# Help | Linear Algebra 1 - isomorphism and invertibility [closed]

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.

## closed as off-topic by Guido A., zipirovich, eranreches, Javi, Jendrik StelznerJun 9 at 11:33

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• After 7 months being a member you should know better how questions should be asked in this site...any idea to add there? – DonAntonio Jun 8 at 23:20

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.
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 $$M_2(F)$$.