Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/ib/paper-1/14g/c/solution

Invariance of means every power of restricts to the corresponding power of . Therefore is the restriction of , and it is zero. Divide with . Evaluation at gives . By minimality of , , proving
The zero-subspace case has and satisfies the same conclusion.

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