V=(x1,x2,x3,x4)|x1+x3-2*x4=0,x1+3*x2-x3=0 是线性空间,求维数和一组基
问题描述：
V=(x1,x2,x3,x4)|x1+x3-2*x4=0,x1+3*x2-x3=0 是线性空间,求维数和一组基
答
考察2X4矩阵:1,0,1,-21,3,-1,0.其秩为2,故V的维数为:4-2 = 2..可取:x3,x4为*未知量,先取:x3=0,x4 =1,求得 x1=2,x2 =-2/3;再取:x3=1,x4=0.求得 x1=-1,x2 =2/3.即得其基:(2,-2/3,0,1),(-1,2/3,1,0).或:(6,-2,0 3) ,...