Existence of two subspaces in vector space $V$

Question

Let $V$ and $W$ be finite dimensional complex vector spaces and let $f: V\xrightarrow[]{}W$ and $g: W\xrightarrow[]{}V$ be linear maps. Suppose that $f(g(w))=w$ holds for any $w\in W$. Show that there exist subspaces $V_0, V_1$ of $V$ that satisfy all of the following three conditions.

1. $V= V_0 \oplus V_1$ that is $V$ is the direct sum of $V_0$ and $V_1$
2. For any $v\in V_0$, $g(f(v))=0$
3. For any $v\in V_1$, $g(f(v))=v$

I tried to think so much to prove the existence but could not get a good idea, here we need to prove the existence of these two subspaces $V_0$ and $V_1$. Does that mean we need to check the three conditions(not empty, closure under addition and scalar multiplication) only? How do we use the given conditions to prove that.

I know from 1 - $V_0 \cap V_1 = \{0\}$

How do we contribute all the things? I would appreciate if anyone could show me the right path

Answer 1

Let $V_1=\ker f$ and let $V_0=g(W)$.

1. If $v\in V_1\cap V_0$, then $v=g(w)$ for some $w\in W$ and $f(v)=0$. But$$0=f(v)=f\bigl(g(w)\bigr)=w$$and therefore $v=g(w)=0$. On the other hand, if $v\in V$, then $v=v-g\bigl(f(v)\bigr)+g\bigl(f(v)\bigr)$. Clearly, $g\bigl(f(v)\bigr)\in V_1$. Furthrmore, $f\left(v-g\bigl(f(v)\bigr)\right)=f(v)-f(v)=0$, and therefore $V=V_0\bigoplus V_1$.
2. If $v\in V_0$, $f(v)=0$ and therefore $g\bigl(f(v)\bigr)=0$.
3. If $v\in V_1$, $v=g(w)$ for some $w\in W$ and therefore$$g\left(f\bigl(v)\right)=g\left(f\bigl(g(w)\bigr)\right)=g(w)=v.$$

Answer 2

Hint

$g$ is bijective $W\to g(W)$ with inverse function $f|_{g(W)}:g(W)\to W$.

Answer 3

We have that $fg= id_W$. Now show that $(gf)^2=gf.$ Hence $gf:V \to V$ is a projection.

Show that $V_0:=ker(gf)$ and $V_1:= im(gf)$ have the desired properties.

Answer 4

Just define $V_0,V_1$ as in 2 and 3, i.e. \begin{align*} V_0&=\ker g\circ f = \{v\in V\mid g(f(v))=0\}\\ V_1&=\ker (g\circ f-id_V)=\{v\in V\mid g(f(v))=v\} \end{align*} Now check that these subspaces satisfy $V=V_0\oplus V_1$. Hint: $$v=v-g(f(v))+g(f(v))$$