线性代数 (一): 证明实对称矩阵特征向量正交

设矩阵 A A A有特征值 λ 1 \lambda_1 λ1及特征向量 u , λ 2 \bold u, \lambda_2 u,λ2及特征向量 v \bold v v

A u = λ 1 u A\bold u = \lambda_1 \bold u Au=λ1u

A v = λ 2 v A\bold v = \lambda_2\bold v Av=λ2v

v T ( A u ) = λ 2 v T u \bold v^T (A \bold u) = \lambda_2\bold v^T\bold u vT(Au)=λ2vTu

( v T A ) u = ( v T A T ) u = ( A v ) T u = λ 1 v T u (\bold v^T A)\bold u = (\bold v^T A^T)\bold u = (A\bold v)^T\bold u = \lambda_1\bold v^T \bold u (vTA)u=(vTAT)u=(Av)Tu=λ1vTu

所以

λ 1 v T u = λ 2 v T u \lambda_1\bold v^T\bold u = \lambda_2\bold v^T \bold u λ1vTu=λ2vTu

( λ 1 − λ 2 ) v T u = 0 (\lambda_1 - \lambda_2)\bold v^T \bold u = 0 (λ1λ2)vTu=0

因为实对称矩阵的特征值互异 (必可对角化),
所以

λ 1 = ̸ λ 2 \lambda_1 =\not \lambda_2 λ1≠λ2

所以

v T u = 0 \bold v^T \bold u = 0 vTu=0

得证.

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