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Let $f_A: \mathrm{Mat}_2(\mathbb R)\rightarrow \mathrm{Mat}_2(\mathbb R)$ be such that $f_A(X)=AX$ be an endomorphism of vector spaces. Clearly if $A$ is invertible, then $f_A$ is invertible and ${(f_A)}^{-1}=f_{A^{-1}}$. How can I prove the following statement?

$f_A\;\textrm{invertible} \Rightarrow A\;\textrm{invertible}$

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Invertible implies surjective. So every matrix is of the form $AX$. Even the identity matrix. – Dylan Moreland Jun 4 '12 at 15:33
thanks... it was so simple! – Dubious Jun 4 '12 at 15:35
up vote 1 down vote accepted

Suppose $\,A\,$ is not invertible, then$$\exists\,\, \mathbf{0}\neq \mathbf{b}:=\begin{pmatrix}b_1\\b_2\end{pmatrix}\in\mathbb{R}^2\,\,s.t.\,\,A\mathbf{x}=\mathbf{b}$$has no solution, so if $\,B\in\operatorname{Mat}_2(\mathbb{R})\,$ is any element with first column equal to $\,\mathbf{b}\,$ , then

$\,\,AX\neq B\,\,\,\forall\,X\in\operatorname{Mat}_2(\mathbb{R})$ , which contradicts $f_A$ is invertible and thus onto...

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