spectral theorem
If \(A\) is a linear transformation from \(V\) to \(V\) and \(A\) is a symmetric matrix, then there is an orthonormal basis consisting of the eigenvectors (see diagonalizable matrix).
1. orthogonal
- Any two eigenvectors with the same eigenvalues can be turned into orthogonal eigenvectors by Gram Schmidt.
- Let \(Ax=\lambda x\) and \(Ay = \lambda y\). And \(y = y_{\perp} + y_{||}\), where \(y_{||} = proj_x(y)\). Then, \(A(y) = A(y_{\perp} + y_{||}) = A(y_{\perp} + cx) = \lambda(y_{\perp} + cx) = \lambda y_{\perp} + \lambda cx\) and \(A(y_{\perp} + y_{||}) = A(y_{\perp}) + Acx = A(y_{\perp}) + c\lambda x\) so \(\lambda y_{\perp} = A(y_{\perp})\). So \(y_{\perp}\) and \(x\) are both eigenvectors.
- Any two eigenvectors \(x,y\) with different eigenvalues \(\lambda,\mu\) are orthogonal. Because \(x^TAy = x^T\mu y\) and \(x^TAy = \lambda x y\). So \((\mu - \lambda)x^Ty = 0\), but \((\mu - \lambda) \neq 0\), so we can divide through by it and get \(x^T y = 0\).