Subspace as graph of unique linear map This question comes from the Grassmann Manifold example from Lee's Introduction to smooth manifold.
Let $\mathbb{R}^n$ be decomposed as a direct sum: $\mathbb{R}^n=V\oplus W$, with $\dim V=k$ and $\dim W=n-k$. Let $L(V,W)$ denotes the space of linear maps from $V$ to $W$. Define the graph of $T \in L(V,W)$ as $G(T)=\{x+Tx : x\in V\}$. It can be seen that for any $T$, $G(T)$ has dimension $k$ and $G(T)\cap W=\{0\}$. Now I want to prove the converse, that any $k$-dimensional subspace $S$ with $S\cap W=\{0\}$, can be seen as a graph of an unique linear map in $L(V,W)$. Any help would be appreciated.
The original question missed the condition with $S$ being $k$-dimensional, and hence ChoF's answer. Sorry about this.
 A: Let $S$ be a subspace of $V\oplus W$ with $\dim S=\dim V=k$ and $S\cap W=\{0\}$.


*

*Since $S\cap W=\{0\}$, the subspace $S+W\subseteq V\oplus W$ has dimension
$$
\dim(S+W)=\dim S+\dim W=\dim V+\dim W=\dim(V\oplus W)
$$
so that $S+W=V\oplus W$.

*For any $v\in V$, there is $s\in S$ and $w\in W$ such that $v=s_v+w_v$, since $S+W=V\oplus W$. Moreover, it is unique since $v=s_v+w_v=s'+w'$ implies
$$
s_v-s' = w'-w_v \in S\cap W=\{0\} \implies s_v=s' \text{ and } w_v=w' 
$$

*Define $T\in L(V,W)$ by $T(v)=-w_v$. Well-definedness of $T$ follows from the uniqueness of $w_v$. It is easy to check that $T$ is linear, since for $u,v\in V$ and $c\in\mathbb{R}$,
$$
\begin{align*}
& s_{u+v}+w_{u+v} = u+v = (s_u+w_u)+(s_v+w_v) \\
\implies\quad & s_{u+v}-(s_u+s_v) = (w_u+w_v)-w_{u+v} \in S\cap W=\{0\} \\
\implies\quad & s_{u+v}=s_u+s_v \quad\text{and}\quad w_{u+v}=w_u+w_v \\
& \\
& s_{cv}+w_{cv} = cv = c(s_v+w_v) = cs_v+cw_v \\
\implies\quad & s_{cv}-cs_v = cw_v-w_{cv} \in S\cap W=\{0\} \\
\implies\quad & s_{cv}=cs_v \quad\text{and}\quad w_{cv}=cw_v
\end{align*} \tag{$*$}
$$

*Note that $v+T(v)=v-w_v=s_v$ for $v\in V$. Thus the mapping $v\mapsto s_v=v+T(v)$ induces an isomorphism from $V$ to $S$.
($\because$) Linearity follows from ($*$) and injectivity follows from: If $s_u=s_v$ for some $u,v\in V$, then $u-v=(u-s_u)-(v-s_v)=w_u-w_v\in V\cap W=\{0\} \implies u=v$. Since $\dim V=\dim S$, the mapping is an isomorphism.

*Now we have $G(T)=\{v+T(v)=s_v\mid v\in V\}=S$.

*(Uniqueness of $T$) Assume that $G(T)=G(T')$ for some other $T'\in L(V,W)$. Then for $v\in V$, we have $v+T(v)=v'+T'(v')$ for some $v'\in V$. It follows that
$$
v-v' = T'(v')-T(v) \in V\cap W=\{0\} \implies v=v' \quad\text{and}\quad T(v)=T'(v')
$$
Therefore $T(v)=T'(v)$ for all $v\in V$, that is, $T=T'$ in $L(V,W)$.
Answer for the original question when there is no assumption on $\dim S$.

There are many linear transformations in $L(V,W)$ of which $S$ can be seen as a graph if $\dim S<k$.

Suppose $V=\mathbb{R}^2$ ($xy$-plane) and $W=\mathbb{R}$ ($z$-axis).
Let $S=\{(x,x)\in V \mid x\in\mathbb{R}\}$ be the line passing through $(1,1)$ in the $xy$-plane. Clearly, $S$ is a subspace of $\mathbb{R}^3=V\oplus W$ satisfying $S\cap W=\{0\}$.
Consider the two linear transformations $T_1(x,y)=0$ and $T_2(x,y)=x-y$ in $L(V,W)$. Since $T_1=T_2=0$ on $S$ we cannot have a unique linear transformation in $L(V,W)$ of which $S$ can be seen as a graph.
