 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-i/3/a/i/solution

T_{A} : x \mapsto A x

R^{n}

, the rank-nullity theorem implies that injectivity and surjectivity are equivalent. Consequently

T_{A}

is invertible if and only if

ker T_{A} = {0}

Using the permitted equivalence between invertibility and a nonzero determinant, we obtain

det A = 0 ⟺ T_{A} is not invertible ⟺ ker T_{A} \neq = {0} ⟺ \exists v \neq = 0 : A v = 0.

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