设x1a1+x2a2+y1b1+y2b2=0,证明x1=x2=y1=y2=0即可.
x1a1+x2a2=-y1b1-y2b2
因为a1,a2分别与b1,b2正交,所以x1a1+x2a2与b1,b2都正交,从而x1a1+x2a2与-y1b1-y2b2也正交,所以x1a1+x2a2=-y1b1-y2b2=0
因为a1,a2线性无关,所以由x1a1+x2a2=0得x1=x2=0.因为b1,b2线性无关,所以由-y1b1-y2b2=0得y1=y2=0
所以,a1,a2,b1,b2线性无关
设x1a1+x2a2+y1b1+y2b2=0,证明x1=x2=y1=y2=0即可.
x1a1+x2a2=-y1b1-y2b2
因为a1,a2分别与b1,b2正交,所以x1a1+x2a2与b1,b2都正交,从而x1a1+x2a2与-y1b1-y2b2也正交,所以x1a1+x2a2=-y1b1-y2b2=0
因为a1,a2线性无关,所以由x1a1+x2a2=0得x1=x2=0.因为b1,b2线性无关,所以由-y1b1-y2b2=0得y1=y2=0
所以,a1,a2,b1,b2线性无关