 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-i/4/a/iii/solution

The definition of an eigenvalue and the kernel criterion for a singular square matrix give

λ is an eigenvalue ⟺ ker (T - λ I) \neq = {0} ⟺ λ I - A is not invertible ⟺ det (λ I - A) = 0 ⟺ χ_{T} (λ) = 0.

For the second equivalence, a linear map from a finite-dimensional vector space to itself has zero kernel exactly when its rank of a linear map is the full dimension, by the rank-nullity theorem, and therefore exactly when it is invertible. This proves both directions.

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