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:




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 $\{0\}$.

  • $\begingroup$ I can't understated why proving that the linear transformation is one-to-one will also prove that it's onto. that's not true, of course, for every linear transformation. the kernel is sufficient example of why subjectivity is not enough to conclude that it's the LT is Isomorphism $\endgroup$ – Jneven Jul 11 '18 at 13:14
  • $\begingroup$ 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. $\endgroup$ – Delta-u Jul 11 '18 at 13:21

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