首先,當(dāng)n > 1,關(guān)于伴隨矩陣的秩,有如下結(jié)果:
若r(A) = n,則r(A*) = n;
若r(A) = n-1,則r(A*) = 1;
若r(A) < n-1,則r(A*) = 0.
證明:當(dāng)r(A) = n,有A可逆,|A| ≠ 0.
于是由A*A = |A|·E可得A* = |A|·A^(-1)也可逆.
當(dāng)r(A) = n-1,A有非零的n-1階子式,故A* ≠ 0,r(A*) ≥ 1.
又A*A = |A|·E = 0,故r(A*)+r(A) ≤ r(A*A)+n = n,即得r(A*) = 1.
當(dāng)r(A) < n-1,A的n-1階子式全為0,故A* = 0,r(A*) = n.
回到原題,由條件A* = A'得r(A*) = r(A') = r(A).
當(dāng)n > 2,根據(jù)前述結(jié)論,只有r(A) = n,故|A| ≠ 0.
對(duì)A*A = |A|·E取行列式得|A*|·|A| = |A|^n.
于是有|A|^2 = |A'|·|A| = |A*|·|A| = |A|^n,解得|A| = 1 (|A|為非零實(shí)數(shù)).
進(jìn)而得A'A = A*A = E,即A為正交矩陣.
n = 1,2時(shí)是有反例的,例如A = 2E.謝謝親。。。。