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$X$ and $Y$ are Banach spaces and suppose that $T$ be a closed subspace of $X$.

$A:X \to Y$, $B:Y\to X$ and $X_{0}:Y\to X$ are linear bounded operators. It is given that the operator $BA$ is invertible on its range space denoted by $R(BA)$, also we have $R(BA)\subseteq T$ and $R(X_0) \subset T$

My question is can we write.

$(BA)^{-1}_{|R(BA)}BA X_{0} = X_{0}$.

Could anyone help me to clear my doubt. I would be very much thankful. I need to confirm this to prove one of my result.

Thanks

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1 Answer 1

Your assumptions allow for a counter example even in finite dimensions: The rank of $C=(BA)^{-1}BA$ is equal to the dimension of $R(BA)$. If $R(BA)$ is a strict subspace of $T$, we have that the rank of $CX_0$ is less than the rank of $X_0$.

A suitable condition on $BA$ would be, that $R(X_0) \cap N(BA) = \{0\}$ and $R(BA)$ closed. That would guarantee that $C$ is the identity on $R(X_0)$.

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Thanks for the answer. I made a mistake in typing. I wanted to ask if we take condition that $R(BA) = R(X_0)$ then can my result holds true? –  mathscrazy Jul 24 '13 at 7:05
    
Let me know if I am wrong. If $R(BA) = R(X_0) = T$ implies that $R(X_0) \cap N(BA) = \{0\}$ and $R(BA)$ is also closed by the assumption on $T$. So according to Guido response, the answer is yes. –  noether Jul 24 '13 at 8:15
    
Maybe I need clarification what you mean by "invertible on its range". My notion is, that a left inverse exists, which maps the range of $BA$ back to its domain. But this does not imply that range and domain are equal. The matrix $\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$ maps into its null space, but you can still define a left inverse. –  Guido Kanschat Jul 24 '13 at 15:08

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