Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 4 / 19I / i

Table of contents
- Solution i

Solution

A traceless skew-Hermitian matrix has the unique form

A = (i x - y + i z y + i z - i x), x, y, z \in R .

This real-linear parametrization is bijective, proving $V ≅ R^{3}$ as real vector spaces. It is not a complex vector space: multiplication by

i

generally turns a skew-Hermitian matrix into a Hermitian one.

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