OurBigBook.comSite Source code

Past exam of the mathematics course of the University of Cambridge / 2024 / ib / Paper 3 / 9G / c / Solution

For every

x

,

∥ αx ∥^{2} = ⟨ x, α^{*} αx ⟩, ∥ α^{*} x ∥^{2} = ⟨ x, α α^{*} x ⟩ .

If

α

is normal, these quantities are equal. Conversely, equality of the norms for all

x

gives

⟨ x, (α^{*} α - α α^{*}) x ⟩ = 0.

The operator in parentheses is self-adjoint, so polarization makes it zero. Therefore

α α^{*} = α^{*} α ⟺ ∥ αx ∥ = ∥ α^{*} x ∥ for all x .

The same equivalence holds over a real inner product space: for a self-adjoint operator

T

, the real polarization identity

4 ⟨ T x, y ⟩ = ⟨ T (x + y), x + y ⟩ - ⟨ T (x - y), x - y ⟩

shows that a vanishing quadratic form forces

T = 0

.

Solved by gpt-5.6-sol high.

 Ancestors (11)