Question

eigen vectors of symmetrical matrix OF TWO UNEQUAL EIGEN VALUES ARE ORTHOGONAL

Answer 1

$Let\ \lambda_1\ and\ \lambda_2\ be\ two\ unequal\ Eigen\ Values\ of\ matrix\ A.\\Let\ X_1\ and\ X_2\ be\ the\ corresponding\ Eigen\ Vectors.\\\\AX_1 =\lambda_1X_1\ Hence\ (A-\lambda_1I)X_1=O\\\\ \{ (A-\lambda_1I)X_1 \}^T = O^T\\\\ \{X_1^T (A-\lambda_1I)^T \}=O\\\\X_1^TA^T-\lambda_1X_1^T=O,\ But\ A^T=A\\\\(X_1^TA-\lambda_1X_1^T)X_2=O,\ Multiplying\ by\ X_2$

$X_1^T\ A\ X_2 - \lambda_1X_1^T\ X_2=O\\\\X_1^T\ \lambda_2\ X_2 - \lambda_1X_1^T\ X_2=O\\\\(\lambda_2-\lambda_1)(X_1^T\ X_2)=O\\\\Since\ \lambda1 \neq \lambda2,\ \ X_1^T\ X_2=O\\\\Or,\ X_2^T\ X_1=O\\Hence,\ X_1\ and\ X_2\ are\ Orthogonal$