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Suppose $\mathcal{X}$ and $\mathcal{Y}$ are two Hilbert spaces. Let $A:\mathcal{X} \mapsto \mathcal{Y}$ be a bounded linear operator. Consider a linear operator equation $Ax=b$. My question is what would be the set of all solutions of this equation provided that it is solvable.

When $A$ is an $m\times n$ real matrix, hence a linear operator from $\mathbb{R}^m$ to $\mathbb{R}^n$, we know that set of all roots of $Ax=b$, if it was solvable, is $$ \{x = A^+b + (I-A^+A)c,\:\text{for all $c\in\mathbb{R}^n$,}\} $$where $A^+$ is the generalized inverse, e.g. the Moore-Penrose pseudoinverse, of $A$.

Do we have similar conclusion for the general linear operator $A$?

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  • $\begingroup$ Since $A$ is linear, the solution set should always be the kernel plus a particular solution. Why do you use pseudoinverse? $\endgroup$ – Troy Woo Aug 31 '14 at 18:27
  • $\begingroup$ @TroyWoo Thanks for your hint. I understand it now. $\endgroup$ – semibruin Aug 31 '14 at 19:11

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