 Solution

ID: past-exam-of-the-algebra-analysis-and-differential-equations-2-course-of-the-university-of-oxford/2011/algebra/1/a/solution

Choose a basis and let

A

represent

T

χ_{T} (t) = det (t I - A) .

S

, the new matrix is

S^{- 1} A S

. Hence

det (t I - S^{- 1} A S) = det (S^{- 1} (t I - A) S) = det (S)^{- 1} det (t I - A) det (S) = det (t I - A) .

Thus the polynomial is independent of the chosen coordinates. It is monic of degree

dim V

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