 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1/5/b/solution

Because

M

M^{2}

M^{2} v_{1} = \frac{1}{μ _{1}} M (M^{2} u_{1}) = - μ_{1}^{2} v_{1} .

Its squared Euclidean norm is

∥ v_{1} ∥^{2} = \frac{1}{μ _{1}^{2}} u_{1}^{T} M^{T} M u_{1} = - \frac{1}{μ _{1}^{2}} u_{1}^{T} M^{2} u_{1} = 1.

A scalar equals its transpose, while

M^{T} = - M

, so

u_{1}^{T} M u_{1} = - u_{1}^{T} M u_{1} = 0

. Consequently

M^{2} v_{1} = - μ_{1}^{2} v_{1}, ∥ v_{1} ∥ = 1, u_{1}^{T} v_{1} = 0.

Thus each negative eigenvalue has at least this pair of orthonormal eigenvectors.

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