# 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.

-
No one has answered yet? – srijan May 19 '12 at 6:36
What is it that you denote by $R(X)$ and $N(X)$ ? – Ewan Delanoy May 20 '12 at 6:57
@EwanDelanoy $R(X)$ and $N(X)$ denots the range and null space of $X$, respectively. – srijan May 20 '12 at 8:29
@EwanDelanoy I am not clear about certain things. I have mentioned them in mmy comments. I just need few clrifications. You deserve your bounty. Thanks for helping me out – srijan May 20 '12 at 18:13
I am sorry i have made few spelling mistakes(Mmy, clrifications) . While it should be my and clarifications. – srijan May 20 '12 at 18:40

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

-
I am very much greatful to you. I have some queries. How to prove this $P_{U} AT \subseteq U$ ? Can you explain please? – srijan May 20 '12 at 17:54
I m not clear about these lines too; Now (1) implies that those two subspaces of ${\mathbb C}^m$ have the same dimension, so they are equal . Explain me please. – srijan May 20 '12 at 18:08
I suppose here : By (2) we have $s\in P_U(AT)$ also: In place of s , u should be there? – srijan May 20 '12 at 18:29
@srijan : First question. $P_U$ is a projector onto $U$, so the image of this projector is $U$ : $P_U({\mathbb C}^m)=U$. Then $P_U(AT) \subseteq P_U({\mathbb C}^m)=U$. – Ewan Delanoy May 21 '12 at 7:19
@srijan : No, you're not right in your last comment. $P_U$ is an application so ${\sf dim}(P_U)$ is meaningless. What happens is that the kernel of $P_U$ is $U^{\perp}=S$, so $P_U$ is injective on $AT$ and hence ${\sf dim} P_U(AT)={\sf dim} (AT)$. – Ewan Delanoy May 21 '12 at 8:42