Tags: linear algebra, quiz-03, spectral theorem, eigenvectors, lecture-04
Consider the matrix:
True or False: The spectral theorem guarantees that \(A\) has 2 orthogonal eigenvectors.
False.
The spectral theorem only applies to symmetric matrices. The matrix \(A\) is not symmetric because \(A^T \neq A\):
Since \(A\) is not symmetric, the spectral theorem does not apply, and we cannot use it to conclude anything about \(A\)'s eigenvectors.
Tags: linear algebra, quiz-03, spectral theorem, eigenvectors, lecture-04
Suppose \(A\) is a symmetric matrix, and \(\vec{u}^{(1)}\) and \(\vec{u}^{(2)}\) are both eigenvectors of \(A\).
True or False: \(\vec{u}^{(1)}\) and \(\vec{u}^{(2)}\) must be orthogonal.
False.
Consider the identity matrix \(I\). Every nonzero vector is an eigenvector of \(I\) with eigenvalue 1. For example, both \((1, 0)^T\) and \((1, 1)^T\) are eigenvectors of \(I\), but they are not orthogonal:
The spectral theorem says that you can find\(n\) orthogonal eigenvectors for an \(n \times n\) symmetric matrix, but it does not say that every pair of eigenvectors is orthogonal.
Aside: It can be shown that if \(\vec{u}^{(1)}\) and \(\vec{u}^{(2)}\) have different eigenvalues, then they must be orthogonal.
Tags: linear algebra, quiz-03, spectral theorem, eigenvectors, diagonalization, lecture-04
Suppose \(A\) is a \(d \times d\) symmetric matrix.
True or False: There exists an orthonormal basis in which \(A\) is diagonal.
True.
By the spectral theorem, every \(d \times d\) symmetric matrix has \(d\) mutually orthogonal eigenvectors. If we normalize these eigenvectors, they form an orthonormal basis.
In this eigenbasis, the matrix \(A\) is diagonal: the diagonal entries are the eigenvalues of \(A\).