From acef4f940d86d958eca5ec8ea5939b6f3895ff03 Mon Sep 17 00:00:00 2001 From: Alex Gregory Date: Mon, 19 Feb 2024 20:58:03 +0000 Subject: [PATCH] Show that symmetric matrix have orthog eigenvectors --- .../Contents/eigenvalues_and_eigenvectors.rst | 30 +++++++++++++++++++ 1 file changed, 30 insertions(+) diff --git a/notes/source/Contents/eigenvalues_and_eigenvectors.rst b/notes/source/Contents/eigenvalues_and_eigenvectors.rst index ba5e50c..8727667 100644 --- a/notes/source/Contents/eigenvalues_and_eigenvectors.rst +++ b/notes/source/Contents/eigenvalues_and_eigenvectors.rst @@ -172,3 +172,33 @@ Since the columns of :math:`A` are linearly independent, some column :math:`v_i` \end{bmatrix} Since the columns of :math:`A` are linearly independent, :math:`\mathbf{AB}_i = \mathbf{e}_i` has a unique solution. Thus we can find each column :math:`\mathbf{B}_i` thus :math:`\mathbf{A}` is invertible. + +.. topic:: Theorem + + A symmetric matrix :math:`\mathbf{A}` with distinct eigenvalues has orthogonal eigenvectors. + + **Proof** + + We only prove for the :math:`2 \times 2` case where :math:`\mathbf{A}` has two eigenvectors. + + .. math:: + + \mathbf{A} \mathbf{x}_1 & = \lambda_1 \mathbf{x}_1 \\ + \mathbf{x}_2^T \mathbf{A} \mathbf{x}_1 & = \lambda_1 \mathbf{x}_2^T \mathbf{x}_1 \\ + + And, + + .. math:: + + \mathbf{A x}_2 & = \lambda_2 \mathbf{x}_2 \\ + \mathbf{x}_1^T \mathbf{A x}_2 & = \lambda_2 \mathbf{x}_1^T \mathbf{x}_2 \\ + \mathbf{x}_2^T \mathbf{A x}_1 & = \lambda_2 \mathbf{x}_2^T \mathbf{x}_1 \\ + + Taking these away from each other, we have, + + .. math:: + + \mathbf{x}_2^T \mathbf{A x}_1 - \mathbf{x}_2^T \mathbf{A x}_1 & = (\lambda_1 - \lambda_2) \mathbf{x}_2^T \mathbf{x}_1 \\ + 0 & = (\lambda_1 - \lambda_2) \mathbf{x}_2^T \mathbf{x}_1 + + Since :math:`\lambda_1` and :math:`\lambda_2` are distinct, their difference is non-zero. Therefore, :math:`\mathbf{x}_2^T \mathbf{x}_1` must be :math:`0`.