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Canonical Forms資料來源：Friedberg, Insel, and Spence, “Linear Algebra”, 2nd ed.,Prentice-Hall. (Chapter 7)
大葉大學 資訊工程系黃鈴玲
Linear Algebra
Jordan_2
Introduction
The advantage of a diagonalizable linear operator is the simplicity of its description.
Not every linear operator is diagonalizable.
Example:T: P2 P2 with T(f) = f’, the derivative of f.The matrix of T with respect to the standard basis {1, x, x2} for P2 is A=
The characteristic polynomial of A is
A has only one eigenvalue (=0)with multiplicity 3. The eigenspace corresponding to =0 is { | rR } A is not diagonalizable
000
200
010
3
00
20
01
0
0
r
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The purpose of this chapter is to consider alternative matrix representations for nondiagonalizable operators.
These representations are called canonical forms.
There are different kinds of canonical forms, there advantages and disadvantages depend on how they applied.
Our focus Jordan canonical form
This form is always available if the underlying field is algebraically closed, that is, if every polynomial with coefficients from the field splits.
Jordan_4
7.1 General Eigenvectors
Let T: VV be a linear operator, A be a matrix representation of T.
Suppose A has eigenvalues 1, 2, …, n, with has n corresponding, linearly independent eigenvectors v1, v2, …, vn. Let B={x1, x2, …, xn} (note that B is a basis for V), then
n
BT
00
00
00
][ 2
1
where [T]B is the diagonal matrix representation of T.
Since A may be not diagonalizable, we will prove that:
Jordan_5
For any linear operator whose characteristic polynomial splits, i.e., characteristic polynomial= there exists an ordered basis B for V such that
)())(( 21 nxxx
k
B
J
J
J
T
00
00
00
][ 2
1
for some eigenvalue j of T. Such a matrix Ji is called a Jordan block corresponding to j, and the matrix [T]B is called the Jordan canonical form of T. The basis B is called a Jordan canonical basis for T.
where Ji is a square matrix of the form [j] or the form
j
j
j
j
0000
1000
0010
0001
Jordan_6
Example 1The 88 matrix
is a Jordan canonical form of a linear operator T:C8C8; that is, there exists a basis B={x1, x2, …, x8} for C8 such that [T]B=J.
00000000
10000000
00300000
00130000
00002000
00000200
00000120
00000012
J
Note that the characteristic polynomial for T and J is det(JI) = (2)4(3)22, and only x1, x4, x5, x7 of B={x1, x2, …, x8} are eigenvectors of T.
Jordan_7
It will be proved that every operator whose characteristic polynomial splits has a unique Jordan canonical form (up to the order of the Jordan blocks).The Jordan canonical form is not completely determined by the characteristic polynomial of the transformation.For example:
00000000
10000000
00300000
00030000
00002000
00001200
00000020
00000012
'J
The characteristic polynomial of J’ is also (2)4(3)22.
Jordan_8
Consider the matrix J and the basis B={x1, x2, …, x8} of Example 1.
T(x2) = x1+2x2 T(x3) = x2+2x3 (T2I)(x2)=x1 (T2I)(x3)=x2
{x1, x2, x3} = {(T2I)2(x3), (T2I) (x3), x3}Note that in these three vectors, only x1=(T2I)2(x3) is an eigenvector.
00000000
10000000
00300000
00130000
00002000
00000200
00000120
00000012
J
Similarly, {x5, x6} = {(T3I)(x6), x6} and {x7, x8} = {T(x8), x8}
The relationship of vectors in basis B
Therefore, if x lies in a Jordan canonical basis of a linear operator T and corresponds to a Jordan block with diagonal entry , then (TI)p(x)=0 for a sufficiently large p.
For example, (TI)(x1)=0, (TI)2(x2)=0, (TI)3(x3)=0.
Jordan_9
DefinitionLet T be a linear operator on a finite-dimensional vector space V.A nonzero vector x in V is called a generalized eigenvector of T if there exists a scalar such that (TI)p(x)=0 for some positive integer p. We say that x is a generalized eigenvector corresponding to .
Generalized Eigenvectors
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DefinitionLet T be a linear operator on a vector space V, and let x be a generalized eigenvector of T corresponding to the eigenvalue .If p denotes the smallest positive integer such that (T)p(x)=0, then the ordered set {(TI)p1(x), (TI)p2(x), …, (TI) (x), x}
is called a cycle of generalized eigenvectors of T corresponding to .
(TI)p1(x) is called the initial vector of the cycle, x is called the end vector of the cycle, andthe length of the cycle is p.
In Example 1, B1={x1, x2, x3}, B2={x4}, B3={x5, x6}, B4={x7, x8} are the cycles of generalized eigenvectors of T that occur in B.
Cycle of Generalized Eigenvectors
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Let T be a linear operator on V, and let be a cycle of generalized eigenvectors of T corresponding to the eigenvalue .
(a) The initial vector of is an eigenvector of T corresponding to the eigenvalue , and no other member of is an eigenvector of T.
(b) is linearly independent.
(c) Let B be an ordered basis for V. Then B is a Jordan canonical basis for V if and only if B is a disjoint union of cycles of generalized eigenvectors of T.
Theorem 7.1
Jordan_12
Generalized EigenspaceDefinitionLet be an eigenvalue of a linear operator T on a vector space V.
The generalized eigenspace of T corresponding to , denoted by K
, is the set
K = { x V : (TI)p(x) = 0 for some positive integer p}.
Theorem 7.2
A subspace W of V is called T-invariant if T(W) W.
Let be an eigenvalue of a linear operator T on a vector space V.
Then K is a T-invariant subspace of V containing E(the eigenspace of T corresponding to ).
Jordan_13
The connection between the generalized eigenspaces and the characteristic polynomial of an operator
Theorem 7.6Let T be a linear operator on a finite-dimensional vector space V such that the characteristic polynomial of T splits. Suppose that are distinct eigenvalues of T with corresponding multiplicities m1, m2, …, mk. Then:
(a) dim (Ki) = mi for all i.
(b) If for each i, Si is a basis for Ki, then the union S = S1∪S2 …∪ ∪Sk is a basis of V.
(c) If B is a Jordan canonical basis for T, then for each i, Bi=B∩ Ki is a basis for Ki.
(d) Ki = Ker((TiI)mi) for all i.
(e) T is diagonalizable if and only if Ei = Ki for all i.
Jordan_14
Example 2Let T: C 3 C 3 be defined by T(x) = Ax, where
411
501
213
A
Solution
Find a basis for each eigenspace and each generalized eigenspace of T.
The characteristic polynomial of T is det(AI) = (t3)(t2)2.
= 3, = 2, with multiplicity m1=1, m2=2
dim(K) = 1, dim(K
) = 2
K = Ker(T3I) = E
= {r(1,2,1)} {r(1,2,1)} is a basis K
= Ker((T2I)2), E = Ker(T2I)
Jordan_15
Example 2 -continued
Since E = {s(1, 3, 1)}, dim(E
)=1< dim(K)=2
The basis of K is a single cycle of length 2.
Choose (1, 3, 1) be the initial vector of the cycle, then a vector v is the end vector if and only if (A2I)v = (1, 3, 1) . Solve (A2I) v= (1, 3, 1), we get v= {(r, 2+3r, r)} { (1, 3, 1), (1, 2, 0)} is a basis for K
.
B={(1,2,1), (1, 3, 1), (1, 2, 0)} is a basis for C3, and
200
120
003
][ BT is a Jordan canonical form of T.
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Example 3Let T: P2
P2 be defined by T(f) = f f’. Find a basis for each ei
genspace and generalized eigenspace of T.
100
210
011
][ BTA
Solution
The characteristic polynomial of T is det(AI) = (t)3.
= 1 with multiplicity m=3
dim(K) = 3 = dim(P2), so any basis of P2, for example, B, is a basis of K
E= {r} {1} is a basis for E
If B={1, x, x2}, then B is an ordered basis for P2 and
Jordan_17
Example 3 -continued
The basis of K is a single cycle of length 3. Choose 1 be the initial vector of the cycle, then a vector v is the end vector if and only if (AI)2vB = (1, , 0) . Solve (AI)2 X= (1, , 0), we get X= {(r, s, 0.5)}, let vB= (0, 0, 0.5), i.e., v = 0.5x2
(A+I) vB = (0, 1, 0) B={ 1, x, 0.5x2} is a basis for P2, and
100
110
011
][ BT is a Jordan canonical form of T.
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Homework
For each of the following linear operators T(x)=Ax, find a basis for each eigenspace and each generalized eigenspace.(a)
(b)
(c) T: P2P2 defined by T(f)=2f f ’.
Find the Jordan canonical forms for above linear operators.
31
11A
013
11821
5411
A
Jordan_19
7.2 Jordan Canonical Form
In this section we will develop a more direct approach for finding the Jordan canonical form and a Jordan canonical basis for a linear operator.
V: n-dim vector spaceT: V V a linear transformation such that its characteristic polynomial splits.Let be distinct eigenvalues of T. B: a Jordan canonical basis for T.Bi: the cycle of B that correspond to i form a basis for Ki
.
Ti: the restriction of T to Ki.
If [Ti]Bi = Ai for all i, then
k
B
AOO
OAO
OOA
JT
2
1
][
Jordan_20
Let Z1, Z2, …Zki be cycles of Bi with length p1 p2 … pki
.
For example:
How to find Ai and Bi?
i
i
i
i
i
i
i
i
i
iA
00000000
00000000
01000000
00000000
00010000
00001000
00000000
00000010
00000001 ki = 4 (4 cycles)p1 = 3 p2 = 3
p3 = 2 p4 = 1
Therefore Ai is entirely determined by the number ki, p1, …, pki.
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The array consists of ki columns (one column for each cycle).
The jth column consists of pj dots that correspond to the members of Zj (1 j ki )
Dot diagram
For example: the dot diagram associated with Bi, where ki = 4,p1 = 3, p2 = 3, p3 = 2, p4 = 1.
Let rj denote the number of dots in the jth row of a dot diagram for Bi. Then (a) r1= dim(V) – rank(TiI).(b) rj= rank((TiI)j1) – rank((TiI)j) if j >1.
Theorem 7.9
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Example 2Let
Find the Jordan canonical form of A and a Jordan canonical basis for the linear transformation TA.
3010
0110
0130
1012
A
The characteristic polynomial of A is det(AI) = (2)3(3) , Thus A has two eigenvalues, 1 =2 and 2 = 3 with multiplicities 3 and 1, respectively. Therefore
Solution
3 ,
200
20
02
21
Ay
x
A
Jordan_23
Example 2 -continued
1120
0000
0000
1120
)2( 2IA
Since r1 = dim (R4) – rank(A2I) = 2, r2 = rank(A2I) – rank((A2I)2) =1 dot diagram:
Now we find the Jordan canonical basis for T: B1={ x1, x2, x3} = {(T2I)(x2), x2, x3}, B2={x4} Note that x1 Ker((T2I)2), but x1 Ker(T2I); and x2, x3 Ker(T2I), x4 Ker(T3I).
2
0
1
0
,
0
2
1
0
,
0
0
0
1
3000
0200
0020
0012
J
K1= Ker((T2I)2)
{ } is a basis for K1.
row 1有 r1個 dot
row 2有 r2個 dot
column 1有 2個 dot故 cycle長為 2
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Example 2 -continued
Then x1 = (T2I)(x2) = (A2I)(x2) =
1010
0110
0110
1010
2IA
2
0
1
0
,
0
2
1
0
Ker(T2I) Let x2=
0
2
1
0
1
1
1
1
Now choose x3 to be an eigenvector which is linearly independent to x1, so let x3 =
0
0
0
1
For =3, it is easily to find an eigenvector. Let x4 =
1
0
0
1
Jordan_25
Example 2 -continued
So B={ } is a Jordan canonical basis for TA.
1
0
0
1
,
0
0
0
1
,
0
2
1
0
,
1
1
1
1
1001
0021
0011
1101
Notice that if Q = then J = Q 1AQ.
Jordan_26
Example 3Let
Find the Jordan canonical form J for A and a matrix Q such that J = Q1AQ.
7362
3322
3102
2242
A
The characteristic polynomial of A is det(AI) = (2)2(4)2 , Thus A has two eigenvalues, 1 =2 and 2 = 4 with multiplicities both 2, respectively. Therefore
Solution
40
4 ,
20
221
yA
xA
Jordan_27
Example 3 -continuedSince r1 = dim (R4) – rank(A2I) = 2, dot diagram for B1:
4000
1400
0020
0002
J
row 1 有 r1個 dot
column 1及 column 2各有 1個 dot
20
021A
Since r1 = dim (R4) – rank(A4I) = 4 3 = 1, r2 = rank(A4I) – rank((A4I)2) =1 dot diagram for B2:
row 1 有 r1個 dot
row 2 有 r2個 dot
column 1有 2個 dot
40
142A
Jordan_28
Example 3 -continued
5362
3122
3122
2240
2IA
Now we find the Jordan canonical basis for T: B1={x1, x2}, B2={x3, x4} = {(T4I)(x4), x4} Note that x1, x2 Ker(T2I); x3 Ker(T4I), x4 Ker((T4I)2) but x4 Ker(T4I).
0
2
1
0
,
2
0
1
2
{ } is a basis for K1.
3362
3122
3142
2242
4IA
1
1
1
0
{ } is a basis for K2.
Jordan_29
Example 3 -continued
1
1
1
0
3362
3122
3142
2242
)4(
d
c
b
a
XIALet
0
1
1
1
4
d
c
b
a
x
Therefore B = { }.
0
1
1
1
,
1
1
1
0
,
0
2
1
0
,
2
0
1
2
0102
1120
1111
1002
Notice that if Q = then J = Q 1AQ.
Jordan_30
Homework 1
Let T be a linear operator on a finite-dimensional vector space V such that the characteristic polynomial of T splits. Let 1 = 2, 2 = 4, and 3 = 3 be the distinct eigenvalues of T, and suppose that the dot diagrams for the restriction of TiI to Ki
(i=1, 2 ,3) are as follows:
Find the Jordan canonical form of T.
1 = 2 2 = 4
3 = 3
Jordan_31
Homework 2
Let T be a linear operator on a finite-dimensional vector space V such that the Jordan canonical form of T is
(a) Find the characteristic polynomial of T.(b) Find the dot diagram corresponding to each eigenvalue of T.(c) For which eigenvalues i, if any, does Eli
= Kli?
3000000
0300000
0020000
0012000
0000200
0000120
0000012
J
Jordan_32
Homework 3
For the following matrix A, find a Jordan canonical form J and a matrix Q such that J = Q1AQ.
211
367
233
A
