Hello all(adsbygoogle = window.adsbygoogle || []).push({});

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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# I A regular matrix <=> mA isomorphism

Have something to add?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**