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Let A be an $m \times n$ regular matrix of rank k. Consider $A$ as a module homomorphism from $\mathbb{R}^n$ into $\mathbb{R}^m$. Since $A$ is regular, there exists a matrix $G:\mathbb{R}^m \rightarrow \mathbb{R}^n$ such that $AGA = A$. Now $AG$ is an idempotent linear map on $\mathbb{R}^m \rightarrow \mathbb{R}^m$ and Range (A) = Range(AG).

By using this fact that "For any idempotent linear map $T : R^m \rightarrow R^m$, Range T is projective", Range(AG) is projective.

How to prove this "For any idempotent linear map $T : R^m \rightarrow R^m$, Range T is projective"?

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    $\begingroup$ An identical question has been closed as “unclear what you're asking”. What makes you think the same question will not follow the same fate? Are you sure you're not confusing $\mathbb{R}$ with $R$, the latter denoting some ring? $\endgroup$
    – egreg
    Commented Sep 23, 2016 at 16:59
  • $\begingroup$ Yes, you are right, this question has been closed when I did not update it. But after editing the question here it was showing the same that it has been closed so I thought of putting it again. $\endgroup$
    – Amanda
    Commented Sep 23, 2016 at 17:03
  • $\begingroup$ $\mathbb R$ is a field, because of this all of its modules are free, and hence also projective. $\endgroup$
    – Asinomás
    Commented Sep 23, 2016 at 17:04
  • $\begingroup$ Well, it makes little sense, as it stands: any vector space is a projective module. $\endgroup$
    – egreg
    Commented Sep 23, 2016 at 17:04
  • $\begingroup$ Yes, $\mathbb{R}$ is a commutative ring. $\endgroup$
    – Amanda
    Commented Sep 23, 2016 at 17:05

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