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No. Over any field $F$, take $V=F^2$, let $T=I$ be the identity operator, and choose the vector subspaceBoth $U$ 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, $L=\mathrm{span}\{e_2\}$ is another nonzero proper vector subspace of $V$, and $T(L)=L$. HenceThe hypotheses do not prevent a direct sum $V=U\oplus L$ whose two summands are invariant subspaces.

The quotient vector space $V/U$ identifies two vectors $v,w$ precisely when $v-w\in U$. Since $U$ is an invariant subspace, $T(U)\subseteq U$. ThusConsequently $T(v)+U=T(w)+U$: the output does not depend on the chosen representative. This proves that the induced linear map on a quotient vector spaceis well-defined. It is a linear map because, for scalars $a,b$,