5 The Rational Canonical Form

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1 5 The Ratonal Canoncal Form Here p s a monc rreducble factor of the mnmum polynomal m T and s not necessarly of degree one Let F p denote the feld constructed earler n the course, consstng of all matrces of the form f(b), f F [x], where B = C(p), the companon matrx of p (We saw that f deg p = n, then F p = {a 0 I n + + a n 1 B n 1 a 0,, a n 1 F } Let f = f(b), where f F [x] Then ths new symbol has the followng propertes: () f + g = f + g; f g = fg; () f = 0 p f; () f = g p (f g); (v) f 1 exsts p does not dvde f Note: If p = x c, then F p = F THEOREM 51 N h, p becomes a vector space over F p f we defne fv = fv = f(t )(v) Frst we must verfy that the above defnton s well defned, that s, ndependent of the partcular polynomal f used to defne the feld element f So suppose f = g Then f = g + kp, k F [x] Hence fv = (g + kp)v = gv + k(pv) = gv + k0 = gv, as v Im p h 1 (T ) Ker p(t ) and consequently pv = 0 The four addton axoms hold as V s already a vector space over F ; The remanng vector space axoms then follow from the left F [x] module axoms: () (f + g) = (f + g)v = (f + g)v = fv + gv = fv + gv; () f(v + w) = f(v + w) = fv + fw = fv + fw; 105

2 () f(gv) = f(gv) = f(gv) = (fg)v = (fg)v; (v) 1v = 1v = v Remark: An F bass for N h, p wll be an F p spannng famly for N h, p, but wll not, n general, be an F p bass for N h, p The precse connecton between F ndependence and F p ndependence s gven by the followng theorem: THEOREM 52 Vectors v 1,, v r form an F p bass for N h, p f and only f the vectors form an F bass for N h, p COROLLARY 51 v 1, T (v 1 ),, T n 1 (v 1 ) v 2, T (v 2 ),, T n 1 (v 2 ) v r, T (v r ),, T n 1 (v r ) ν h, p = dm Fp N h, p = 1 deg p dm F N h, p = ν(ph (T )) ν(p h 1 (T )) deg p The exposton for p = x c now goes over to general p, wth small changes We agan have the decreasng sequence of dmensons: where ν 1, p = dm Fp Ker p(t ) = Also ν 1, p ν b, p 1, ν(p(t )) deg p ν 1, p + + ν b, p = ν(pb (T )) deg p, (24) where p b m T There s a correspondng dot dagram where the number of dots n the h th row from the bottom represents the nteger ν h, p We also have a smlar theorem to an earler one, n terms of the conjugate partton e 1 e γ 1 of the partton (24) above, where γ = ν 1, p = dm Fp Ker p(t ) = ν(p(t )) deg p 106

3 THEOREM 53 Vectors v 1,, v γ V can be found wth the property that p e1 1 v 1,, p eγ 1 v γ form an F p bass for Ker p(t ) Moreover () m T, vj = p e j ; () Ker p b (T ) = C T, v1 C T, vγ In concluson, f m T = p b 1 1 p bt t, we now have the drect sum decomposton t γ V = C T, vj, =1 j=1 where m T, vj = p e j and e 1 = b e γ form the conjugate partton for the dot dagram correspondng to p Here γ = ν(p (T )) deg p Takng T cyclc bases β j for C T, vj, then gves a bass for V Moreover β = t γ =1 j=1 γ =1 j=1 β j t [T ] β β = C(p e j ) The matrx on the rght s sad to be n ratonal canoncal form If nstead, we take the followng bass β j for C T, v j β j : v j, T (v j ),, T n 1 (v j ) p (T )(v j ), T p (T )(v j ),, T n 1 p (T )(v j ) p e j 1 (T )(v j ), T p e j 1 (T )(v j ),, T n 1 p e j 1 (T )(v j ), 107

4 (wth n = deg p ) whch reduces to the Jordan bass when p = x c, t s not dffcult to verfy that we get a correspondng matrx H(p e j ) called a hypercompanon matrx, whch reduces to the elementary Jordan matrx J ej (c ) when p = x c : H(p e j ) = C(p ) 0 0 N C(p ) 0 0 N 0 0 N C(p ) where there are e j blocks on the dagonal and N s a square matrx of same sze as C(p ) whch s everywhere zero, except n the top rght hand corner, where there s a 1 The overall effect s an unbroken subdagonal of 1 s We then get the correspondng ratonal canoncal form: [T ] β β = t γ H(p e j ) =1 j=1 Computatonal Remark: We can do our computatons completely over F, wthout gong nto F p, as follows Suppose v 1,, v r form an F spannng famly for N h, p Then we could, n prncple, perform the LRA over F p on ths spannng famly and fnd an F p bass v c1,, v cr A lttle thought reveals that f we had nstead appled the LRA algorthm over F to the expanded sequence:, v 1, T (v 1 ),, T n 1 (v 1 ); ; v r, T (v r ),, T n 1 (v r ), we would have obtaned the F bass for N h, p : v c1, T (v c1 ),, T n 1 (v c1 ); ; v cr, T (v cr ),, T n 1 (v cr ) from whch we select the desred F p bass v c1,, v cr Let A = M 6 6 (Z 3 ) Here m A = p 2, p = x 2 + x + 2 F [x], F = Z 3 108

5 p(a) = , ν(p(a)) = 4, ν 1, p = ν(p(a)) deg p = 2 p 2 (A) = 0, ν(p 2 (A)) = 6, ν 2, p = ν(p2 (A)) ν(p(a)) deg p Hence we have a correspondng F p dot dagram: N 2, p N 1, p = = 1 We have to fnd an F p bass p(a)v 11 for N 2, p and extend ths to an F p bass p(a)v 11, v 12 for N(p(A)) An F bass for N(p 2 (A)) s E 1,, E 6 Then N 2, p = p(a)e 1,, p(a)e 6 and the LRA gve p(a)e 2 as an F p bass for N 2, p so we can take v 11 = E 2 We fnd the columns of the followng matrx form an F bass for N(p(A)): We place p(a)e 2 n front and then pad the resultng matrx to get The frst four columns p(a)e 2, Ap(A)E 2, E 1, AE 1 of ths matrx form a LR F bass for N(p(A)) and hence p(a)e 2, E 1 form an F p bass for N(p(A)) So we can take v 12 = E 1 109

6 Then V 6 (Z 3 ) = N(p 2 (A)) = C TA, v 11 C TA, v 12 Then jonng hypercompanon bases for C TA, v 11 and C TA, v 12 : v 11, Av 11, p(a)v 11, Ap(A)v 11 and v 12, Av 11 gves a bass v 11, Av 11, p(a)v 11, Ap(A)v 11 ; v 12, Av 11 for V 6 (Z 3 ) Fnally f P s the non sngular matrx whose columns are these vectors, we transform A nto drect sum of hypercompanon matrces: P 1 AP = H(p 2 ) H(p) = Explctly, we have P = Unqueness of the Ratonal Canoncal Form Suppose that T : V V s a lnear transformaton over F and that β s a bass for V such that t γ [T ] β β = C(p e j ) (25) where =1 j=1 e 1 e γ 1 (26) and p 1,, p t are dstnct monc rreducble polynomals We show that the polynomals p and the sequences (26) are determned by the transformaton T Frst, t s not dffcult to show that β = t γ =1 j=1 110 β j,

7 where β j : v j, T (v j ),, T n j 1 (v j ) and n j = deg p e j and m T, vj = p e j Then we have the drect sum decomposton t γ V = C T, vj Also f we wrte b = e 1, we have and hence =1 j=1 γ Ker p b (T ) = V = t =1 j=1 C T, vj Ker p b (T ) Then from equaton (25) above, t follows that m T = lcm p e j = p b 1 1 p bt t, thereby determnng p 1,, p t up to order Then t can be shown that f 1 h b, then N h, p has F p bass p e 1 1 v 1,, p e j h 1 v jh, where e 1,, e jh are the ntegers not less than h There are consequently dm Fp N h, p = ν h, p such ntegers and hence the number of ntegers e 1,, e γ equal to h s equal to ν h, p ν h+1, p, whch depends only on T In other words, for each, the sequence e 1,, e γ depends only on T 52 Deductons from the Ratonal Canoncal Form THEOREM 54 where p a ch T, and p b m T ν(p b (T )) = a deg p 111

8 Note that ths determnes b we may evaluate ν(p h (T )) deg p for h = 1, 2, untl we get a value of a Then that h = b PROOF a bass for V such that A = [T ] β β = t γ C(p e j ) =1 j=1 So t γ ch T = =1 j=1 ch B,j where, for brevty, we wrte B,j = C(p e j ) Hence ch T = = = t γ t p =1 j=1 p e j =1 j=1 γ e j ν(p t b (T )) deg p p =1 as requred THEOREM 55 ch T = m T a bass β for V such that [T ] β β = C(pb 1 1 ) C(p bt t ) where p 1,, p t are dstnct monc rreducbles and b 1 b t 1 Note that f ch A = m A (e matrx A s non-derogatory PROOF T = T A n the above), we say that the 112

9 ch T = t =1 ch C(p b ) = t =1 p b, m T = lcm (p b 1 1,, p bt t ) = pb 1 1 p bt t = ch T Suppose that ch T = m T We deduce that the dot dagram for each p conssts of a sngle column of b dots, where p b m T ; that s, Observe that for t may be wrtten dm Fp N h,p = 1 for h = 1, 2,, b = j=1 ν(p h (T )) N, deg p h ν(p j (T )) ν(pj 1 (T )) deg p h dm Fp N j,p N j=1 Then, for each = 1, 2,, t we have the followng sequence of postve ntegers: 1 ν(p (T )) < ν(p2 (T )) deg p deg p But a = b here, as we are assumng that ch T t follows that < < ν(pb (T )) = a deg p ν(p h (T )) = h for h = 1, 2,, b deg p = m T In partcular, and h = 1 gves ν(p (T )) = 1 = γ deg p So the bottom row of the -th dot dagram has only one element; t looks lke ths: b 113

10 and we get the secondary decomposton Ker p b (T ) = C T,v 1 Further, f β = β 11 β t1, where β 1 s the T cyclc bass for C T,v1, then t γ [T ] β β = C(p e j ) as requred = =1 j=1 t =1 C(p b ) = C(p b 1 1 ) C(p bt t ) THEOREM 56 m T = p 1 p 2 p t, a product of dstnct monc rreducbles, f and only f a bass β for V such that [T ] β β = C(p 1) C(p 1 ) }{{} γ 1 tmes C(p t ) C(p t ) }{{} (27) γ t tmes Note: Ths s a generalzaton of an earler result, namely that a transformaton s dagonable f and only f ts mnmum polynomal splts nto a product of dstnct lnear factors PROOF Assume β such that (27) holds Then m T = lcm (p 1,, p }{{} 1,, p t,, p t ) }{{} γ 1 γ t = lcm (p 1,, p t ) = p 1 p 2 p t Assume m T = p 1 p t Then b = 1 for = 1,, t (e the -th dot dagram has heght 1) and β such that as e j = 1, j γ t [T ] β β = C(p ), =1 j=1 114

11 53 Elementary dvsors and nvarant factors 531 Elementary Dvsors DEFINITION 51 The polynomals p e j occurrng n the ratonal canoncal form of T are called the elementary dvsors of T Smlarly the elementary dvsors of a matrx A M n n (F ) are the polynomals p e j occurrng n the ratonal canoncal form of A THEOREM 57 Lnear transformatons T 1, T 2 : V V have the same elementary dvsors f and only f there exsts an somorphsm L : V V such that T 2 = L 1 T 1 L PROOF only f Suppose that T 1 and T 2 have the same elementary dvsors Then bases β, γ for V such that Then we have the equatons Hence so or φ β T 1 φ 1 β [T 1 ] β β = [T 2] γ γ = A φ β T 1 = T A φ β φ γ T 2 = T A φ γ = T A = φ γ T 2 φ 1 γ, φ 1 γ φ β T 1 φ 1 β φ γ = T 2, L 1 T 1 L = T 2, where L = φ 1 β φ γ s an somorphsm f Suppose that L 1 T 1 L = T 2 Then m T1 = m T2 = p b 1 1 p bt t, say; also for all and h, because p h (T 2 ) = p h (L 1 T 1 L) = L 1 p h (T 1 )L, 115

12 we have ν(p h (T 2 )) = ν(p h (T 1 )) Hence for each p, the correspondng dot dagrams for T 1 and T 2 are dentcal and consequently the elementary dvsors for T 1 and T 2 are dentcal COROLLARY 52 Let A, B M n n (F ) Then A s smlar to B f and only f A and B have the same elementary dvsors PROOF A s smlar to B P non sngular, wth P 1 AP = B P non sngular, wth T 1 P T AT P = T B L an somorphsm, wth L 1 T A L = T B 532 Invarant Factors THEOREM 58 Let T : V V be a lnear transformaton over F Then there exst non constant monc polynomals d 1,, d s F [x], such that () d k dvdes d k+1 for 1 k s 1; () vectors v 1,, v s V exst such that where m T, vk = d k V = s C T, vk, Remark: If β s the bass for V obtaned by strngng together the T cyclc bases for each C T, vk, we obtan the matrx drect sum k=1 s [T ] β β = C(d k ) k=1 Ths matrx s also sad to be n ratonal canoncal form PROOF 116

13 Let s = max (γ 1,, γ t ) and f 1 t and γ < j s, defne e j = 0 and v j = 0, the zero vector of V Now arrange the polynomals, 1 t; 1 j s as a t s rectangular array: p e j Let p e 1s 1 p e 11 1 p ets t p e t1 t d 1 = p e 1s 1 p ets t,, d s = p e 11 1 p e t1 t be the products along columns of the array, from left to rght d 1,, d s are monc non constant polynomals and Then d 1 d 2 d s Also C T, vj = {0} f v j = 0, so V s the drect sum of the followng ts T cyclc subspaces: C T, v1s C T, v11 C T, vts C T, vt1 Then by Problem Sheet 5, Queston 15(b), f we let v 1 = v 1s + v ts,, v s = v 11 + v t1, we have m T, v1 = d 1,, m T, vs = d s and C T, v1 = C T, v1s C T, vts C T, vs = C T, v11 C T, vt1 Consequently V = C T, v1 C T, vs DEFINITION 52 Polynomals d 1,, d s satsfyng the condtons of the above theorem are called nvarant factors of T There s a smlar defnton for matrces: f A M n n (F ) s smlar to a drect sum s C(d k ), k=1 117

14 where d 1,, d s are non constant monc polynomals n F [x] such that d k dvdes d k+1 for 1 k s 1, then d 1,, d s are called nvarant factors of A So the nvarant factors of A are the nvarant factors of T A THEOREM 59 The nvarant factors of a lnear transformaton T : V V are unquely defned by T PROOF Reverse the constructon n the proof of the above theorem usng Queston 15(a) of Problem Sheet 5, thereby recapturng the rectangular array of elementary dvsors, whch n turn s unquely determned by T EXAMPLE 51 Suppose T : V V has elementary dvsors Form the rectangular array p 2 1, p 3 1, p 3 1; p 2, p 2 2, p 2 2, p 4 2; p 3, p 3, p 4 3, p 5 3, p p 2 1 p 3 1 p p 2 p 2 2 p 2 2 p 4 2 p 3 p 3 p 4 3 p 5 3 p 5 3 Then the nvarant factors of T are obtaned by respectvely multplyng along columns: d 1 = p 3 d 2 = p 2 p 3 d 3 = p 2 1p 2 2p 4 3 d 4 = p 3 1p 2 2p 5 3 d 5 = p 3 1p 4 2p 5 3 THEOREM 510 If d 1,, d s are the nvarant factors of T : V V, then () m T = d s ; () ch T = d 1 d s 118

15 PROOF Suppose B = [T ] β β = s k=1 C(d k) s the canoncal form correspondng to the nvarant factors d 1,, d s of T Then Also m T = m B = lcm (m C(d1 ),, m C(ds)) = lcm (d 1,, d s ) = d s ch T = ch B = s ch C(dk ) = k=1 s d k We shall soon see that the nvarant factors of a lnear transformaton or matrx are of ndependent nterest For example the nvarant factors allow us to calculate the dmenson of the vector space Z L, M consstng of all lnear transformatons N : U V whch satsfy the equaton MN = NL, where L : U U and M : V V are gven lnear transformatons over F It turns out that there s a more drect way of fndng the nvarant factors of T To ntroduce ths algorthm, we need to dscuss an nterestng equvalence relaton on M m n (F [x]), whch n turn leads to the so-called Smth canoncal form of a matrx over F [x] k=1 119

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