Stable Subspaces of Linear Transformations (11.2.9)

Dummit and Foote Abstract Algebra, section 11.2, exercise 9:

Let $φ∈\text{End}(V)$, and let $W \subseteq V$ be a $\varphi$-stable subspace. Show that $\varphi$ induces linear tranformations $\varphi_{|W}$ and $\overline{\varphi}$ on the spaces $W$ and $V/W$. Show that if $\varphi_{|W}$ and $\overline{\varphi}$ are nonsingular then $\varphi$ is nonsingular. Show the converse holds when $V$ is finite dimensional, but not necessarily when $V$ is infinite dimensional.

Proof: $\varphi_{|W}$ is clearly a linear transformation by the stability of $W$, and define $\overline{\varphi}(\overline{v})=\overline{\varphi(v)}$. For well definedness, suppose $\overline{v_1}=\overline{v_2}$; then $v_1-v_2 \in W$ so $v_1=v_2+w$ and $\overline{\varphi(v_1)}=\overline{\varphi(v_2+w)}=\overline{\varphi(v_2)}$.

Assume these two are nonsingular, and now assume $φ(v)=0$. If $v∈W$, then since $φ_{|W}$ is nonsingular we have $v=0$. If $v∉W$, then $\overline{v}≠0$ and since $\overline{φ}$ is nonsingular we have $φ(v)∉W$ so a fortiori $φ(v)≠0$, a contradiction.

For the converse, since $φ$ is nonsingular we naturally have $φ_{|W}$ is nonsingular. Assume $V$ is finite dimensional to show $\overline{φ}$ is nonsingular; now that we may assume $W$ is finite dimensional, assume $φ(v)∈W$. Then since $φ_{|W}$ is nonsingular and thus surjective, we may obtain $w∈W$ such that $φ(w)=φ(v)$, implying $v=w$ so that in particular $v∈W$ and $\overline{v}=0$.

Now assume $V$ is the direct sum of the countably infinite number of copies of $\mathbb{R}$. Letting $φ$ be the right shift operator with $W$ the subspace consisting of vectors whose first coordinate is zero, we see $φ$ is nonsingular yet $\overline{φ}$ is manifestly not nonsingular by $\overline{φ}(\overline{e_1})=0$.