Problems tagged with quadratic forms

Problems tagged with "quadratic forms"

Problem #48

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 \(\vec{x}\cdot A\vec{x}\).

Part 1)

What unit vector \(\vec{x}\) maximizes \(\vec{x}\cdot A\vec{x}\)?

Solution

\(\vec{u}^{(1)} = \frac{1}{3}(1, 2, 2)^T\) 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 \(\vec{u}^{(1)}\).

Part 2)

What is the maximum value of \(\vec{x}\cdot A\vec{x}\) over all unit vectors?

Solution

\(4\) The maximum value equals the largest eigenvalue. We can verify:

\[\vec{u}^{(1)}\cdot A\vec{u}^{(1)} = \vec{u}^{(1)}\cdot(4\vec{u}^{(1)}) = 4(\vec{u}^{(1)}\cdot\vec{u}^{(1)}) = 4 \cdot 1 = 4 \]

Part 3)

What unit vector \(\vec{x}\) minimizes \(\vec{x}\cdot A\vec{x}\)?

Solution

\(\vec{u}^{(3)} = \frac{1}{3}(2, -2, 1)^T\) 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 \(\vec{u}^{(3)}\).

Part 4)

What is the minimum value of \(\vec{x}\cdot A\vec{x}\) over all unit vectors?

Solution

\(-10\) The minimum value equals the smallest eigenvalue. We can verify:

\[\vec{u}^{(3)}\cdot A\vec{u}^{(3)} = \vec{u}^{(3)}\cdot(-10\vec{u}^{(3)}) = -10(\vec{u}^{(3)}\cdot\vec{u}^{(3)}) = -10 \cdot 1 = -10 \]