Need help to understand a theorem I have been reading a theorem related with the existence of the outer generalized inverse of a matrix where i have certain difficulties to understand the theorem.
Theorem is as follows.
Let $A\in\mathbb{C}^{m\times n}$, rank$(A) = r$, and let $T$ and $S$ be a subspace of $\mathbb{C^n}$ and $\mathbb{C^m}$, respectively, with $\dim (T )= \dim (S^\perp)=t\leq r$. 
Then, $A$    has a $\{2\}$ - inverse X such that $R(X) = T$ and $N(X) = S$ iff one of the following condition is satisfied (where $R(X)$ and $N(X)$ denots the range and null space of $X$, respectively)
$AT\oplus S$ = $C^{m}$
$P_{S}{^\perp} AT = S^{\perp}$
$A^*S^\perp\oplus T^\perp$ = $C^{n}$
$P_T~ A^*S^\perp = T$
$\{2\}$ - inverse  of a matrix $A$ is a $n\times m$ matrix $X$ satisfying matrix equation $XAX = X$. 
All above conditions are equivalent.
$P_{L.M}$ stands for the projection on to the space $L$ parallel to $M$ while $P_{L}$ stands for orthogonal projection onto sub space $L$ parallel to $L^\perp$.
Earlier i have posted same theorem where i was not clear about $AT$. Now that is cleared to me by answer given by David mitra .
I need a proper interpretation of these terms $P_{S}{^\perp} AT = S^{\perp}$, $P_T~ A^*S^\perp = T$, $A^*S^\perp$. it is given in the theorem that  $AT\oplus S$ = $C^{m}$ that means that there must exist projction operator $P_{AT}{S}$ .E projection onto subspace $AT$ parallel to $S$.
Also we have $dim (AT) = \dim S^\perp$.
I don't need proof.  It has some connection with direct sum of sub spaces and projection associated with that.
I just need their interpretation. I have to use this theorem for my own work. But how can i use if the things are not cleared to me? I really need help so that i can proceed further.
Heartily thanks for giving me your precious time.
 A: Let us put $U=S^{\perp}$. I number your four statements from (1) to (4) :
$ {\rm (1) \ } AT\oplus S = {\mathbb C}^{m}$
$ {\rm (2) \ } P_{U} AT = U$
${\rm (3) \ }  A^*U\oplus T^\perp = {\mathbb C}^{n}$
${\rm (4) \ } P_T~ A^*U = T$
Let us show that those statements are all equivalent. It will suffice to show
that $(1) \Leftrightarrow (2)$, $(3) \Leftrightarrow (4)$ and  $ (1)  \Leftrightarrow (3) $.
$(1) \Rightarrow (2)$ : Suppose that (2) is true. Since $P_U$ is a projection onto $U$, we already have $P_{U} AT \subseteq U$. Now (1) implies that those two subspaces of
${\mathbb C}^m$ have the same dimension, so they are equal and (2) follows.
$(2) \Rightarrow (1)$ : Suppose that (2) is true. Then 
$${\sf dim}(U)={\sf dim}(T) \geq {\sf dim}(AT) \geq {\sf dim}(P_UAT)={\sf dim} (U)$$
So all those dimensions are equal to ${\sf dim}(U)$. Let $c\in{\mathbb C}^m$. Then 
we can write $c=u+s$, with $u\in U,s\in S$. By (2) we have $u\in P_U(AT)$ also,
so there is a $t\in T$ such that $P_U(At)=u$. Then $At=u+s'$, for some $s'\in S$. 
So $c=At+s-s'$, and we have shown that ${\mathbb C}^m=AT+S$. This sum must in fact be direct because of the dimensions, and (1) follows.
We have thus shown that $(1) \Leftrightarrow (2)$. The proof of $(3) \Leftrightarrow (4)$ is similar, replacing $(A,T,U)$ with $(A^{*},U,T)$.
$(1) \Rightarrow (3)$ : Suppose that (1) is true. Let $v\in A^{*}S \cap T^{\perp}$. We have an $s\in S$ such that $v=A^{*}s$. Then $s\in S \cap (AT)=\lbrace 0 \rbrace$, so $v=0$. So $A^{*}S \cap T^{\perp}=\lbrace 0 \rbrace$, and those two subspaces of ${\mathbb C}^m$
must supplement each other because of the dimensions. So (3) follows.
$(3) \Rightarrow (1)$ : Suppose that (3) is true. If the subspace $AT+ S$ is not the whole of ${\mathbb C}^{m}$, there is a nonzero vector $w$ that's orthogonal to this subspace. Then $w$ is orthogonal to both $AT$ and $S$, and $A^{*}w \in T^{\perp} \cap (A^{*}S^{\perp})$. So $A^{*}w=0$. Since $w\in S^{\perp}$, we deduce ${\sf dim} (A^{*}S^{\perp})<{\sf dim} (S^{\perp})$. But then ${\sf dim} (A^{*}S^{\perp})+{\sf dim}(T^{\perp})<n$, and the sum in (3) cannot be direct.
This shows that $AT+ S={\mathbb C}^{m}$, and the sum must be direct because of the dimensions. QED
