证明设k1β1+k2β2+···+krβr=0,则(k1+k2+···+kr)α1+(k2+···+kr)α2+···+krαr,=0.由于α1,α2···,αr线性无关,所以从而k 1 =k 2 =···=k r =0,因此,β1,β2,···,βr线性无关.