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I came upon a trivial-seeming claim that I can't prove myself: Let $H_1$, $H_2$ and $H_3$ be finite-dimensional Hilbert spaces, let $A:H_1\rightarrow H_2$ and $B,C:H_3\rightarrow H_1$ be linear operators between these Hilbert spaces. Then $A^*AB = A^*AC$ if and only if $AB=AC$ (where $A^*$ denotes the adjoint of $A$).

The "if" part is obvious: If $AB=AC$, then $A^*AB = A^*AC$, no matter what $A^*$ is. As for the "only if" part, I tried the following among others:

$$A^*AB = A^*AC\ \Leftrightarrow\ A^*(AB-AC)=0\ \Rightarrow\ \langle A^*(AB-AC)w,u\rangle=0\ \forall u\in H_1,w\in H_3\ \Leftrightarrow\ \langle(AB-AC)w,Au\rangle = 0\quad \forall u\in H_1,w\in H_3\ \Leftrightarrow\ \mathrm{im}(AB-AC)=(\mathrm{im}\ A)^\bot = \mathrm{ker}\ A^*$$

and I'm stuck. I'm sure one has to exploit the definition and properties of adjoints in some way. I feel the answer is far from complicated yet I failed so far. Any help would be appreciated.

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Let $x \in \ker(A^*A)$, then $$ 0 = \left<A^*Ax,x\right> = \left<Ax,Ax\right> = \|Ax\|^2 $$ so $x \in \ker A$. This proves $\ker(A^*A) = \ker(A)$. Hence if $A^*(AB - AC) = 0$ then $\mathrm{im}\, (B-C) \subseteq \ker(A^*A) = \ker(A)$, hence we have $AB - AC = 0$.

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