设α₁=(1,1,1,1)^T,α₂=(0,0,2,-6)^T,将α₁α₂标准正交化并补足为R^4的一组标准正交基.把之前的分数也补上
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![设α₁=(1,1,1,1)^T,α₂=(0,0,2,-6)^T,将α₁α₂标准正交化并补足为R^4的一组标准正交基.把之前的分数也补上](/uploads/image/z/12474717-69-7.jpg?t=%E8%AE%BE%CE%B1%26%238321%3B%3D%EF%BC%881%2C1%2C1%2C1%EF%BC%89%5ET%2C%CE%B1%26%238322%3B%3D%EF%BC%880%2C0%2C2%2C-6%EF%BC%89%5ET%2C%E5%B0%86%CE%B1%26%238321%3B%CE%B1%26%238322%3B%E6%A0%87%E5%87%86%E6%AD%A3%E4%BA%A4%E5%8C%96%E5%B9%B6%E8%A1%A5%E8%B6%B3%E4%B8%BAR%5E4%E7%9A%84%E4%B8%80%E7%BB%84%E6%A0%87%E5%87%86%E6%AD%A3%E4%BA%A4%E5%9F%BA.%E6%8A%8A%E4%B9%8B%E5%89%8D%E7%9A%84%E5%88%86%E6%95%B0%E4%B9%9F%E8%A1%A5%E4%B8%8A)
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设α₁=(1,1,1,1)^T,α₂=(0,0,2,-6)^T,将α₁α₂标准正交化并补足为R^4的一组标准正交基.把之前的分数也补上
设α₁=(1,1,1,1)^T,α₂=(0,0,2,-6)^T,将α₁α₂标准正交化并补足为R^4的一组标准正交基.
把之前的分数也补上
设α₁=(1,1,1,1)^T,α₂=(0,0,2,-6)^T,将α₁α₂标准正交化并补足为R^4的一组标准正交基.把之前的分数也补上
β1 = α1 = (1,1,1,1)'
β2 = α2 -[(α2,β1)/(β1,β1)]β1 = (1,1,3,-5)'
设x = (x1,x2,x3,x4) 与 β1,β2 正交.
则 x1+x2+x3+x4 = 0
x1+x2+3x3-5x4 = 0
解得基础解系:α3=(1,-1,0,0)',α4=(4,0,-3,-1)
将α3,α4正交化得:
β3 = (1,-1,0,0)',β4 = (2,2,-3,-1)'
将β1,β2,β3,β4单位化:
γ1 = β1/||β1|| = (1/2,1/2,1/2,1/2)'
γ2 = β2/||β2|| = (1/6,1/6,3/6,-5/6)'
γ3 = β3/||β3|| = (1/√2,-1/√2,0,0)'
γ4 = β4/||β4|| = (2/3√2,2/3√2,-3/3√2,-1/3√2)'
γ1,γ2,γ3,γ4 即为R^4的一组标准正交基.