 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-1-course-of-the-university-of-oxford/2010/2/b/solution

tr (A B) = \sum_{i} \sum_{j} a_{ij} b_{ji} = \sum_{j} \sum_{i} b_{ji} a_{ij} = tr (B A) .

Hence $φ (A, B) = φ (B, A)$ . Combined with part (a), this makes

φ

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