(1) 两个子空间的和是直和只需要证明它们的交只有零向量.
设Y∈ker(A)∩im(A),则AY = 0且存在X使Y = AX.
∵A² = A,∴Y = AX = A²X = A(AX) = AY = 0.即ker(A)∩im(A) = {0},二者的和为直和.
(2) 充分性:对X∈ker(A),AX = 0.∴A(BX) = BAX = 0,BX∈ker(A).ker(A)是B的不变子空间.
而对Y∈im(A),存在X使Y = AX,∴BY = BAX = A(BX)∈im(A).im(A)也是B的不变子空间.
必要性:ker(A)的维数为n-r(A),im(A)的维数为r(A).已证二者的和是直和,于是V = ker(A)+im(A).
对X∈ker(A),有AX = 0,∴BAX = 0.∵ker(A)是B的不变子空间,∴BX∈ker(A),∴ABX = 0 = BAX.
而对Y∈im(A),存在X使Y = AX,∴AY = A²X = AX = Y,∴BAY = BY.
∵im(A)是B的不变子空间,∴存在Z使BY = AZ,∴ABY = A²Z = AZ = BY = BAY.
AB与BA在ker(A)和im(A)上的限制相等.又∵V = ker(A)+im(A),∴在V上有AB = BA.
