0 02E 即令B= (）a-(日-"）-(6)o-(--） /E O /E0】 0 则BBE-A0 必要性 0E+A -e+利(45)-*-对( 02E --20 所以E-A2=0即A2=E. 例7证明：A1=4-1,其中A是n×n矩阵(n≥2). 证由AA”=4E得AA|=AA|=IAE=4P·E=Am (1)当14≠0时.41==14-1： ②)当1A=0时，)A=O时，A=0,于是A1=1An-1:)r(A)>0时，AA°=1AE=0.因r(A)+ r(A)≤n,故r(4)<n,即A1=0,也有A1=4n- 例8设A=(a)s×n,B=()nxm,证明；r(AB)2r(A)+r(B)-n 运明法-设r利=n=a4国=r存在可证矩珠PQ度PAQ-(8）令gB- Bm,因为PQ可逆，所以 B(m-n)xm r=rAB=rP4QQB.而P4AQQB=（E0）(Bx）=（Bxm） 00 B 0 于是r(Bxm）=n但rQB=,这说明Bm-xm中线性无关的行数为2-n,而总行数为n-n, 故r2-r≤n-n,即r之n+2-n ((：)-(8） c-(5品）则g≥r+,又@=n+ra.所+ra2+@ 例9设A是一个n阶矩阵且秩r(A)=1. ()证明：A=aB,其中a,B为n维行向量，且当8a≠0时，A相似于对角阵， (②)若A的第一行和第一列的元素全为1，求A及A10, 证明()令A B2 ,其中，…，品为A的行向量组.因为r(A)=1,所以存在每个，不妨 1 设B=≠0，且对任意月，存在数a使得3=a8,2≤j≤n.于是A 令a=(1,a2,…,n 第3页(4) ø©5 E − A 0 0 E + A ! −−−−→ c2 + c1 E − A E − A 0 E + A ! −−−−→ r2 + r1 E − A E − A E − A 2E ! −−−−−−−−−−−−−−→ r1 − 1 2 (E − A) × r2 0 0 E − A 2E ! −−−−−−−−−−−−−−→ c1 − c2 × 1 2 (E − A) 0 0 0 2E ! =-P1 = E 0 E E ! , P2 = E − 1 2 (E − A) 0 E ! , Q1 = E E 0 E ! , Q2 = E 0 − 1 2 (E − A) E ! . KP2P1 E − A 0 0 E + A ! Q1Q2 = 0 0 0 2E ! , §±r(A + E) + r(A − E) = n. 7 á5 E − A 0 0 E + A ! −−−−→ r1 + r2 E − A E + A 0 E + A ! −−−−→ c2 + c1 E − A 2E 0 E + A ! −−−−−−−−−−−−−−→ r2 − 1 2 (E + A) × r1 E − A 2E − E−A 2 2 − A 0 ! −−−−−−−−−−−−−−→ c1 − c2 × 1 2 (E − A) 0 2E − E−A 2 2 0 ! . §±E − A2 = 0, =A2 = E. ~7 y²: |A∗ | = |A| n−1 , Ÿ•A¥n × n› (n ≥ 2). y dAA∗ = |A|E|A||A∗ | = |AA∗ | = ||A|E| = |A| n · |E| = |A| n. (1)|A| 6= 0û,|A∗ | = |A| n |A| = |A| n−1 ; (2)|A| = 0û, (i)A = Oû,A∗ = O, u¥|A∗ | = |A| n−1 ;(ii)r(A) > 0û,AA∗ = |A|E = O.œr(A) + r(A∗ ) ≤ n,r(A∗ ) < n,=|A∗ | = 0,èk|A∗ | = |A| n−1 ~8 A = (aij )s × n, B = (bij )n×m, y²; r(AB) ≥ r(A) + r(B) − n. y² {ò r(A) = r1, r(B) = r2, r(AB) = r,3å_› P, Q¶P AQ = Er1 0 0 0 ! , -Q−1B = Br1×m B(n−r1)×m ! , œèP, Qå_, §± r = r(AB) = r(P AQQ−1B), P AQQ−1B = Er1 0 0 0 ! Br1×m B(n−r1)×m ! = Br1×m 0 ! , u¥r(Br1×m) = r, r(Q−1B = r2, ˘`²B(n−r1)×m•Ç5Ã'1Íèr2 − r, o1Íèn − r1, r2 − r ≤ n − r1, =r ≥ r1 + r2 − n. { E 0 0 AB ! −−−−−−−−→ r2 + r1 × A E 0 A AB ! −−−−−−−−→ c2 − c1 × B E B A 0 ! . -C = E 0 0 AB ! . Kr(C) ≥ r(A) + r(B), qr(C) = n + r(AB), §±n + r(AB) ≥ r(A) + r(B). ~9 A¥òán› Öùr(A) = 1. (1) y²: A = α 0β,Ÿ•α, βènë1ï˛, Öβα0 6= 0û, AÉquÈ ; (2) eA1ò1⁄1òÉè1, ¶A9A100 . y² (1) -A =   β1 β2 . . . βn   , Ÿ•β1, · · · , βnèA1ï˛|. œèr(A) = 1, §±3záβi , ÿî β = β1 6= 0, ÖÈ?øβj , 3Íaj¶βj = ajβ, 2 ≤ j ≤ n. u¥A =   1 a2 . . . an   β, -α = (1, a2, · · · , an), 1 3 ê