Let $T$ be linear operator, show there is a linear operator $U$ such that $UT(v) = v$, $TU(w)=w$ for $v \in R(T^*), w \in R(T)$ Let $V$ and $W$ be finite dimensional inner product spaces over the complex number field $\mathbb{C}$. Let $T:V\rightarrow W$ be a linear operator and $T^* : W \rightarrow V$ denote the adjoint operator of $T$ defined by $<T(v), w> = <v, T^*(w)>$ for $v\in V$ and $w\in W$.
I want to show that there is a linear operator $U: W \rightarrow V$ such that
\begin{align}
    UT(v) = v, \quad \forall v\in R(T^*), \quad TU(w) = w, \quad \forall w\in R(T).  
  \end{align}
where $N(T)$ denotes the null space (kernel) of $T$ and $R(T)$ denotes the range (image) of $T$.

My trial :
$v \in R(T^*)$, so $v= T^{*} w$ for some $w \in W$. And the first condition gives $UT(v) = UTT^*w=T^*w$, similar way the second condition gives $TUTv=Tv$ for some $v \in V$.
Basically I am tring to show $U:W \rightarrow V$ is a linear operator. so $w_1, w_2 \in W$, $U(w_1 + w_2) = U(w_1) + U(w_2)$, $U(cw_1) = c U(w_1)$.
I know if $T$ is a linear operator then $T^*$ is also a linear operator.
 A: You didn't show the existence of $U$, and that's the main problem.

Claim: $$V = R(T^*) \oplus_\perp \ker T$$ $$W = R(T) \oplus_\perp \ker T^*$$

Proof:
$\forall v \in \ker T$ and $\forall w \in W$
$$
\langle v \mid T^* w \rangle = \langle T v \mid w \rangle = 0,
$$
so $\ker T \perp R(T^*)$. Analogously, $\ker T^* \perp R(T)$.
Let $v \in \left(\ker T \oplus_\perp R(T^*)\right)^\perp$. Suppose $v \neq 0$. Then
$$
0 \neq \langle Tv \mid Tv \rangle = \langle T^*Tv \mid v \rangle = 0.
$$
Hence $\left(\ker T \oplus_\perp R(T^*)\right)^\perp = 0$, so $V = R(T^*) \oplus_\perp \ker T$. Analogously, $W = R(T) \oplus_\perp \ker T^*$.  Q.E.D.
Therefore, $S := T|_{R(T^*)}: R(T^*) \to R(T)$ is a linear bijection. Moreover, $T = S \oplus 0$. Take $U = S^{-1} \oplus 0 : R(T) \oplus_\perp \ker T^* =  W \to V$ (so it is linear since $S^{-1}$ is linear). Then it is obvious that $\forall v \in R(T^*)$ $UT v = v$ and $\forall w \in R(T)$ $TU w = w$.
For example, if $w \in R(T)$, then $$TU w = T\left(U|_{R(T)}\right) w = \left(T|_{R(T^*)}\right)\left(U|_{R(T)}\right) w = SS^{-1} w = w.$$
