Prove that eigenvectors of a symmetric matrix corresponding to different eigenvalues are orthogonal?

Answer 1

Let ( A ) be a symmetric matrix, and let ( \mathbf{u} ) and ( \mathbf{v} ) be eigenvectors corresponding to distinct eigenvalues ( \lambda_1 ) and ( \lambda_2 ), respectively. By definition, we have ( A\mathbf{u} = \lambda_1 \mathbf{u} ) and ( A\mathbf{v} = \lambda_2 \mathbf{v} ). Taking the inner product of ( A\mathbf{u} ) with ( \mathbf{v} ), we get ( \langle A\mathbf{u}, \mathbf{v} \rangle = \lambda_1 \langle \mathbf{u}, \mathbf{v} \rangle ). Since ( A ) is symmetric, ( \langle A\mathbf{u}, \mathbf{v} \rangle = \langle \mathbf{u}, A\mathbf{v} \rangle = \lambda_2 \langle \mathbf{u}, \mathbf{v} \rangle ). Equating the two expressions yields ( \lambda_1 \langle \mathbf{u}, \mathbf{v} \rangle = \lambda_2 \langle \mathbf{u}, \mathbf{v} \rangle ). Since ( \lambda_1 \neq \lambda_2 ), this implies ( \langle \mathbf{u}, \mathbf{v} \rangle = 0 ), proving that the eigenvectors are orthogonal.

Answer 2

To prove that eigenvectors of a symmetric matrix corresponding to different eigenvalues are orthogonal, let ( A ) be a symmetric matrix, and let ( \mathbf{v_1} ) and ( \mathbf{v_2} ) be eigenvectors associated with distinct eigenvalues ( \lambda_1 ) and ( \lambda_2 ) respectively. We have ( A\mathbf{v_1} = \lambda_1 \mathbf{v_1} ) and ( A\mathbf{v_2} = \lambda_2 \mathbf{v_2} ). Taking the inner product of the first equation with ( \mathbf{v_2} ) gives ( \langle A\mathbf{v_1}, \mathbf{v_2} \rangle = \lambda_1 \langle \mathbf{v_1}, \mathbf{v_2} \rangle ), and using the symmetry of ( A ), we can also express this as ( \langle \mathbf{v_1}, A\mathbf{v_2} \rangle = \lambda_2 \langle \mathbf{v_1}, \mathbf{v_2} \rangle ). Equating both expressions leads to ( \lambda_1 \langle \mathbf{v_1}, \mathbf{v_2} \rangle = \lambda_2 \langle \mathbf{v_1}, \mathbf{v_2} \rangle ), and since ( \lambda_1 \neq \lambda_2 ), we conclude that ( \langle \mathbf{v_1}, \mathbf{v_2} \rangle = 0 ), proving that the eigenvectors are orthogonal.