# Help in Proving a theorem

For the last few days I am trying to prove Result 2 which I have written below that uses the concepts of matrix decompostions to write matrix $A$ in the block form. I need help to prove this theorem. I would be very much thankful for any kind of sugestions and help .

Let $X$ and $Y$ denote arbitrary Banach spaces and $B(X, Y)$ be the set of all bounded linear operators from $X$ to $Y$. $T$ and $S$, respectively, be closed subspaces of $X$ and $Y$. Then the following statements are equivalent:

Result 1:(a) A has a $(2)$ inverse $B\in B( Y, X)$ such that $R(B) = T$ and $N(B) = S$.

(b) $T$ is a complemented subspace of $X$, $A|_T:T \rightarrow A(T)$ is invertible and $A(T)\oplus S = Y$.

In the case when $(a)$ or $(b)$ holds, $B$ is unique and is denoted by $A^2_{T,S}$.

Result 2: Suppose that the conditions of Result 1 are satisfied. If we take $T_1 = N(A^2_{T,S} A)$ then $X = T \oplus T_1$ holds and A has the following matrix form:

$A =\left( \begin{array}{cc} A_1 & 0\\ 0 & A_2 \\ \end{array} \right) :\left( \begin{array}{c} T \\ T_1 \\ \end{array} \right)$ $\rightarrow$ $\left( \begin{array}{c} A(T) \\ S \\ \end{array} \right)$

where $A_1$ is invertible.

Some of the definitions and results are as follows:

1: A matrix $X \in \mathbb{C}^{n\times m}$ is called a $\{2\}$-inverse of $A$ with the prescribed range $T$ and null space $S$, denoted by $A^{(2)}_{T,S}$, if the following conditions are satisfied: \begin{eqnarray*} XAX = X, ~~~ R(X) = T ~ and~~ N(X) = S. \end{eqnarray*} same definition can be extended to banach spaces.

2: $(A A^{(2)}_{T,S})^2 = A A^{(2)}_{T,S}$, $(A^{(2)}_{T,S}A)^2 = A^{(2)}_{T,S}A$

Here is the link of the research article where I found this resultt (Lemma 1.1 and Lemma 1.2). The iterative methods for computing the generalized inverse $A^{(2)}_{T,S}$ of the bounded linear operator between Banach spaces

Result 2. is a pure consequence of Result 1.: the matrix form means basically that $A|_T:T\to A(T)$, $\ Y=A(T)\oplus S$, and that $A(T_1)\subseteq S$. Then $A_1:=A|_T$ and $A_2:=A|_{T_1}$ will do the job. Note that by Result 1. we already know that $A_1$ is isomorphism.

Let's call $B:=A^{(2)}_{T,S}$, and recall that $T_1=N(BA)$ and $S=N(B)$. So, for $x\in T_1$, we have $BAx=0$, that is, $Ax\in N(B)=S$, as wished.

One basic thing to use is that if an operator $U:X\to X$ is idempotent ($U^2=U$), then it is a projection: $X=N(U)\oplus R(U)$, and $U|_{R(U)}=id$.

Result 1. (a)$\Rightarrow$(b): Assume that $BAB=B$, and $R(B)=T$, $N(B)=S$. Then both $AB:Y\to Y$ and $BA:X\to X$ are idempotent. We have $$By=0 \implies ABy=0 \implies By=BABy=0\,,$$ so, $S=N(B)=N(AB)$. We also have $R(AB)=A(R(B))=A(T)$.

Similarly, $T_1=N(BA)$ and $R(BA)=T$, because $R(B)=T$ implies $R(BA)\subseteq T$, and for all $t\in T$ there is an $y$ such that $By=t$, then $t=By=BA(By)$.

• Thank you very much for your reply. Could you please tell me why other two blocks are zero only diagonal blocks are non zero? Apr 17 '13 at 15:05
• Yes: the first column of the matrix corresponds to the (restricted) mapping $T\to A(T)\oplus S$, and the $S$-component here is always $0$, because $A(T)\subseteq A(T)$. Similarly for the second column as $A|_{T_1}:T_1\to A(T)\oplus S$. $$\pmatrix{A_1&0\\0&A_2}\pmatrix{t\\t_1}=\pmatrix{A_1(t)\\A_2(t_1)}\,.$$ Apr 17 '13 at 16:02
• Million times thanks to you. I was trying to prove this theorem for the last many days but couldn't get any clue. Thanks again . :) Apr 17 '13 at 17:29