【证明】实对称矩阵特征向量正交

性质 1设 λ1,λ2\lambda_1,\lambda_2λ1​,λ2​ 是对称矩阵 A\boldsymbol{A}A 的两个特征值，p1,p2\boldsymbol{p}_1,\boldsymbol{p}_2p1​,p2​ 是对应的特征向量。若 λ1≠λ2\lambda_1 \ne \lambda_2λ1​=λ2​，则 p1\boldsymbol{p}_1p1​ 与 p2\boldsymbol{p}_2p2​ 正交。

证明因为 A\boldsymbol{A}A 为对称矩阵，所以有

λ1p1T=(λ1p1)T=(Ap1)T=p1TAT=p1TA(1)\lambda_1 \boldsymbol{p}_1^T = (\lambda_1 \boldsymbol{p}_1)^T = (\boldsymbol{A} \boldsymbol{p}_1)^T = \boldsymbol{p}_1^T \boldsymbol{A}^T = \boldsymbol{p}_1^T \boldsymbol{A} \tag{1} λ1​p1T​=(λ1​p1​)T=(Ap1​)T=p1T​AT=p1T​A(1)

于是将 (1)(1)(1) 式代入，有

λ1p1Tp2=(p1TA)p2=p1T(Ap2)=p1Tλ2p2=λ2p1Tp2\lambda_1 \boldsymbol{p}_1^T \boldsymbol{p}_2 = (\boldsymbol{p}_1^T \boldsymbol{A}) \boldsymbol{p}_2 = \boldsymbol{p}_1^T (\boldsymbol{A} \boldsymbol{p}_2) = \boldsymbol{p}_1^T \lambda_2 \boldsymbol{p}_2 = \lambda_2 \boldsymbol{p}_1^T \boldsymbol{p}_2 λ1​p1T​p2​=(p1T​A)p2​=p1T​(Ap2​)=p1T​λ2​p2​=λ2​p1T​p2​

即

(λ1−λ2)p1Tp2=0(\lambda_1 - \lambda_2) \boldsymbol{p}_1^T \boldsymbol{p}_2 = \boldsymbol{0} (λ1​−λ2​)p1T​p2​=0

因为 λ1≠λ2\lambda_1 \ne \lambda_2λ1​=λ2​，所以 p1Tp2=0\boldsymbol{p}_1^T \boldsymbol{p}_2 = \boldsymbol{0}p1T​p2​=0，即 p1\boldsymbol{p}_1p1​ 与 p2\boldsymbol{p}_2p2​ 正交。得证。