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{SECT 0 {EXCHG {PARA 4 "" 0 "" {TEXT 204 23 "Jordan Canonical Form I" 
}}{PARA 0 "" 0 "" {TEXT 205 0 "" }}{PARA 0 "" 0 "" {TEXT 205 0 "" }}
{PARA 0 "" 0 "" {TEXT 205 16 "By Russell Blyth" }}{PARA 0 "" 0 "" 
{TEXT 205 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "restart: w
ith(LinearAlgebra):" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 206 32 "Jordan C
anonical Form - Examples" }}{EXCHG {PARA 0 "" 0 "" {TEXT 205 158 "For \+
an n x n matrix  A  over a field F, the Jordan canonical form can be f
ound as long as the characteristic polynomial of A splits. Let's see a
 few examples." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "A := Matr
ix([[-2, 0, 2], [-2, -1, 3], [0, 0, -4]]);\nfactor(" }{MPLTEXT 1 0 33 
"CharacteristicPolynomial(A,t));\n" }{MPLTEXT 1 0 14 "JordanForm(A);" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT 205 100 "A has three distinct eigenval
ues and hence is diagonalizable. The Jordan canonical form is diagonal
." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "B := Matrix([[1, 1, 1]
, [0, 1, 0], [0, 0, 0]]);\n" }{MPLTEXT 1 0 40 "factor(CharacteristicPo
lynomial(B,t));\n" }{MPLTEXT 1 0 14 "JordanForm(B);" }}}{EXCHG {PARA 0
 "" 0 "" {TEXT 205 164 "Thus B is has two Jordan blocks, one 1x1  bloc
k corresponding to the eigenvalue 0, and a 2x2 block corresponding to \+
the eigenvalue 1 - thus B is not diagonalizable." }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 60 "C := Matrix([[1,1,1,1], [0,1,0,0], [0,0,1,0],
 [0,0,0,0]]);\n" }{MPLTEXT 1 0 40 "factor(CharacteristicPolynomial(C,t
));\n" }{MPLTEXT 1 0 14 "JordanForm(C);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT 205 75 "Here there are three Jordan blocks - two corresponding t
o the eigenvalue 1." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "E :=
 Matrix([[1,1,1,1], [0,1,0,0], [0,0,1,0], [0,0,0,1]]);\n" }{MPLTEXT 1 
0 40 "factor(CharacteristicPolynomial(E,t));\n" }{MPLTEXT 1 0 14 "Jord
anForm(E);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 205 65 "E has three Jordan
 blocks, all corresponding to the eigenvalue 1." }}{PARA 0 "" 0 "" 
{TEXT 205 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "F := Matri
x([[1,1,1,1], [0,1,0,0], [0,0,2,1], [0,0,0,2]]);\n" }{MPLTEXT 1 0 40 "
factor(CharacteristicPolynomial(F,t));\n" }{MPLTEXT 1 0 14 "JordanForm
(F);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 205 27 "F has two 2x2 Jordan blo
cks" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 104 "G := Matrix([[1,1,1
,1,1,1], [0,1,0,0,0,1], [0,0,1,1,1,0], [0,0,0,1,1,1],[0,0,0,0,2,1],[0,
0,0,0,0,2]]);\n" }{MPLTEXT 1 0 40 "factor(CharacteristicPolynomial(G,t
));\n" }{MPLTEXT 1 0 14 "JordanForm(G);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT 205 128 "G has a 2x2 Jordan block corresponding to the eigenvalu
e 2, and a 3x3  and a 1x1 Jordan block corresponding to the eigenvalue
 1." }}}{EXCHG {PARA 0 "" 0 "" {TEXT 205 0 "" }}{PARA 0 "" 0 "" {TEXT 
205 140 "Note that G and H (below) have the same characteristic polyno
mial, but different Jordan canonical forms. (Thus the Jordan canonical
 form is " }{TEXT 203 3 "not" }{TEXT 205 162 " determined by the chara
cteristic polynomial alone - the diagonal entries of the Jordan canoni
cal form, however, are determined by the characteristic polynomial.)" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 104 "H := Matrix([[1,1,1,1,1,
1], [0,1,1,0,0,1], [0,0,1,1,1,0], [0,0,0,1,1,1],[0,0,0,0,2,1],[0,0,0,0
,0,2]]);\n" }{MPLTEXT 1 0 40 "factor(CharacteristicPolynomial(H,t));\n
" }{MPLTEXT 1 0 16 "JordanForm(H);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
205 95 "Now find a basis with respect to which the linear operator T =
 L_B is in Jordan canonical form." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 4 "B;\n" }{MPLTEXT 1 0 38 "factor(CharacteristicPolynomia
l(B,t));" }}}{EXCHG {PARA 204 "" 0 "" {TEXT 207 12 "We know that" }
{TEXT 200 185 " K_0(T) has dimension 1, and K_1(T) has dimension 2, ma
tching the multiplicities of the eigenvalues. We know that K_0(T) = N(
T) and K_1(T) = N((T-I)^2). We find bases for these spaces. " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "id3 := IdentityMatrix(3);\n"
 }{MPLTEXT 1 0 21 "GES0 := NullSpace(B);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "beta01 := GES0[1];" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "GES1 := NullSpace((B - id3)^2);" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 20 "beta11 := GES1[1];\n" }{MPLTEXT 1 0 20 "beta1
2 := GES1[2];\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 205 34 "There are thr
ee basis vectors for " }{XPPEDIT 2 0 "Typesetting:-mrow(Typesetting:-m
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tingGI(_syslibGF'6%-I#miGF$6#Q!F'-I*mverbatimGF$6#Q26#*$)I\"RG6\"\"\"$
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rGQ%trueF'/%'lspaceGQ$0emF'/%'rspaceGQ/thickmathspaceF'/%)stretchyGF</
%*symmetricGF</%(maxsizeGQ)infinityF'/%(minsizeGQ\"1F'/%(largeopGF</%.
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205 369 ", but we need to make sure we have bases for each generalized
 eigenspace that are cycles of generalized eigenvalue. Certainly beta0
1 is fine for the (generalized) eigenspace corresponding to the eigenv
alue 0. For the eigenvalue 1 we need to find a vector in the generaliz
ed eigenspace that is not killed by the (p-1) = (2-1) = 1st power of (
T-I). Try the vector beta11:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 21 "v2 := (B-id3).beta11;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 205 279 "
Since v2 is not zero, beta11 is an eigenvector, is a good choice for t
he final vector of the cycle of generalized eigenvectors corresponding
 to the eigenvalue 1, with the initial vector then being v2 (=beta12) \+
(v2 is an eigenvector of T corresponding to the eigenvalue 1). Check:"
 }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "(B-id3).beta12;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT 205 116 "Since beta11 and v2 (= beta12 in
 this case) are independent, our Jordan basis is \{beta01, v2, beta12
\}. Try it out:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "Q := <be
ta01|v2|beta11>;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 205 59 "This matrix \+
then transforms B to its Jordan canonical form." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 11 "Q^(-1).B.Q;" }}}{SECT 1 {PARA 212 "" 0 "" 
{TEXT 204 9 "Exercises" }}{PARA 0 "" 0 "" {TEXT 200 40 "1. Find a Jord
an basis for the matrix F." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 
"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 200 40 "2. Find a Jordan basis for \+
the matrix H." }}}{EXCHG {PARA 0 "> " 0 "" {XPPEDIT 19 1 "" "%#%?G" }}
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}
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