E1  N (( A  I ))
x 
 2  3
 x   1  2  3  x 


( A  I )  y    1
2
3    y   0   y   s  1   t  0 
 z 
 0   1 
 z   0
0
0   z 
x 
 2  3
 x   1  2  3  x 


( A  I ) 2  y    1
2
3    y   0   y   s  1   t  0  , ∴ E1  K 1 in this case,
 z 
 0   1 
 z   0
0
0   z 
 2  3  1 
1 0 0
      
   1 ,  0 ,  1   and J  0 1 0  D is a Diagonal form
 0   1   0  
0 0 2
      
J1
0
Jordan Form and Block: J= 


0
 j
0
0  0

0
J 2   
, in which Jj= 
  0



 0 Jk 

 0
0
j 1
 
0 j 
0
 is called

  

j 1 

   0  j 
1
0
 
the Jordan block corresponding to λj.
Eg. Let
J
J  1
0
0
 2 1 
3.5 1 
, where J1= 
and J2= 


 . And then we have
J2 
 0  2
 0 3.5
0
0
 2 1
 0 2 0
0 

J=
.
0
0 3.5 1 


0
0 3.5
0
 2 4 2
 2 0 1
Eg. For A= 
 2  2 3

 2  6 3
2
3
, find Q such that Q-1AQ=J.
3

7
(Sol.) det( A  I )  (  2) 2 (  4) 2 , 1  2,  2  4
2 0
 J 1  

0 2
2.  2 : r1 =4–Rank(A-4I)=4–3=1 and r2 =Rank(A-4I)–Rank( ( A  4 I ) 2 )=3-2=1
2 0 0 0
0 2 0 0 

4 1 


∴ Dot diagram:
 J 2  

J

J

J

1
2

0 0 4 1 

0 4


0 0 0 4
1. 1 : r1 =4-Rank(A-2I)=4–2=2= m1 , ∴ Dot diagram:  
Bases for K 1 and K 2 :
x   0
 y   2
For 1 , ( A  2 I )    
 z   2
  
 w   2
  2  0  
 4 2 2  x 
    



 2 1 3  y 
1 1 

 0  1   ,   
 2 1 3  z 
 0   2  
  
 2 0  
 6 3 5   w
 x    2  4 2 2  x 
 x  0 
 y    2  4 1 3  y 
   
     0   y   1 
( A  4I )   
 z    2  2  1 3  z 
 z  1 
  
  
   
w   2  6 3 3   w

 w 1 
For  2 ,
 x    2  4 2 2   x  0 
x  1 
 y   2  4 1 3  y  1 
 y   1








( A  4I )



  
 z   2  2  1 3  z  1 
 z   1
  
    
   
 w  2  6 3 3  w 1 
 w  0 
2
1
∴ Q
0

2
1
1 1  1
fulfills Q-1AQ=J.
2 1  1

0 1 0
0 0

 0   1  

    

 1 1 
   2   ,   

1   1 

1   0  

0  2  3
Eg. For A= 1 3
3  , find its Jordan canonical form (or diagonal form).

0 0
1 
(Sol.) det(A–λI ) =  (  1) 2 (  2) =0, 1  1(double ), 2  2 .
1 0
  J 1  

0 1 
For λ1, ∵ r1 =n-Rank(A-1I)=3-1=2, dot diagram: 
 D  J 1  J 2
1 0 0
 0 1 0  J  .
0 0 2
Eg. Let T be a linear operator on V defined by T(f)=
Jordan canonical basis for T.
(Sol.)
0 1 0 0 0
0 0 0 2 0

0 0 0 0 0
A
0 0 0 0 0
0 0 0 0 0

0 0 0 0 0
f
, and a basis α={1,x,y,x2,y2,xy}. Find a
x
0
0

1
6
 , det( A  I )    0 , λ=0, 0, 0, 0, 0, 0.
0
0

0
r1 =6–Rank(A-0I) =6-3=3
r2 =Rank((A-0I))–Rank( ( A  0I ) 2 )=3–1=2
r3 =Rank( ( A  0 I ) 2 )–Rank( ( A  0 I ) 3 )=1
0
0

1   
0
∴ dot diagram: 2  
J 
0
3 
0

0
1 0 0 0 0
0 1 0 0 0
0 0 0 0 0

0 0 0 1 0
0 0 0 0 0

0 0 0 0 0
∵ The 1st column of dot diagram consists of 3 dots, ∴
2
( x1 )  0
x 2
∴ x1  x 2 , (T  I )( x1 )  2 x , (T  I ) 2 ( x1 )  2 .
∵ The 2nd column of dot diagram consists of 2 dots, ∴
x2  xy , (T  I )( x2 )  y . Finally, x3  y 2
 β={2,2x,x2,y,xy,y2} is a Jordan canonical basis for T.

( x2 )  0
x
2  4
1  2
  
Eg. Suppose that the dot diagrams of A are as follows:
 
 
,


,
3  3
 
. Find the

Jordan canonical form.
(Sol.)
2
0

0

0
0













1 0 0 0
2 1 0 0
0 2 0 0
0 0 2 1 0
0 0 0 2 0
0 0 2 0 0 0 0
4 1 0 0
0 4 1 0
0 0 4 0
4
0








0
0
0
0

0
0
0
0

0
0

3 0 
 3


Nipotent: If Ap=0 for some positive integer P.
Theorem (a) A is nipotent  A has only zero eigenvalue. (b) Any diagonalizable nipotent matrix
is always equal to the zero matrix.
 1  1
5
Eg. For B= 
 , calculate B .
4
3


1 1 1 1  1  1  1 1 
1 1 1 0 0 1
(Sol.) J  



 Q 1 BQ , J  






DM
0 1 2 1  4 3   2  1
0 1 0 1 0 0
0 1 0 1 0 0
 M3  0  M 4  0
M2 





0 0 0 0 0 0
5
4
1 0
1 0 0 1
1 5
 5
0000  
∴ J  (D  M )  





0 1 
0 1 0 0
0 1
5
5
 1 1  1 5 1 1  9  5
 B 5  QJ 5 Q 1  



.
 2  1 0 1 2 1  20 11 
6-2 Minimum Polynomials
Minimum polynomial: A monic polynomial of least positive degree for which p(A) = 0.
Theorem
(a) If g(λ) is any polynomial for g(A)=0, then p(λ) divides g(λ). In particular, p(λ) divides the
characteristic polynomial of A.
(b) The minimum polynomial and the characteristic polynomial have the same zeros.
(c) The minimum polynomial is unique.
(Proof) (a) If g(λ)=p(λ)q(λ)+r(λ), deg(r(λ))<deg(p(λ)). Since g(A)=p(A)q(A)+r(A)=0 and
p(A)=0  r(A) = 0, but p(λ) is the minimum-degree polynomial and deg(r(A)) < deg(p(A))
∴ They are contradictory to each other, ∴ r(λ)=0
(b) Let characteristic polynomial=g(λ)=p(λ)q(λ). If p(λ)=0, then g(λ)=0.
x
eigenvector
Let
be the
of A, then p(λ)x=g(λ)x=0.
eigenvalue

For x  0 , 0  p( A)( x)  p( ) x  p( )  0 , ∴ p(λ) and g(λ) has the same zeros.
3  1 0 
Eg. For A= 0 2 0 , find the minimum polynomial for A.
1  1 2
(Sol.) ∵ det(A-  I) =-(λ-2)2(λ-3), ∴ the minimum polynomial for A is either (λ-2)(λ-3) or (λ-2)2(λ-3).
Substitute A into (λ-2)(λ-3) and shows that p(A) = (A-2I)(A-3I) = 0.
Theorem A is a diagonalizable matrix.  The minimum polynomial for A is of the form
p(A)=(A-λ1I)…(A-λkI).
Theorem Let λ1, λ2, …, λk be the distinct eigenvalues of A, and pi is the order of the largest
Jordan block corresponding
p(λ)=(λ-λ1)p1(λ-λ2)p2…(λ-λk)pk.
to
λi,
and
then
the
minimum
polynomial
of
A is
 2 1 0 0 0 0 0
 0 2 1 0 0 0 0


 0 0 2 0 0 0 0


Eg. If the Jordan canonical form of A is 0 0 0 2 1 0 0 , find the minimum polynomial.
 0 0 0 0 2 0 0


 0 0 0 0 0 3 1
0 0 0 0 0 0 3
(Sol.) For λ1=2, p1=3. For λ2=3, p2=2, ∴ p(A)=(A-2I)3(A-3I)2, but the characteristic polynomial of A is
(λ-2)5(λ-3)2.
2  4
1  2
  
Eg. If the dot diagrams of A are as follows:
 
 
,


,
3  3
 
, find the minimum

polynomial.
(Sol.) p1 is 3, p2 is 3, and p3 is 1, ∴ p(A)=(A-2I)3(A-4I)3(A+3I).
Application of the Minimal Polynomials: Computing matrix functions
 3 0 1
Eg. For A=  2 2 2, find A3-3A2+3A.
 1 0 1 
1 0 1
(Sol.) f ( )  (  2) 3  0 , 1  2, m1  3 , A  2I   2 0 2  , rank(A-2I)=1
 1 0  1
r1 =3–rank(A-2I)=3–1=2, dot diagram:
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2 1 0
r2  1  J  0 2 0  The minimal polynomial is (λ-2)2, ∴ ( A  2I ) 2  A2  4 A  4I  0
0 0 2
g ( A)  A3  3 A 2  3 A  aA  bI , g ' ( A)  3 A 2  6 A  3I  aI
g ( A  2I )  8I 3  12 I  6I  2I  (2a  b) I  2a+b=2
5 0 3
g ' ( A  2I )  12I 2  12I  3I  aI  a=3  b=-4, ∴ A3  3 A 2  3 A  3 A  4 I   6 2 6  .
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 3 0  1