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

Table of contents
- Solution 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.

 Ancestors (12)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version