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

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 $a,b$,

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.

Write $K_j=\ker T^j$, where $T^0=I$ and $K_0=\{0\}$. These kernels satisfy $K_j\subseteq K_{j+1}$. Also $T(K_{i+1})\subseteq K_i$ and $T(K_i)\subseteq K_{i-1}$: for example, $T^i(Tv)=T^{i+1}v=0$ when $v\in K_{i+1}$. Thus the proposed linear map between quotient vector spaces is well-defined and has the formulaIf $v+K_i$ belongs to its kernel, then $Tv\in K_{i-1}$. Therefore $T^{i}v=T^{i-1}(Tv)=0$, so $v\in K_i$ and $v+K_i$ is zero. We have provedThese are the injections between successive kernel quotients. In finite dimension, they imply $\dim K_{i+1}-\dim K_i\le \dim K_i-\dim K_{i-1}$.

Surjectivity need not hold. Take a nonzero vector space $V$ and the zero linear map $T=0$. At $i=1$, we have $K_2=K_1=V$ and $K_0=0$, so