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204 0 "" }} {PARA 0 "" 0 "" {TEXT 204 16 "By Russell Blyth" }}{PARA 0 "" 0 "" {TEXT 204 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "restart: w ith(LinearAlgebra):" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 205 34 "Rational Canonical Form - Examples" }}{EXCHG {PARA 0 "" 0 "" {TEXT 204 114 "Fo r an n x n matrix A over a field F, the rational canonical form can \+ always be found. Let's see a few examples." }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 52 "A := Matrix([[-2, 0, 2], [-2, -1, 3], [0, 0, -4]]); " }{MPLTEXT 1 0 30 "CharacteristicPolynomial(A,t);" }{MPLTEXT 1 0 9 " \nfactor(" }{MPLTEXT 1 0 31 "CharacteristicPolynomial(A,t));" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 204 333 "Next, compute the set K_phi(A) f or each factor phi of the characteristic polynomial. In this case, we will simply be computing the eigenspace for each eigenvalue, and henc e arrive at the rational canonical form in diagonal form - the diagona l entries correspong to 1x1 companion matrices with (respectively) ent ries -4, -2 and -1." }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 272 "Note t hat the built-in Maple command RationalCanonicalForm gives a different answer, an alternate version of the rational canonical form. Our vers ion has the advantage that when the linear operator or matrix is diago nalizable, then the rational canonical form is diagonal." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "RationalCanonicalForm(A);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "B := Matrix([[1, 1, 1], [0, \+ 1, 0], [0, 0, 0]]);\n" }{MPLTEXT 1 0 31 "factor(CharacteristicPolynomi al" }{MPLTEXT 1 0 7 "(B,t));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 111 "We know that the companion matrix for the factor t is just a 1x1 bloc k with entry 0. What about for the factor " }{XPPEDIT 18 0 "Typesettin g:-mrow(Typesetting:-mi(\"\"), Typesetting:-mrow(Typesetting:-mi(\"\") , Typesetting:-msup(Typesetting:-mrow(Typesetting:-mo(\"(\", form = \" prefix\", fence = \"true\", separator = \"false\", lspace = \"thinmath space\", rspace = \"thinmathspace\", stretchy = \"true\", symmetric = \+ \"false\", maxsize = \"infinity\", minsize = \"1\", largeop = \"false \", movablelimits = \"false\", accent = \"false\", font_style_name = \+ \"2D Comment\", size = \"12\", foreground = \"[0,0,0]\", background = \+ \"[255,255,255]\"), Typesetting:-mrow(Typesetting:-mi(\"t\"), Typesett ing:-mo(\"−\", form = \"infix\", fence = \"false\", separator = \+ \"false\", lspace = \"mediummathspace\", rspace = \"mediummathspace\", stretchy = \"false\", symmetric = \"false\", maxsize = \"infinity\", \+ minsize = \"1\", largeop = \"false\", movablelimits = \"false\", accen t = \"false\", font_style_name = \"2D Comment\", size = \"12\", foregr ound = \"[0,0,0]\", background = \"[255,255,255]\"), Typesetting:-mn( \"1\")), Typesetting:-mo(\")\", form = \"postfix\", fence = \"true\", \+ separator = \"false\", lspace = \"thinmathspace\", rspace = \"verythin mathspace\", stretchy = \"true\", symmetric = \"false\", maxsize = \"i nfinity\", minsize = \"1\", largeop = \"false\", movablelimits = \"fal se\", accent = \"false\", font_style_name = \"2D Comment\", size = \"1 2\", foreground = \"[0,0,0]\", background = \"[255,255,255]\")), Types etting:-mn(\"2\"), superscriptshift = \"0\"), Typesetting:-mi(\"\")), \+ Typesetting:-mi(\"\"));" "-I%mrowG6#/I+modulenameG6\"I,TypesettingGI(_ syslibGF'6%-I#miGF$6#Q!F'-F#6%F+-I%msupGF$6%-F#6%-I#moGF$63Q\"(F'/%%fo rmGQ'prefixF'/%&fenceGQ%trueF'/%*separatorGQ&falseF'/%'lspaceGQ.thinma thspaceF'/%'rspaceGFE/%)stretchyGF?/%*symmetricGFB/%(maxsizeGQ)infinit yF'/%(minsizeGQ\"1F'/%(largeopGFB/%.movablelimitsGFB/%'accentGFB/%0fon t_style_nameGQ+2D~CommentF'/%%sizeGQ#12F'/%+foregroundGQ([0,0,0]F'/%+b ackgroundGQ.[255,255,255]F'-F#6%-F,6#Q\"tF'-F763Q(−F'/F;Q&infixF '/F>FBF@/FDQ0mediummathspaceF'/FGFjo/FIFBFJFLFOFRFTFVFXFenFhnF[o-I#mnG F$6#FQ-F763Q\")F'/F;Q(postfixF'F=F@FC/FGQ2verythinmathspaceF'FHFJFLFOF RFTFVFXFenFhnF[o-F^p6#Q\"2F'/%1superscriptshiftGQ\"0F'F+F+" }{TEXT 204 112 "? Just like for the Jordan form, we check the nullity of (B-I ), and we know that (B-I)^2 has nullity 2 for sure." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "id3 := IdentityMatrix(3);\nN" }{MPLTEXT 1 0 18 "ullSpace(B - id3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 58 "Sinc e the nullity of (B-I) is one, we have the dot diagram" }}{PARA 0 "" 0 "" {TEXT 204 0 "" }}{PARA 0 "" 0 "" {TEXT 204 4 "\245" }}{PARA 0 "" 0 "" {TEXT 204 4 "\245" }}{PARA 0 "" 0 "" {TEXT 204 0 "" }}{PARA 0 "" 0 "" {TEXT 204 49 "This gives us one 2x2 block for the factor phi = " } {XPPEDIT 18 0 "Typesetting:-mrow(Typesetting:-mi(\"\"), Typesetting:-m row(Typesetting:-mi(\"\"), Typesetting:-msup(Typesetting:-mrow(Typeset ting:-mo(\"(\", form = \"prefix\", fence = \"true\", separator = \"fal se\", lspace = \"thinmathspace\", rspace = \"thinmathspace\", stretchy = \"true\", symmetric = \"false\", maxsize = \"infinity\", minsize = \+ \"1\", largeop = \"false\", movablelimits = \"false\", accent = \"fals e\", font_style_name = \"2D Comment\", size = \"12\", foreground = \"[ 0,0,0]\", background = \"[255,255,255]\"), Typesetting:-mrow(Typesetti ng:-mi(\"t\"), Typesetting:-mo(\"−\", form = \"infix\", fence = \+ \"false\", separator = \"false\", lspace = \"mediummathspace\", rspace = \"mediummathspace\", stretchy = \"false\", symmetric = \"false\", m axsize = \"infinity\", minsize = \"1\", largeop = \"false\", movableli mits = \"false\", accent = \"false\", font_style_name = \"2D Comment\" , size = \"12\", foreground = \"[0,0,0]\", background = \"[255,255,255 ]\"), Typesetting:-mn(\"1\")), Typesetting:-mo(\")\", form = \"postfix \", fence = \"true\", separator = \"false\", lspace = \"thinmathspace \", rspace = \"verythinmathspace\", stretchy = \"true\", symmetric = \+ \"false\", maxsize = \"infinity\", minsize = \"1\", largeop = \"false \", movablelimits = \"false\", accent = \"false\", font_style_name = \+ \"2D Comment\", size = \"12\", foreground = \"[0,0,0]\", background = \+ \"[255,255,255]\")), Typesetting:-mn(\"2\"), superscriptshift = \"0\") , Typesetting:-mi(\"\")), Typesetting:-mi(\"\"));" "-I%mrowG6#/I+modul enameG6\"I,TypesettingGI(_syslibGF'6%-I#miGF$6#Q!F'-F#6%F+-I%msupGF$6% -F#6%-I#moGF$63Q\"(F'/%%formGQ'prefixF'/%&fenceGQ%trueF'/%*separatorGQ &falseF'/%'lspaceGQ.thinmathspaceF'/%'rspaceGFE/%)stretchyGF?/%*symmet ricGFB/%(maxsizeGQ)infinityF'/%(minsizeGQ\"1F'/%(largeopGFB/%.movablel imitsGFB/%'accentGFB/%0font_style_nameGQ+2D~CommentF'/%%sizeGQ#12F'/%+ foregroundGQ([0,0,0]F'/%+backgroundGQ.[255,255,255]F'-F#6%-F,6#Q\"tF'- F763Q(−F'/F;Q&infixF'/F>FBF@/FDQ0mediummathspaceF'/FGFjo/FIFBFJF LFOFRFTFVFXFenFhnF[o-I#mnGF$6#FQ-F763Q\")F'/F;Q(postfixF'F=F@FC/FGQ2ve rythinmathspaceF'FHFJFLFOFRFTFVFXFenFhnF[o-F^p6#Q\"2F'/%1superscriptsh iftGQ\"0F'F+F+" }{TEXT 204 262 ", namely the companion matrix for this factor. Now we depart from the JCF process to compute the rational ca nonical basis. For this factor, we need to choose a vector in K_phi(T) which is not in the null space of (B-I) which acts as the generator o f the T-cycle." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "B1 := Nul lSpace((B-id3)^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "v := \+ B1[1];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "Bv := B.v;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 204 159 "\{v, Bv\} is the B-cyclic basis \+ for the first factor. We also need a basis vector for the 1x1 companio n matrix for the factor t. (The dot diagram is just \245)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "B0 := NullSpace(B);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "w := B0[1];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 127 "This vector alone is the B-cyclic basis for the sec ond factor. Now we make the change of basis matrix Q from the basis ve ctors:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "Q := ;" } }}{EXCHG {PARA 0 "" 0 "" {TEXT 204 145 "Check to see that the similari ty results in the rational canonical form we expect. Convince yourself that the rational canonical form is correct!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "Q^(-1).B.Q;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 43 "Repeat for the following (familiar) matrix:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 193 "C := Matrix([[-41, -21, -11, 126, 63, -19 ], [587, 118, 46, -726, -363, 292], [-60, -11, -7, 66, 33, -30], [35, \+ 20, 10, -118, -60, 15], [117, -3, -6, 8, 6, 63], [94, 43, 23, -258, -1 29, 44]]);\n" }{MPLTEXT 1 0 38 "factor(CharacteristicPolynomial(C,t)); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 332 "As before, C has two distin ct irreducible (linear) factors. Let phi_1 = t+3 and phi_2 = t-2. We w ish to find the dot diagram for C corresponding to each factor. Comput e a basis for the null space of powers of phi_i(C) until the nullity m atches the multiplicity of the factor phi_i. For the factor phi_1 we a re looking for nullity 2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "id6 := IdentityMatrix(6);\nN" }{MPLTEXT 1 0 18 "ullSpace(C+3*id6); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 83 "The fact that this first nul lspace has dimension 1 tells us that the dot diagram is" }}{PARA 0 "" 0 "" {TEXT 204 0 "" }}{PARA 0 "" 0 "" {TEXT 204 4 "\245" }}{PARA 0 "" 0 "" {TEXT 204 4 "\245" }}{PARA 0 "" 0 "" {TEXT 204 0 "" }}{PARA 0 "" 0 "" {TEXT 204 111 "Compute a basis for the space K_phi_1(C), and choo se an appropriate vector to generate the T-cycle of length 2." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "CB1 := NullSpace((C+3*id6)^2 );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 182 "Pick a vector in CB1 not \+ lying in the eigenspace (if CCv1 computes to the zero vector, change t he subscript in the following command). This is the generator of a C-c ycle of length 2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "Cv1 := CB1[2];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 18 "Compute the cycle:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "CCv1 := C.Cv1;" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 204 149 "We now turn to the second factor phi_2 = t-2. First find the dot diagram in the usual fashion. We expe ct to find a four-dimensional subspace K_phi_2." }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 24 "E2:= NullSpace(C-2*id6);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 108 "This tells us that there are two cycles, and henc e two companion matrices, corresponding to the factor phi_2" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "NullSpace((C-2*id6)^2);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 204 159 "Here we have a dimension three s ubspace, still not enough to span the set K_phi_2. However, we are jus t one vector short, so we know the next step will suffice" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "CB2 := NullSpace((C-2*id6)^3);" }}} {EXCHG {PARA 203 "" 0 "" {TEXT 206 74 "As we found when computing the \+ Jordan form, the dot diagram for K_phi_2 is" }}{PARA 0 "" 0 "" {TEXT 204 0 "" }}{PARA 0 "" 0 "" {TEXT 204 15 "\245 \245" }}{PARA 0 "" 0 "" {TEXT 204 4 "\245" }}{PARA 0 "" 0 "" {TEXT 204 4 "\245" }}{PARA 0 "" 0 "" {TEXT 204 0 "" }}{PARA 0 "" 0 "" {TEXT 204 0 "" }}{PARA 0 "" 0 "" {TEXT 200 133 "Find a vector in this basis which will generate a cycle of length three, that is find one which is not in the null spac e of (A-2*I)^2" }}{PARA 0 "" 0 "" {TEXT 204 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 21 "((C-2*id6)^2).CB2[1];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 91 "(If necessary, replace the index 1 with another index until the vector computed is nonzero)" }}{PARA 0 "" 0 "" {TEXT 204 0 "" }}{PARA 0 "" 0 "" {TEXT 204 46 "Compute the cycle vectors generated from this:" }}{PARA 0 "" 0 "" {TEXT 204 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "Cv2 := CB2[1];\n" }{MPLTEXT 1 0 16 "CCv2 := C.Cv2 ;\n" }{MPLTEXT 1 0 19 "C2Cv2 := (C^2).Cv2;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 428 "We still have one vector to find, for the remaining cyc le of length 1. We should be able to find a vector in the basis for th e eigenspace (the nullspace of C-2I) which is not in the span of the c ycle we already have computed. Check by finding the rank of the matrix with columns the vectors from the known cycle and final column a vect or from the basis for the eigenspace. (Change the index of E2 until th e rank of testmat is 4)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 " testmat := ;\nR" }{MPLTEXT 1 0 13 "ank(testmat); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 83 "We have found a vector for o ur cycle of length one (change the index as necessary):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "Cv3 := E2[2];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 43 "We can now put together our basis beta for " } {XPPEDIT 2 0 "Typesetting:-mrow(Typesetting:-mi(\"\"), Typesetting:-mv erbatim(\"6#*$)I\"RG6\"\"\"'\"\"\"\"), Typesetting:-mo(\";\", form = \+ \"infix\", fence = \"false\", separator = \"true\", lspace = \"0em\", \+ rspace = \"thickmathspace\", stretchy = \"false\", symmetric = \"false \", maxsize = \"infinity\", minsize = \"1\", largeop = \"false\", mova blelimits = \"false\", accent = \"false\", font_style_name = \"2D Math \", size = \"12\", foreground = \"[0,0,0]\", background = \"[255,255,2 55]\"));" "-I%mrowG6#/I+modulenameG6\"I,TypesettingGI(_syslibGF'6%-I#m iGF$6#Q!F'-I*mverbatimGF$6#Q26#*$)I\"RG6\"\"\"'\"\"\"F'-I#moGF$63Q\";F '/%%formGQ&infixF'/%&fenceGQ&falseF'/%*separatorGQ%trueF'/%'lspaceGQ$0 emF'/%'rspaceGQ/thickmathspaceF'/%)stretchyGF " 0 "" {MPLTEXT 1 0 36 "QC := ;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 61 "This matri x then transforms C to its rational canonical form." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "QC^(-1).C.QC;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 84 "Check the companion matrices to see that they are correc t. The following are useful:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "expand((t+3)^2);\n" }{MPLTEXT 1 0 16 "expand((t-2)^3);" }}} {SECT 1 {PARA 211 "" 0 "" {TEXT 203 8 "Exercise" }}{PARA 0 "" 0 "" {TEXT 200 117 "1. Find a rational canonical basis for the matrix E, us ing an analysis similar to the one performed for the matrix C." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 246 "E := Matrix([[1, 1, -25, -1 , 0, 0, 0, 14], [3, -2, 28, -3, 1, 1, 3, -18], [0, 0, 1, 1, 0, 0, -1, \+ 0], [0, -3, 27, 1, 1, 0, 0, -18], [9, -9, 84, 0, 4, 3, 0, -54], [0, -3 , 75, 3, 0, 1, 1, -42], [0, 0, -2, 0, 0, 0, 1, 1], [0, 0, 0, 2, 0, 0, \+ -2, 1]]);" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 0 "" }}{PARA 0 "" 0 "" {TEXT 204 263 "So far all of the examples have had linear irrreducible factors - that is, t he characteristic polynomial splits. Next we tackle an example in whi ch the irreducible factors are not all linear. We have to modify our i nterpretation of the dot diagram just a little." }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 157 "F := Matrix([[0, 0, 4, -1, 0, 0], [1, 0, 26, \+ -2, 0, -2], [-2, 1, 12, -3, -2, 0], [-8, 4, 57, -14, -8, 0], [0, 0, 5, 1, 0, -1], [-17, 9, 117, -27, -17, -1]]);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 38 "factor(CharacteristicPolynomial(F,t));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 174 "Here we have one irreducible factor phi _1 = t^2 + t + 1. There are several possible dot diagrams with three d ots (the power of phi_1). We determine the dot diagram as before." }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "NF := NullSpace(F^2 + F + id 6);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 445 "Hence we already have fo ur vectors in a basis for the nullspace of phi_1(F). *Note* that this \+ corresponds to *two* vectors in the first row of the dot diagram, sinc e phi_i is a *quadratic* factor, each dot contributes *two* to the dim ension of the appropriate space. We see later how to deal with this si tuation. We go on to compute the next nullspace, but already knowing w hat the dot diagram must look like, we expect a basis with six vectors ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "FB1 := NullSpace(((F^2 + F + id6)^2));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 73 "Note that we have just the standard basis here. The dot diagram is thus:" }} {PARA 0 "" 0 "" {TEXT 204 0 "" }}{PARA 0 "" 0 "" {TEXT 204 13 "\245 \+ \245" }}{PARA 0 "" 0 "" {TEXT 204 4 "\245" }}{PARA 0 "" 0 "" {TEXT 204 0 "" }}{PARA 0 "" 0 "" {TEXT 204 112 "This dot diagram tells us th at we have two companion matrices, one corresponding to (phi_1)^2, and one to phi_1." }}{PARA 0 "" 0 "" {TEXT 204 0 "" }}{PARA 0 "" 0 "" {TEXT 204 228 "We continue as before, but realizing that each dot corr esponds to *two* vectors now, so we are looking for a vector which gen erates an F-cycle of length 4. First, choose a vector from FB1 which \+ is not in the nullspace of phi_1." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "v1 := FB1[1];\n" }{MPLTEXT 1 0 19 "(F^2 + F + id6).v1 ;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 134 "OK, we have a good choice \+ for v1 (if not, choose a different vector in FB1). Now compute the cyc le of length four which accompanies v1" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "Fv1 := F.v1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "F2v1 := F.Fv1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "F3v 1 := F.F2v1;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 281 "These are obvio usly linearly independent vectors. Next we must find the second cycle. We must find a vector in the null space of phi_1 whose F-cycle is ind ependent of the cycle we just computed. Try various choices until we s ucceed. (We want the matrix Ftestmat to have full rank.)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "v2 := NF[1];\n" }{MPLTEXT 1 0 14 "F v2 := F.v2;\n" }{MPLTEXT 1 0 41 "Ftestmat := ;\nR" }{MPLTEXT 1 0 14 "ank(Ftestmat);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 106 "As soon as we have succeeded, we have found our rationa l canonical basis. Test the change of basis matrix:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "FQ := ;\n" } {MPLTEXT 1 0 13 "FQ^(-1).F.FQ;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 204 83 "Check that the companion matrices are as desired, those of (phi_1) ^2, and of phi_1." }}}{SECT 1 {PARA 212 "" 0 "" {TEXT 203 8 "Exercise" }}{PARA 0 "" 0 "" {TEXT 200 118 "2. Find a rational canonical basis \+ for the matrix G, using an analysis similar to the one performed for t he matrix F." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 217 "G := Matrix ([[12, -6, -3, 4, -20, 6, 0], [1, 0, 0, 0, -4, 0, 0], [0, 1, 3, 0, 0, \+ 0, 0], [-54, 27, 36, -18, 112, -27, -6], [3, -3/2, 0, 1, -5, 3/2, 0], \+ [25, -12, -18, 8, -58, 12, 4], [9, -11/2, 0, 7/2, -17, 11/2, 1]]);" } {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}} {EXCHG {PARA 209 "" 0 "" {TEXT 207 0 "" }}}} {MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 15 10 1804 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }