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The Jordan Canonical Form of a Nilpotent Matrix

Math 422

Schur"s Triangularization Theorem tells us that every matrixAis unitarily similar to an upper triangular

matrixT. However, the only thing certain at this point is that the the diagonal entries ofTare the eigenvalues

Decomposition of a singular matrixAof indexkproduces a block diagonal matrix C0 0L similar toAin whichCis non-singular,rank(C) =rankAk;andLis nilpotent of indexk. Is it possible are predictable? The goal of this lecture is to do exactly this for nilpotent matrices. LetLbe annnnilpotent matrix of indexk:ThenLk16= 0andLk= 0:Let"s compute the eigenvalues ofL:Supposex6= 0satis...esLx=x;then0 =Lkx=Lk1(Lx) =Lk1(x) =Lk1x=Lk2(Lx) =

2Lk2x==kx;thus= 0is the only eigenvalue ofL.

Now ifLis diagonalizable, there is an invertible matrixPand a diagonal matrixDsuch thatP1LP=D. Since the diagonal entries ofDare the eigenvalues ofL;and= 0is the only eigenvalue ofL, we have D= 0:SolvingP1LP= 0forLgivesL= 0:Thus a diagonalizable nilpotent matrix is the zero matrix, or the “simpli...ed"form ofLwill be non-zero. LetLbe a non-zero nilpotent matrix. SinceLis triangularizable, there exists an invertiblePsuch that P 1LP=2 6

66640

0 0 0 003 7 7775:

The simpli...cation procedure given in this lecture produces a matrix similar toLwhose non-zero entries lie

exclusively on the superdiagonal ofP1LP:An example of this procedure follows below. De...nition 1LetLbe nilpotent of indexkand de...neL0=I:Forp= 1;2;:::;k;letx2Cnsuch that y=Lp1x6= 0:The Jordan chain onyof lengthpis the set

Lp1x;:::;Lx;x.

Exercise 2LetLbe nilpotent of indexk:Ifx2Cnsatis...esLk1x6= 0;prove thatLk1x;:::; Lx;xgis linearly independent.

Example 3LetL=2

40 1 2

0 0 3

0 0 03

5 ;thenL2=2

40 0 3

0 0 0

0 0 03

5 andL3= 0:Letx=2 4a b c3 5 ;then Lx=2

40 1 2

0 0 3

0 0 03

52
4a b c3 5 =2

4b+ 2c

3c 03 5 andL2x=2

40 0 3

0 0 0

0 0 03

52
4a b c3 5 =2 43c
0 03 5

Note that

the Jordan chain onyis8< :2 49
0 03 5 ;2 48
9 03 5 ;2 41
2 33
59=

Form the matrix

P=L2xjLxjx=2

49 8 1

0 9 2

0 0 33

5 1 then

J=P1LP=1243

2

42724 7

0 2718

0 0 813

52

40 1 2

0 0 3

0 0 03

52

49 8 1

0 9 2

0 0 33

5 =2

40 1 0

0 0 1

0 0 03

5 is the “Jordan form"ofL: Since= 0is the only eigenvalue of annnnon-zero nilpotentL;the eigenvectors ofLare exactly the non-zero vectors inN(L). ButdimN(L)< nsinceLis not diagonalizable and possesses anincomplete

set of linearly independent eigenvectors. So the process by which one constructs the desired similarity

transformationP1LPinvolves appropriately extending a de...cient basis forN(L)to a basis forCn. This process has essentially two steps:

Construct a somewhat special basisBforN(L):

ExtendBto a basis forCnby building Jordan chains on the elements ofB: De...nition 4A nilpotent Jordan blockis a matrix of the form 2 6

66666640 1 00

0 0 1 ...0 0 0 0 ...0 ............1

0 0 003

7

7777775:

A nilpotent Jordan matrixis a block diagonal matrix of the form 2 6 6664J
100

0J2...0

.........0

0 0Jm3

7

7775;(1)

where eachJiis a nilpotent Jordan block. When the context is clear we refer to a nilpotent Jordan matrix as a Jordan matrix. Theorem 5Everynnnilpotent matrixLof indexkis similar to annnJordan matrixJin which the number of Jordan blocks isdimN(L); the size of the largest Jordan block iskk; for1jk;the number ofjjJordan blocks is rank

Lj12rankLj+rankLj+1;

the ordering of theJi"s is arbitrary. Whereas the essentials of the proof appear in the algorithm below, we omit the details.

De...nition 6IfLis a nilpotent matrix, a Jordan formofLis a Jordan matrixJ=P1LP:The JordanstructureofLis the number and size of the Jordan blocks in every Jordan formJofL.

Theorem 5 tells us that Jordan form is unique up to ordering of the blocksJi. Indeed, given any prescribed

ordering, there is a Jordan form whose Jordan blocks appear in that prescribed order. 2 De...nition 7The Jordan Canonical Form(JCF) of a nilpotent matrixLis the Jordan form ofLin which the Jordan blocks are distributed along the diagonal in order of decreasing size. Example 8Let us determine the Jordan structure and JCF of the nilpotent matrix L=2 6

6666641 12 0 11

3 1 5 11 3

21 0 01 0

2 1 0 0 1 0

531111

3211 013

7

777775row-reduce

!2 6

6666641 0 023

43
13

0 1 043

53
23

0 0 113

23
23

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 03

7

777775;

the number of Jordan blocks isdimN(L) = 3:Then L 2=2 6

6666646 3 3 1 1 2

633112

0 0 0 0 0 0

0 0 0 0 0 0

633112

6331123

7

777775row-reduce

!2 6

6666641

12 12 16 16 13

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 0 0 03

7

777775

has rank1andL3= 0;therefore theindex(L) = 3and the size of the largest Jordan block is33. Let r

0=rankL0= 6

r

1=rankL1= 3

r

2=rankL2= 1

r

3=rankL3= 0

r

4=rankL4= 0;

then the numberniofiiJordan blocks is n

1=r02r1+r2= 1

n

2=r12r2+r3= 1

n

3=r22r3+r4= 1:

Obtain the JCF by distributing these blocks along the diagonal in order of decreasing size. Then the JCF of

Lis2 6

6666640 1 00 00

0 0 10 00

0 0 00 00

0 0 00 10

0 0 00 00

0 0 00 00

3 7

777775:

This leaves us with the task of determining a non-singular matrixPsuch thatP1LPis the JCF ofL:

As mentioned above, this process has essentially two steps. A detailed algorithm follows our next de...nition

and a key exercise. De...nition 9LetEbe any row-echelon form of a matrixA:Letc1;:::;cqindex the columns ofEcontaining leading1"s and letyidenote thecthicolumn ofA:The columnsy1;:::;yqare called the basic columns ofA. Exercise 10LetB= [b1j jbp]be annpmatrix with linearly independent columns:Prove that

multiplication byBpreserves linear independence, i.e., iffv1;:::;vsgis linearly independent inCp, then

fBv1;:::;Bvsgis linearly independent inCn. 3

Nilpotent Reduction to JCF

Given a nilpotent matrixLof indexk;seti=k1and letSk1=fy1;:::;yqgbe the basic columns of L k1:

1. ExtendSk1[ [ Sito a basis forRLi1\N(L)in the following way:

(a) Letfb1;:::;bpgbe the basic columns ofLi1and letB= [b1j jbp]: (b) SolveLBx= 0forxand obtain a basisfv1;:::;vsgforN(LB);thenfBv1;:::;Bvsgis a basis forRLi1\N(L)by Lemma 11 below. (c) Form the matrix[y1j jyqjBv1j jBvs];its basic columnsn y

1;:::;yq;Bv1;:::;Bvjo

form a basis forRLi1\N(L)containingSk1[ [ Si:Let S i1=n Bv

1;:::;Bvjo

(d) Decrementi;letfy1;:::;yqgbe the basic columns ofLi;and repeat step 1 untili= 0. Then S k1[ [ S0=fb1;:::;btgis a basis forN(L).

2. For eachj;ifbj2 Si;...nd a particular solutionxjofLix=bjand build a Jordan chainLixj;:::;Lxj;xj:

Set p j=Lixjj jLxjjxj; the desired similarity transformation is de...ned by the matrix

P= [p1j jpt]:

Lemma 11In the notation of the algorithm above,fBv1;:::;Bvsgis a basis forRLi1\N(L). Proof.Note thatBvj2R(B)andLBvj= 0impliesBvj2N(L):ButR(B) =RLi1implies that Bv j2R(B)\N(L) =Mi1for allj:The setfBv1;:::;Bvsgis linearly independent by Exercise 10. To show thatfBv1;:::;BvsgspansMi1, lety2R(B)\N(L):Sincey2R(B);there is someu2Cn such thaty=Bu;sincey2N(L);0 =Ly=LBu:Thusu2N(LB)and we may expressuin the basis fv1;:::;vsgasu=c1v1++csvs:Thereforey=Bu=B(c1v1++csvs) =c1Bv1++csBvsand

fBv1;:::;Bvsgspans.In particular, supposeLis annnnilpotent matrix of indexn:Choose a non-zero vectorx1such that

L n1x16= 0:Then x

12N(Ln)rNLn1

x

2=Ln1x1)x22N(L)

x

3=Ln2x1)x32NL2rN(L)

x n=Lx1)xn2NLn1rNLn2: Thusfx2;:::;xn;x1g=Ln1x1;:::;Lx1;x1is linearly independent inCnandfx2gis a basis for the

1-dimensional subspaceN(L):ThusLn1x1;:::;Lx1;x1is a Jordan chain onx2of lengthnand the

Jordan form ofLconsists of onennJordan blockJ. LetP=Ln1x1j jLx1jx1;then

J=P1LP=2

6

6666640 1 00

0 0 10

0 0 0 ............1

0 0 003

7

777775:

4

Example 12Let

L=2 6

640 1 2 3

0 0 4 5

0 0 0 6

0 0 0 03

7 75:
Then L 2=2 6

640 0 4 17

0 0 0 24

0 0 0 0

0 0 0 03

7

75;L3=2

6

640 0 0 24

0 0 0 0

0 0 0 0

0 0 0 03

7

75;L4=2

6

640 0 0 0

0 0 0 0

0 0 0 0

0 0 0 03

7 75
andLis nilpotent of index4:Letx1=e4;then L

3x1= [24;0;0;0]T

L

2x1= [17;24;0;0]T

Lx

1= [3;5;6;0]T

and L3x1;L2x1;Lx1;x1is a Jordan chain ony=L3x1. Form the matrix

P=L3x1jL2x1jLx1jx1=2

6

6424 17 3 0

0 24 5 0

0 0 6 0

0 0 0 13

7 75;
Then

J=P1LP=2

6

6424 17 3 0

0 24 5 0

0 0 6 0

0 0 0 13

7 7512
6

640 1 2 3

0 0 4 5

0 0 0 6

0 0 0 03

7 752
6

6424 17 3 0

0 24 5 0

0 0 6 0

0 0 0 13

7 75=2
6

640 1 0 0

0 0 1 0

0 0 0 1

0 0 0 03

7 75:
Example 13Let"s construct the matrixPthat produces the Jordan form ofLin Example 8. SinceLhas index3;we row-reduceL2and ...nd one basic column y

1= [6;6;0;0;6;6]T:

ThenS2=fy1gcontains the basic column ofRL2;seti= 2:

1. ExtendS2to a basis forR(L)\N(L):

(a) Row-reducing, we ...nd that the basic columns ofLare its ...rst three columns. Thus we let B=2 6

6666641 12

3 1 5 21 0
2 1 0 531
3213
7

777775:

(b) ComputeLBand solveLBx= 0to obtain a basis forN(LB): LB=2 6

6666641 12 0 11

3 1 5 11 3

21 0 01 0

2 1 0 0 1 0

531111

3211 013

7

7777752

6

6666641 12

3 1 5 21 0
2 1 0 531
3213
7

777775=2

6

6666646 3 3

633
0 0 0 0 0 0 633
6333
7

777775

5 2 6

6666646 3 3

633
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