Tags: optimization, linear algebra, quiz-03, eigenvalues, eigenvectors, lecture-04
Let \(A\) be a symmetric matrix with eigenvalues \(6\) and \(-9\).
True or False: The maximum value of \(\|A\vec{x}\|\) over all unit vectors \(\vec{x}\) is \(6\).
False. The maximum is \(9\).
The maximum of \(\|A\vec{x}\|\) over unit vectors is achieved when \(\vec{x}\) is an eigenvector corresponding to the eigenvalue with the largest absolute value.
Here, \(|-9| = 9 > 6 = |6|\), so the maximum is achieved at the eigenvector with eigenvalue \(-9\).
If \(\vec{u}\) is a unit eigenvector with eigenvalue \(-9\), then:
Tags: optimization, linear algebra, quiz-03, eigenvalues, eigenvectors, quadratic forms, lecture-04
Let \(A\) be a \(3 \times 3\) symmetric matrix with the following eigenvectors and corresponding eigenvalues: \(\vec{u}^{(1)} = \frac{1}{3}(1, 2, 2)^T\) has eigenvalue \(4\), \(\vec{u}^{(2)} = \frac{1}{3}(2, 1, -2)^T\) has eigenvalue \(1\), and \(\vec{u}^{(3)} = \frac{1}{3}(2, -2, 1)^T\) has eigenvalue \(-10\).
Consider the quadratic form \(\vec{x}\cdot A\vec{x}\).
What unit vector \(\vec{x}\) maximizes \(\vec{x}\cdot A\vec{x}\)?
\(\vec{u}^{(1)} = \frac{1}{3}(1, 2, 2)^T\) The quadratic form \(\vec{x}\cdot A\vec{x}\) is maximized by the eigenvector with the largest eigenvalue. Among \(4\), \(1\), and \(-10\), the largest is \(4\), so the maximizer is \(\vec{u}^{(1)}\).
What is the maximum value of \(\vec{x}\cdot A\vec{x}\) over all unit vectors?
\(4\) The maximum value equals the largest eigenvalue. We can verify:
What unit vector \(\vec{x}\) minimizes \(\vec{x}\cdot A\vec{x}\)?
\(\vec{u}^{(3)} = \frac{1}{3}(2, -2, 1)^T\) The quadratic form is minimized by the eigenvector with the smallest eigenvalue. Among \(4\), \(1\), and \(-10\), the smallest is \(-10\), so the minimizer is \(\vec{u}^{(3)}\).
What is the minimum value of \(\vec{x}\cdot A\vec{x}\) over all unit vectors?
\(-10\) The minimum value equals the smallest eigenvalue. We can verify: