Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2011/ii/paper-4/19i/i/solution

A traceless skew-Hermitian matrix has the unique form
This real-linear parametrization is bijective, proving as real vector spaces. It is not a complex vector space: multiplication by generally turns a skew-Hermitian matrix into a Hermitian one.

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