#### Math_QED

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Let ##m_A: \mathbb{K^n} \rightarrow \mathbb{K^n}: X \mapsto AX## and ##A \in M_{m,n}(\mathbb{K})##

(I already proved that this function is linear)

I want to prove that:

A regular matrix ##\iff m_A## is an isomorphism.

So, here is my approach. Can someone verify whether this is correct?

##\Rightarrow##

##A## is a regular matrix, thus ##A^{-1}## exists

Then is ##m_A## an isomorphism, because we can find an inverse function ##m_A^{-1}: X \mapsto A^{-1}X## It's straightforward to see that this function is the inverse. (##m_A \circ m_A^{-1} = 1_{\mathbb{K^n}} ## and ##m_A^{-1} \circ m_A = 1_{\mathbb{K^n}}##)

##\Leftarrow##

##m_A## is an isomorphism, thus ##m_A## has an inverse ##m_A^{-1}: X \mapsto BX##.

When we apply the definition of inverse function, we would deduce that:

##AB = I_n## and ##BA = I_n##. Then it follows, by definition of the inverse matrix, that ##B = A^{-1}##. Thus ##A^{-1}## must exist, because the inverse is unique and there is only one possibility to construct this inverse. Therefore, A is a regular matrix.

Thanks in advance.