Question

线性代数 求一个四维单位向量,使之与(1,1,-1,1)，(1,-1,-1,1),(2,1,1

3)都正交。

Answer 1

设所求向量是 (x1, x2, x3, x4)^T, 则
x1+x2-x3+x4 = 0
x1-x2-x3+x4 = 0
2x1+x2+x3+3x4 = 0
系数矩阵 A =
[1 1 -1 1]
[1 -1 -1 1]
[2 1 1 3]
行初等变换为
[1 1 -1 1]
[0 -2 0 0]
[0 -1 3 1]
行初等变换为
[1 0 -1 1]
[0 1 0 0]
[0 0 3 1]
r(A) = 3, 方程组同解变形为
x1-x3 = -x4
x2 = 0
3x3 = -x4
取 x4 = -3， 得基础解系 (4, 0, 1, -3)^T
该向量与已知 3 个向量都正交，
单位化，得 (4/√26, 0, 1/√26, -3/√26)^T