n阶方阵A对任意n维向量x,满足x^TAx=0,充要条件为AT=-A;证明:充分性:f=x^TAx,显然有f=x^T(A^T)x,所以f= x^T(-A)x即有:x^T(-A)x= x^TAx所以 x^TAx=0必要性:x^TAx=0有x^T(A^T)x=0所以 x^T(A+ A^T)x=0
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![n阶方阵A对任意n维向量x,满足x^TAx=0,充要条件为AT=-A;证明:充分性:f=x^TAx,显然有f=x^T(A^T)x,所以f= x^T(-A)x即有:x^T(-A)x= x^TAx所以 x^TAx=0必要性:x^TAx=0有x^T(A^T)x=0所以 x^T(A+ A^T)x=0](/uploads/image/z/8753854-22-4.jpg?t=n%E9%98%B6%E6%96%B9%E9%98%B5A%E5%AF%B9%E4%BB%BB%E6%84%8Fn%E7%BB%B4%E5%90%91%E9%87%8Fx%2C%E6%BB%A1%E8%B6%B3x%5ETAx%3D0%2C%E5%85%85%E8%A6%81%E6%9D%A1%E4%BB%B6%E4%B8%BAAT%3D-A%EF%BC%9B%E8%AF%81%E6%98%8E%EF%BC%9A%E5%85%85%E5%88%86%E6%80%A7%EF%BC%9Af%3Dx%5ETAx%2C%E6%98%BE%E7%84%B6%E6%9C%89f%3Dx%5ET%EF%BC%88A%5ET%EF%BC%89x%2C%E6%89%80%E4%BB%A5f%3D+x%5ET%EF%BC%88-A%EF%BC%89x%E5%8D%B3%E6%9C%89%EF%BC%9Ax%5ET%EF%BC%88-A%EF%BC%89x%3D+x%5ETAx%E6%89%80%E4%BB%A5+x%5ETAx%3D0%E5%BF%85%E8%A6%81%E6%80%A7%EF%BC%9Ax%5ETAx%3D0%E6%9C%89x%5ET%EF%BC%88A%5ET%EF%BC%89x%3D0%E6%89%80%E4%BB%A5+x%5ET%EF%BC%88A%2B+A%5ET%EF%BC%89x%3D0)
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