已知A和B都是n阶矩阵,且E-AB是可逆矩阵,证明E-BA可逆
推荐回答(5个)
你好!你说的对,α≠0不能得出Aα≠0,这个证法不对。下图是正确的做法,结论也更一般。经济数学团队帮你解答,请及时采纳。谢谢!
反证,若E-BA不可逆,则存在X不为0,使(E-BA)X=0(方和有非零解)->X=BAX
则(E-AB)AX=AX-ABAX=AX-AX=0
也即(E-AB)Y=0有非零解(其中Y=AX),与题专设矛盾,所以E-BA可逆,但属这种证法不能求其逆的具体表示。
例如:
假设E-BA不可逆,则(E-BA)X = 0 有非零解,则可得 X=BAX。
又 (E-AB)AX = AX - ABAX = AX-AX = 0,即AX为(E-AB)Y = 0的一个非零解,由此可证
因为E-AB可逆,则存在可逆阵C使得C(E-AB)=E,则C-CAB=E
左乘B右乘A,有BCA-BCABA=BA
有BCA=(E+BCA)BA推出(BCA+E)-E=(E+BCA)BA,整理有(BCA+E)(E-BA)=E,根所定义知E-BA可逆
扩展资料;
设σ是线性空间V的一个线性变换,称:
Ker(σ)= {α∈V|σ(α)=0}
为σ的核;称:
Im(σ) =σ(V) = {σ(α)|α∈V}
为σ的像(或值域),Ker(σ)与σ(V)都是V的子空间,且:
dim Ker(σ) + dimσ(V) =n.
证明:容易看出Ker(σ)是V的子空间。证明:σ(V)也是V的子空间。
参考资料来源:百度百科-可逆线性变换
简单计算一下即可,答案如图所示
Aα=0的话,BAα也就等于0,BAα=α=0,与α不等于0矛盾,所以Aα肯定不等于0
你Aα是0向量那你BAα能=α?α可是一个零向量啊,上面那个最佳回答也是逆天,想都不想就啊对对对,只会搬运答案
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