 Solution

ID: past-exam-of-the-algebra-1-and-differential-equations-1-course-of-the-university-of-oxford/2015/2/b/iii/solution

For

v \in V_{λ}

, use commutation to obtain

T (S v) = S (T v) = S (λ v) = λ S v .

Thus

S v \in V_{λ}

, proving that every $T$ -eigenspace is an $S$ -invariant subspace. This does not assert that an arbitrary eigenvector of

T

is already an eigenvector of

S

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