042 jordex

4.2 Two Jordan Canonical Form Examples 4.2.1 Example (a): LetA=2 6644 0 1 0 2 2 3 0 1 0 2 0 4 0 1 23 7752M44(Q). We ndchT= (x2)2(x3)2=p2 1p2 2, wherep1=x2; p2=x3. CASE 1,p1=x2: p1(A) =A2I4=2 6642 0 1 0 2 0 3 0 1 0 0 0 4 0 1 03 775!2 6641 0 0 0 0 0 1 0 0 0 0 0 0 0 0 03 775; so(p1(A)) = 1= 2. Hence b1= 1 and the corresponding dot diagram has height 1, width 2, with associated Jordan blocks J1(2)J1(2):

N1;x2 We ndv11=2 6640 1 0 03 775andv12=2 6640 0 0 13 775form a basis for Ker p1(TA) = N(A2I4) andmTA;v11=mTA;v12=x2. Also Ker (pb1 1(TA)) =N(p1(A)) =N(A2I4) =CTA;v11CTA;v12: Note thatCTA;v11andCTA;v12have Jordan bases 11:v11and12:v12 respectively. CASE 2,p2=x3: p2(A) =A3I4=2 6641 0 1 0 21 3 0 1 01 0 4 0 113 775!2 6641 0 01 3 0 1 01 3 0 0 11 3 0 0 0 03 775; so(p2(A)) = 1 = 2; also(p2 2(A)) = 2. Hence b2= 2 and we get a corresponding dot diagram consisting of two vertical dots, with associated Jordan block J2(3):
N2;x3
N1;x3 71 We have to nd a basis of the form p2(TA)(v21) = (A3I4)v21for Kerp2(TA) = N(A3I4). To ndv21we rst get a basis for N(A3I4)2. We have p2 2(A) = (A3I4)2=2 6640 0 0 0 3 14 0 0 0 0 0 1 0 2 13 775!2 6641 021 0 1103 0 0 0 0 0 0 0 03 775 and we nd X1=2 6642 10 1 03 775andX2=2 6641 3 0 13 775is such a basis. Then we have N2;p2=hp2X1; p2X2i =hp2(A)X1; p2(A)X2i=h(A3I4)X1;(A3I4)X2i =*2 6643 3 3 93 775;2 6641 1 1 33 775+ =*2 6643 3 3 93 775+ : Hence we can take v21=X1. ThenmTA;v21= (x3)2. Also Kerpb2 2(TA)) =N(p2 2(A)) =N(A3I4)2=CTA;v21: MoreoverCTA;v21has Jordan basis 21:v21;(A3I4)v21. Finally we have V4(Q) =CTA;v11CTA;v12CTA;v21and=11[12[ 21is a basis for V4(Q). Then with P= [v11jv12jv21j(A3I4)v21] =2 6640 0 2 3 1 0 103 0 0 13 0 1 0 93 775 we have P1AP= [TA] =J1(2)J1(2)J2(3) =2 6642 0 0 0 0 2 0 0 0 0 3 0 0 0 1 33 775: 72 4.2.2 Example (b): LetA2M66(F) have the property that chA=x6; mA=x3and (A) = 3; (A2) = 5;((A3) = 6): Next, with h;x= dimFNh;xwe have 1;x=(A) = 3 = 1; 2;x=(A2)(A) = 53 = 2; 3;x=(A3)(A2) = 65 = 1: Hence the dot diagram corresponding to the (only) monic irreducible factor xofmAis
N3;x

N2;x


N1;x Hence we read o that 9a non-singular P2M66(F) such thatP1AP= J3(0)J2(0)J1(0). To nd such a matrix Pwe proceed as follows: (i) First nd a basis for N3;x. We do this by rst nding a basis for N(A3):X1; X2; X3; X4; X5; X6. Then N3;x=hA2X1; A2X2; A2X3; A2X4; A2X5; A2X6i: We now apply the LRA (left{to{right algorithm) to the above spanning family to get a basis A2v11forN3;x, whereA2v11is the rst non{zero vector in the spanning family. (ii) Now extend the linearly independent family A2v11to a basis for N2;x. We do this by rst nding a basis Y1; Y2; Y3; Y4; Y5forN(A2). Then N2;x=hAY1; AY 2; AY 3; AY 4; AY 5i: We now attach A2v11to the head of this spanning family: N2;x=hA2v11; AY 1; AY 2; AY 3; AY 4; AY 5i and apply the LRA to nd a basis for N2;xwhich includes A2X1. This will have the form A2v11; Av 12, whereAv12is the rst vector in the list AY1;:::;AY 5which is not a linear combination of A2v11. (iii) Now extend the linearly independent family A2v11; Av 12to a basis forN1;x=N(A). We do this by rst nding a basis Z1; Z2; Z3forN(A). 73 Then place the linearly independent family A2v11; Av 12at the head of this spanning family: N1;x=hA2v11; Av 12; Z1; Z2; Z3i: The LRA is then applies to the above spanning family selects a basis of the formA2v11; Av 12; v13, wherev13is the rst vector among Z1; Z2; Z3which is not a linear combination of A2v11andAv12. ThenmTA;v11=x3; mTA;v12=x2; mTA;v13=x. Also Kerpb1 1(TA) =N(A3) =CTA;v11CTA;v12CTA;v13: Finally, if we take Jordan bases 11:v11; Av 11; A2v11; 12:v12; Av 12; 13:v13 for the three T{cyclic subspaces CTA;v11; CTA;v12; CTA;v13, respectively, we then get the basis =11[12[13 =v11; Av 11; A2v11;v12; Av 12;v13 forV6(F). Then if P= [v11jAv11jA2v11jv12jAv12jv13] we have P1AP= [TA] =J3(0)J2(0)J1(0) =2 66666640 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 03 7777775: 74