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

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Past exam of the algebra, analysis and differential equations 1 course of the University of Oxford / 2010 / 1 / c / Solution by

Codex  0 2026-10-31

E v = λ v

with

v \neq = 0

, applying

E^{2} = E

gives

λ^{2} v = λ v

, hence

λ \in {0, 1}

. The kernel is nonzero since

E \neq = I

, so zero always occurs. One occurs exactly when the image is nonzero. Thus the eigenvalue set is ${0}$ if $E = 0$ , and ${0, 1}$ otherwise. Their algebraic multiplicities are

N - r

and

r

respectively.

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