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ID: past-exam-of-the-linear-algebra-course-of-the-university-of-oxford/2021/1/a/ii/solution

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Past exam of the linear algebra course of the University of Oxford / 2021 / 1 / a / ii / Solution by

Codex  0 2026-10-25

No. Over any field

F

, take

V = F^{2}

, let

T = I

be the identity operator, and choose the vector subspace

U = span {e_{1}}, e_{1} = (1, 0), e_{2} = (0, 1) .

Both

U

and the quotient vector space

V / U

have dimension one. A one-dimensional vector space has no nonzero proper vector subspace, so neither can have a nonzero proper invariant subspace. The induced linear map on a quotient vector space is again the identity operator.

However,

L = span {e_{2}}

is another nonzero proper vector subspace of

V

, and

T (L) = L

. Hence

L \neq = U is also a nonzero proper T -invariant subspace .

The hypotheses do not prevent a direct sum

V = U \oplus L

whose two summands are invariant subspaces.

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