厦门大学高等代数教案网站IP地址:59.77.1.116;域名: gdjpkc. xmu. edu.cn §6.3极小多项式与 Cayley- Hamilton定理 教学目的与要求理解极小多项式的含义,了解极小多项式的存在,唯一性, 掌握极小多项式的性质,熟练掌握 Cayley- Hamilton定理. 零化多项式 定义1设A∈K×n,f(x)=a3x3+a3-1x3-1+…+a1x+a0∈K[,若成立 f(A)=aA+a-1A-1+…+a1A+a0I=0, 则称A适合多项式f(x)或称∫(x)为A的零化多项式 例1设A 01 则(A-I)2=0.令f(x)=x2-2x+1,则f(x)是 A的零化多项式 下面的结论说明矩阵总存在零化多项式 命题1对于A∈Kmn×n,存在∫(x)∈K[a,使得∫(A)=0. 证明由于 dimknxn=n2,故,A,A2,…,An2必线性相关,即存在不全为 零的a1,0≤i≤n2,使 an2A2+an2-1A-1+…+a1A+a0I=0 令f(x)=an2x2+an2-1xm2-1+…+a1x+ao,则有f(A)=0.口 注1矩阵的零化多项式不唯一.在上面的例子中,f(x)=x3-x2-x+1 也是A的零化多项式 在同构意义下,我们有 定义设φ是数域K上n维线性空间V的线性变换.设f(x)=a3x a4-1x3-1+…+a1x+ao∈K[r].若成立 f(9)=a92+a3-1y9-1+…+a1y+ aids=0 则称φ适合多项式∫(x),或称f(x)为φ的零化多项式
a, }G - IP 6 59.77.1.116; *e gdjpkc.xmu.edu.cn §6.3 =$w( Cayley-Hamilton R DPKBUQL RJ=$w 5 YJ=$w + .=$w 8{V. Cayley-Hamilton R [9$w CS 1 t A ∈ Kn×n , f(x) = asx s + as−1x s−1 + · · · + a1x + a0 ∈ K[x], p U f(A) = asA s + as−1A s−1 + · · · + a1A + a0I = 0, , A y7$w f(x) ; f(x) A [9$w G 1 t A = 1 1 0 1 , , (A − I) 2 = 0. ℄ f(x) = x 2 − 2x + 1, , f(x) x A [9$w I^~dL1>+[9$w JM 1 "& A ∈ Kn×n , + f(x) ∈ K[x], v f(A) = 0. VI #& dimKn×n = n 2 , 3 I, A, A2 , · · · , An 2 4?+ m [ ai , 0 ≤ i ≤ n 2 , v an2A n 2 + an2−1A n 2−1 + · · · + a1A + a0I = 0. ℄ f(x) = an2 x n 2 + an2−1x n 2−1 + · · · + a1x + a0, ,$ f(A) = 0. 2 W 1 L1 [9$w +r T=9 f(x) = x 3 − x 2 − x + 1 x A [9$w +2 Æb$ CS’ t ϕ x|) K r n PC V :t f(x) = asx s + as−1x s−1 + · · · + a1x + a0 ∈ K[x]. p U f(ϕ) = asϕ s + as−1ϕ s−1 + · · · + a1ϕ + a0idV = 0, , ϕ y7$w f(x), ; f(x) ϕ [9$w 1
注2设p是数域K上n维线性空间V的线性变换.51,52,……,5n是V的 一组基.φ在此基下的矩阵为A,即 g?(51,……,5n)=(51,……,5n)A. 设∫(x)∈K,则f(x)是A的零化多项式的充分必要条件是f(x)是的零化 多项式 命题设φ是数域K上n维线性空间V的线性变换,则存在f(x)∈K[x], 使得∫(y)=0 下面给出另一种证明方法 证明对任意0≠1∈V,y51,y251,…,p"s1∈V,则1;,y51,y251,…,p"s1 必线性相关,故有不全为零的a0,a1,…,a1n使a1ny"51+a1n-1yn-151+…+ a11951+a1051=(a1ny2+a1n-19n-1+……+a119+a101v)51=0. 令f1(x)=a1nxn+a1n-1xmn-1+…+a11x+a10,则0≤degf1(x)≤n,且 f1(y)1=0.设51,…,5n为V的一组基,对每个,有f(x)使f(yp)k1=0, 1≤i≤n,0≤degf(x)≤n,令f(x)=f(x)…fn(x),则0≤degf(x)≤n2,且 f(y)5:=0,1≤i≤n.所以f(y)=0.口 极小多项式 定义2若A∈Kx(或φ∈L(V)适合非零首一多项式m(x)且m(x)是 A(或q)所适合的多项式中次数最小者,则称m(x)是A(或φ)的一个极小多项 式.记作mA(x)(或m(x) 命题2(1)设A∈K1×,0≠f(x)∈K可,且f(A)=0,则mA(x)|f(x). (2)矩阵A的极小多项式唯一 (3)相似矩阵有相同的极小多项式 证明(1)由带余除法,f(x)=mA(x)q(x)+r(x)且degr(x)< degna(x) 将A代入上式,得∫(A)=mA(4)q(4)+r(4).因f(4)=0,mA(4)=0.故 r(4)=0.若r(x)≠0,则A适合一个次数低于mA(x)的多项式r(x),矛盾.故 r(x)=0,即mA(x)|f(x). (2)若mA(x),9(x)都是A的极小多项式,由(1)知mA(x)|g(x),且g(x) mA(x).故mA(x)=cg(x),0≠c∈K.又mA(x),g(x)均首1,从而c=1,即
W 2 t ϕ x|) K r n PC V : ξ1, ξ2, · · · , ξn x V ?< ϕ +< L1 A, ? ϕ(ξ1, · · · , ξn) = (ξ1, · · · , ξn)A. t f(x) ∈ K[x], , f(x) x A [9$w Æ+Dx f(x) x ϕ [9 $w JM’ t ϕ x|) K r n PC V :,+ f(x) ∈ K[x], v f(ϕ) = 0. .\ :3d)' VI "n 0 6= ξ1 ∈ V, ϕξ1, ϕ2 ξ1, · · · , ϕn ξ1 ∈ V , , ξ1, ϕξ1, ϕ2 ξ1, · · · , ϕn ξ1 43$ m [ a10, a11, · · · , a1n v a1nϕ n ξ1 + a1n−1ϕ n−1 ξ1 + · · · + a11ϕξ1 + a10ξ1 = (a1nϕ n + a1n−1ϕ n−1 + · · · + a11ϕ + a101V )ξ1 = 0. ℄ f1(x) = a1nx n + a1n−1x n−1 + · · · + a11x + a10, , 0 ≤ degf1(x) ≤ n, i f1(ϕ)ξ1 = 0. t ξ1, · · · , ξn V ?<"`- ξi , $ fi(x) v fi(ϕ)ξi = 0, 1 ≤ i ≤ n, 0 ≤ degfi(x) ≤ n, ℄ f(x) = f1(x)· · · fn(x), , 0 ≤ degf(x) ≤ n 2 , i f(ϕ)ξi = 0, 1 ≤ i ≤ n. f(ϕ) = 0. 2 &=$w CS 2 p A ∈ Kn×n (; ϕ ∈ L(V )) y7*[z $w m(x) i m(x) x A(; ϕ) y7 $w9|/, m(x) x A(; ϕ) -=$ wAA mA(x)(; mϕ(x)). JM 2 (1) t A ∈ Kn×n , 0 6= f(x) ∈ K[x], i f(A) = 0, , mA(x) | f(x). (2) L1 A =$w (3) L1$ =$w VI (1) #'' f(x) = mA(x)q(x) + r(x) i degr(x) <degmA(x). E A orw f(A) = mA(A)q(A) + r(A). ! f(A) = 0, mA(A) = 0. 3 r(A) = 0. p r(x) 6= 0, , A y7 -| & mA(x) $w r(x), _#3 r(x) = 0, ? mA(x) | f(x). (2) p mA(x), g(x) !x A =$w# (1) 4 mA(x) | g(x), i g(x) | mA(x). 3 mA(x) = cg(x), 0 6= c ∈ K. % mA(x), g(x) Nz 1, % c = 1, ? 2
(a)=g(a (3)i A=P-BP, mA(A)=0, mA(B)=mA(PAP-)=PmA(A)P-I 0.故mB(x)|mA(x).同理mA(x)|mB(x).因mA(x),mB(x)首1,故mA(x) 命题3设A Al 这里A1,A2为方阵,则mA(x)=[mA1(x),mA(x) 证明因0=mA(A) mA(41) 所以mA(A2)=0, 故mA(x)mA(x),1≤i≤2.另一方面,设g(x)∈K[x],mA(x)lg(x),1≤i≤ 则g(A)=0,1≤i≤2,故g(A) g(A1) g(A2) 0.从而mA(x)|9(x) 故有mA(x)={mA1(x),mA2(x)].口 弓理1设是A的特征值,mA(x)是A的极小多项式,则(x-)0)|mA(x) 证明由余数定理mA(x)=(x-0)q(x)+r,r∈K.将A代入上式, 0=mA(A)=(A-A01)q(A)+rIn则rln=(M0-A)q(A).两边同取行列式, 则r=|I-AlyA)=0.故r=0. 推论1设A1,A2,…,A。为A的互异特征值,mA(x)是A的极小多项式, 则(x-A1)(x-A2)…(x-As)|mA(x) 三. Cayley- Hamilton定理 引理2若A= 0入2 ,则fA(A)=0 证明设e1,e2,…,en是K上标准单位向量,则 Ae1= Arel Ae2=a12e1+入2 入 Aen=ane1+a2ne2+.+Ane, 令∫(x)=(x-A1)(x-A2)…(x-An).而(A-AD)(A-入,D=(A-入,D 3
mA(x) = g(x). (3) t A = P −1BP, mA(A) = 0, , mA(B) = mA(P AP −1 ) = PmA(A)P −1 = 0. 3 mB(x) | mA(x). R mA(x) | mB(x). ! mA(x), mB(x) z 1, 3 mA(x) = mB(x). 2 JM 3 t A = A1 A2 , 0S A1, A2 )1, mA(x) = [mA1 (x), mA2 (x)]. VI ! 0 = mA(A) = mA(A1) mA(A2) , mA(Ai) = 0, 1 ≤ i ≤ 2. 3 mAi (x) | mA(x), 1 ≤ i ≤ 2. \ ) t g(x) ∈ K[x], mAi (x) | g(x), 1 ≤ i ≤ 2. , g(Ai) = 0, 1 ≤ i ≤ 2, 3 g(A) = g(A1) g(A2) = 0. % mA(x) | g(x). 3$ mA(x) = [mA1 (x), mA2 (x)]. 2 TF 1 t λ0 x A 25mA(x) x A =$w, (x−λ0) | mA(x). VI #'| R mA(x) = (x − λ0)q(x) + r, r ∈ K. E A orw 0 = mA(A) = (A − λ0I)q(A) + rIn , rIn = (λ0I − A)q(A). WlZw , r n = |λ0I − A||q(A)| = 0. 3 r = 0. 2 NH 1 t λ1, λ2, · · · , λs A 8 25 mA(x) x A =$w , (x − λ1)(x − λ2)· · ·(x − λs) | mA(x). q Cayley-Hamilton R TF 2 p A = λ1 a12 · · · a1n 0 λ2 · · · a2n · · · · · · · · · · · · 0 0 · · · λn , , fA(A) = 0. VI t e1, e2, · · · , en x Kn r< X, Ae1 = λ1e1 Ae2 = a12e1 + λ2e2 · · · · · · Aei = a1ie1 + a2ie2 + · · · + λiei · · · · · · Aen = a1ne1 + a2ne2 + · · · + λnen ℄ f(x) = (x − λ1)(x − λ2)· · ·(x − λn). % (A − λiI)(A − λjI) = (A − λjI) 3
(A-入1D),,等f(A)e1=(A-A+1)…(A-AnD)(A-A1)…(A-AD)e;=0, 1≤i≤n.故f(A)=0,即fA(4)=0.口 Cayley- Hamilton间理对任意A∈Knxn,fA(A)=0. 练结对任意A∈Knxn∈C"×n,由§6.2关,必存在P∈Cnx使P1AP U,U∈K×n,U上三角阵由对理关f(U)=0,而fA(x)=f(x),故fA(4)= fG(4)=f(PUP-1)=Pf(U)P-1=0.0 推论2mA(x)|fA(x) 推论3在不计或数的情况下,fA(x)与mA(x)多相同的根 推论4设φ∈L(V),高f(y)=0 四.例极 教2设A是n阶非奇定阵,求故A-1=9(4),其化g(x)是的个n-1次多 存式 练结设fA(x)=x2+a1xn-1+…+an-1x+an,因|4≠0,故an=fA(0) (-1)|A≠0.又fA(4)=0,即An+a1A-1+…+an-1A+anln=0,,等 In=-a(A"+a1A-1+…+an-14)=A-a(4-1+a14n-2+…+an-1l)小从而 A-1=-4(4-1+a14-2+…+an-1D).令g(x) 即可 教3举例说明:构含相同的矩阵称必相,,极小多存式相同的矩阵称必相 解41=(01 与A2=(00)的构含相同,但不相, 0100 0100 0000 0000 B 0001 与B2 0000 的极小多存式都是x2,但 0000 0000 它们不相 教4设A,B为n阶矩阵,fA(x),mA(x)分别为A的构多存式和极小多存 式,mB(x)是B的极小多存式.若(mA(x),mB(x)=1,求故:(f4(x),mB(x)
(A − λiI), f(A)ei = (A − λi+1I)· · ·(A − λnI)(A − λ1I)· · ·(A − λiI)ei = 0, 1 ≤ i ≤ n. 3 f(A) = 0, ? fA(A) = 0.2 Cayley-Hamilton CF "n A ∈ Kn×n , fA(A) = 0. VI "n A ∈ Kn×n ⊆ C n×n , # §6.2 4+ P ∈ C n×n v P −1AP = U, U ∈ Kn×n ,U rqF1#"R4 fU (U) = 0, % fA(x) = fU (x), 3 fA(A) = fU (A) = fU (P UP −1 ) = P fU (U)P −1 = 0.2 NH 2 mA(x) | fA(x). NH 3 + ;| jQ fA(x) ( mA(x) $ / NH 4 t ϕ ∈ L(V ), , fϕ(ϕ)=0. T= G 2 t A x n H*h 1k3 A−1 = g(A), g9 g(x) x - n − 1 $ w VI t fA(x) = x n + a1x n−1 + · · ·+ an−1x + an. ! |A| 6= 0, 3 an = fA(0) = (−1)n |A| 6= 0. % fA(A) = 0, ? An + a1An−1 + · · · + an−1A + anIn = 0, In = − 1 an (An+a1An−1+· · ·+an−1A) = A·[− 1 an (An−1+a1An−2+· · ·+an−1I)]. % A−1 = − 1 an (An−1+a1An−2+· · ·+an−1I). ℄ g(x) = − 1 an x n−1− a1 an x n−2−· · ·− an−1 an ?O 2 G 3 MT~d25 L1 =$w L1 E A1 = 0 1 0 0 ( A2 = 0 0 0 0 25 B1 = 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 ( B2 = 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 =$w!x x 2 , b 2 G 4 t A, B n HL1fA(x), mA(x) + A 2$w6=$ wmB (x) x B =$wp (mA(x), mB(x)) = 1, k3(fA(x), mB(x)) = 1. 4
证明与证法.若(fA(x),mB(x)=d(x)≠1,则fA(x),mB(x)在C上至少 有一个公阵根a,即a既是fA(x)的根,也是mB(x)的根.由题设及推论3知, a必是A的一个特征值,同时也是B的一个特征值.相又(x-a)|fA(x),且 (x-a)|mB(x),进又(x-a)|mA(x),因此(mA(x),mB(x)≠1,矛由,口 例5设A,B为n论矩阵,fA(x),mA(x)在别为A的特征多项式和极小多 项式,mB(x)是B的极小多项式,若(mA(x),mB(x)=1.求证它fA(B)是非 异阵 证明由例4知存在u(x),v(x)使得fA(x)u(x)+mB(x)v(x)=1.解B代入 上式两同,因mB(B)=0,故有fA(B)u(B)=I,相又fA(B)可逆.口 作业P2462,3,7 为充题1设n论可逆阵A的极小多项式是mA(x)=xm+a1xm-1+…+am 求A-1及A*的极小多项式 为充题2证明它若n论矩阵A可对理化,则A的极小多项式是mA(x) ∏=1(x-入),其中,A,…,入是A的所有互异特征值 选做题P2468 5
VI (3'p (fA(x), mB(x)) = d(x) 6= 1, , fA(x), mB(x) + C r7s $ -01/ a, ? a Bx fA(x) /x mB(x) /#t>^ 3 4 a x A -25ux B -25% (x − a) | fA(x), i (x − a) | mB(x), K% (x − a) | mA(x), ! (mA(x), mB(x)) 6= 1, _# 2 G 5 t A, B n HL1 fA(x), mA(x) + A 2$w6=$ w mB(x) x B =$wp (mA(x), mB(x)) = 1. k3 fA(B) x* 1 VI #T 4 4+ u(x), v(x) v fA(x)u(x) + mB(x)v(x) = 1. E B o rwW! mB(B) = 0, 3$ fA(B)u(B) = I, % fA(B) Of 2 YR P246 2, 3, 7 Æ 1 t n HOf1 A =$wx mA(x) = x m +a1x m−1 +· · ·+am. k A−1 > A∗ =$w Æ 2 3dp n HL1 A O"F9, A =$wx mA(x) = Qs i=1(x − λi), g9 λ1, λ2, · · · , λs x A $8 25 OXM P246 8. 5