线性代数证明题设x1,x2,x3,x4线性无关,且x1=y1-y2-y3-y4,x2=y1+y2-y3-y4,x3=-y1-y2+y3-y4,x4=-y1-y2-y3+y4.证明：y1,y2,y3,y4线性无关

Question

线性代数证明题
设x1,x2,x3,x4线性无关,且x1=y1-y2-y3-y4,x2=y1+y2-y3-y4,x3=-y1-y2+y3-y4,x4=-y1-y2-y3+y4.证明：y1,y2,y3,y4线性无关

Answer 1

证明:由已知得
(x1,x2,x3,x4)=(y1,y2,y3,y4)P
其中矩阵P =
1 1 -1 -1
-1 1 -1 -1
-1 -1 1 -1
-1 -1 -1 1
因为行列式 |P|=-8≠0,所以P可逆.
即有 (x1,x2,x3,x4)P^-1=(y1,y2,y3,y4)
即y1,y2,y3,y4可由x1,x2,x3,x4线性表示.
故向量组x1,x2,x3,x4与y1,y2,y3,y4等价.
所以 r(y1,y2,y3,y4)
= r(x1,x2,x3,x4) [等价的向量组秩相同]
= 4 [已知x1,x2,x3,x4线性无关]
所以 y1,y2,y3,y4 线性无关.