# For given operators $a$ and $b$, solve the equation $xb=a$

Let $$H$$ be a Hilbert space. For given operators $$a$$ and $$b$$ on $$H$$, how can we find all solutions of the equation $$xb=a$$?

• What type of answer are you looking for? – Luca Ghidelli Sep 11 at 20:56
• Any type. Of course, all types will be perfect. – Ali Bagheri Sep 11 at 21:07

I will assume that you want $$x$$ to be a bounded operator on $$H$$. Then, a necessary condition for such an $$x$$ to exist, is that there must be some constant $$C>0$$ satisfying $$\Vert a(\eta)\Vert \leq C\Vert b(\eta)\Vert$$ for all $$\eta\in H$$.
Conversely, suppose that $$a,b\in B(H)$$ and $$C>0$$ satisfy these conditions. Let $$V_0={\rm ran}(b)=bH$$ (which need not be closed), and note that the restriction of $$b$$ to $$\ker(b)^\perp$$ is a linear bijection $$b_1: \ker(B)^\perp \to V_0$$.
Let $$x_0 = ab_1^{-1}: V_0 \to H$$ (this is well-defined and linear by the previous remark), and note that $$x_0b =a$$ as functions $$H\to H$$.
Given $$\xi\in V_0$$ there exists $$\eta\in\ker(B)^\perp$$ such that $$\xi=b_1(\eta) = b(\eta)$$, and so $$x_0(\xi)=a(\eta)$$. Hence, by the 2nd condition, we have $$\Vert x_0(\xi)\Vert \leq C\Vert b(\eta) \Vert \leq C\Vert\xi\Vert$$ and so $$x_0$$ is bounded linear. Therefore, if we let $$V=\overline{V_0}$$, basic functional analysis tells us that $$x_0$$ has a unique continuous extension to a bounded operator $$x_1: V \to H$$.
Finally, we can define $$x:H\to H$$ to be $$x_1y$$ for any idempotent $$y\in B(H)$$ satisfying $$yH= V$$. (For example, we could take the orthogonal projection from $$H$$ onto $$V$$, but there are other possibilities.) Then $$xb=x_1b=a$$, as required; and all solutions arise this way.