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take B ∈ $$M_n(F) $$ Define a linear map $$T:M_n(F)\rightarrow M_n(F)$$ by : $$T(A)=A·B$$ Show that T is linear. Prove that T is isomorphism iff B is invertible.

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    $\begingroup$ After 7 months being a member you should know better how questions should be asked in this site...any idea to add there? $\endgroup$ – DonAntonio Jun 8 at 23:20
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Let $B\in M_n(F)$. We wish to show that the linear map $T:M_n(F)\to M_n(F)$ defined by $T(A)=A\cdot B$ is an isomorphism if and only if $B$ is invertible.

To do so, first suppose that $T$ is an isomorphism. This means that $T$ is surjective, so there exists an $A\in M_n(F)$ such that $T(A)=I_n$. It follows that $A\cdot B=I_n$ so that $A=B^{-1}$.

Conversely, suppose that $B$ is invertible. Then $A\in \operatorname{Ker}(T)$ implies that $T(A)=O_n$ so that $A\cdot B=O_n$ and right-multiplying by $B^{-1}$ then implies that $A=O_n\cdot B^{-1}=O_n$. This means that $T$ is injective. Since $T$ is an endomorphism, it follows that $T$ is an isomorphism.

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Try showing that $\det T = \det B$. This will show your desired result.

Edit: You can try in the case of $n=2$ first. Consider the action of $T$ of the standard basis of $M_2(F)$.

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