# Let $\|LA-AR\|_{F}\geq\|A\|_{F}$, as $L\in M_{m\times m} ,R\in M_{n\times n},A\in M_{m\times n}$, prove: $LX-XR=Y$ as $X,Y\in M_{m\times n}$

Let $$L\in M_{m\times m}(\mathbb{R}), R\in M_{n\times n}(\mathbb{R})$$ be some matrices, such that for every matrix $A\in M_{m\times n}(\mathbb{R})$:

$$\|LA-AR\|_{F}\geq\|A\|_{F}$$ as $\|A\|_{F}=\sqrt{\sum_{i=1}^{m}\sum_{i=1}^{n}|a_{ij}|^{2}}$

Prove: for every $Y\in M_{m\times n}(\mathbb{R})$ there exist $X\in M_{m\times n}(\mathbb{R})$ such that:

$$LX-XR=Y$$

Consider $$\phi: M_{m\times n}(\mathbb{R}) \to M_{m\times n}(\mathbb{R})$$ $$X \mapsto LX-XR$$ This is a linear operator from a vector space of finite dimension to itself.
Thus to show the surjectivity it is sufficient to show the injectivity i.e that $$\ker(\phi)=\{0\}$$
and you can use the inequality to show that the kernel is indeed $\{0\}$.
• Indeed it is not true for any linear transformation, but the important point here is that $\phi:E \to F$ with $\dim E = \dim F = d< + \infty$. So using $\dim(\ker(\phi))+\dim(\phi(E))=d$ you can prove that such linear transformation is onto iff it is one to one. – Delta-u Jul 11 '18 at 13:21