Linear Algebra Question of the Day

1. Maximize $\mathbf{x}^T H \mathbf{x}$ subject to $\|\mathbf{x}\| = 1$ and $\mathbf{x} \perp \mathbf{u}_1$. (correct answer)
2. Minimize $\mathbf{x}^T H \mathbf{x}$ subject to $\|\mathbf{x}\| = 1$ and $\mathbf{x} \perp \mathbf{u}_3$.
3. Maximize $\mathbf{x}^T H^{-1} \mathbf{x}$ subject to $\|\mathbf{x}\| = 1$ and $\mathbf{x} \perp \mathbf{u}_1$.
4. Minimize $\mathbf{x}^T H \mathbf{x}$ subject to $\|\mathbf{x}\| = 1$ and $\mathbf{x} \perp \text{span}\{\mathbf{u}_1, \mathbf{u}_3\}$.

Explanation: By the Spectral Theorem and variational characterization of eigenvalues, maximizing x^T H x subject to ||x|| = 1 gives x = u₁ (largest eigenvalue). When we add the constraint x ⊥ u₁, we maximize over the orthogonal complement, and the maximum is achieved at u₂ (second-largest eigenvalue). Choice B seeks a minimum with wrong constraint. Choice C involves H^(-1), which has eigenvalues 1/λᵢ in reverse order. Choice D over-constrains the problem to a one-dimensional space.