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Let be $m<n$ and $M_m(m\times n, \mathbb{R})$ be the set of the matrices $m \times n$ of full rank $m$.

I want to show that $M_m(m\times n,\mathbb{R})$ is an open subset of $M(m\times n,\mathbb{R})$

I reasoned as follows. Let $A$ be an element in $M_m(m\times n,\mathbb{R})$. Then $A$ has some $m\times m$ submatrix whose determinant is non zero. Let be $j_1,\dots,j_m$ the columns of one such submatrix of $A$, and let $\pi\colon M(m\times n,\mathbb{R})\to M(m,\mathbb{R})$ be the map who takes a matrix $B$ and associates the $m\times m$ submatrix obtained by taking the columns $j_1,\dots, j_m$. Then $\pi$ is smooth, and then continuous. I already know that $GL(m,\mathbb{R})$ is open in $M(m,\mathbb{R})$. So I conclude that $\pi^{-1}(GL(m,\mathbb{R}))$ is an open subset of $M(m\times n,\mathbb{R})$ contained in $M_m(m\times n,\mathbb{R})$ and containing $A$.

So $M_m(m\times n,\mathbb{R})$ is a neighbor of all its points, then it is open.

Is my proof correct?

Minato
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1 Answers1

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Obviously you are right.

I give here something different. I show $S:=$ set of all matrices of rank strictly less than $m$ , is a closed subset of $M(m\times n, \mathbb{R})$ . So let $\{A_p\}_{p\in \Bbb N}$ be a sequence in $S$ converging to some $A\in M(m\times n, \mathbb{R})$. If possible let $A\in M_m(m\times n, \mathbb{R})$. Then we have a matrix $B\in M_m(n\times m, \mathbb{R})$ such that $AB=I_m$. Now $\{A_pB\}_{p\in \Bbb N}$ converges to $AB$. But notice that $det(A_pB)=0$ for each $p$ , since $rank(A_pB)≤rank(A_p)<m$ , also $rank(AB)=rank(I_m)=1$. Now $det$ is a continuous function i.e. $0=det(A_pB)\rightarrow det(AB)=1$ as $p\rightarrow \infty$, contradiction. Therefore $S$ is closed in $M(m\times n, \mathbb{R})$ i.e. $ M_m(m\times n, \mathbb{R})$ is open in $ M(m\times n, \mathbb{R})$.

Existence of $B\in M_m(n\times m, \mathbb{R})$ for $A\in M_m(m\times n, \mathbb{R})$ with $AB=I_m$ can be given as : $ dim(column\ space(A))=dim(row\ space (A))=m$ , so that j-th column $e_j$ of $I_m$ can be written as $Ab_j=e_j$ where $b_j$ is column matrix, for each $j=1,....,m$. Therefore define $B=[b_1,....,b_m]$.

Sumanta
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  • Thank-you. It's very beautiful. Let me only check a few things:

    1. ${A_pB }_{p\in\mathbb{N}}$ converges to $AB$ by the continuity of the row-column product, right?

    2. I have that the columns $A_1,\dots,A_n$ of $A$ are $n$ elements of $\mathbb{R}^m$ and they contain a basis of $\mathbb{R}^m$ since they contain $m$ linear indipendent vectors. Now $e_j$ is an element of $\mathbb{R}^m$, so I can write $e_j=b_j^1 A_1+\dots+ b_j^nA_n$. Now the product $Ab_j$ with $b_j=$ the column $(b_j^1,\dots, b_j^n)$ is exactly the linear combination which gives $e_j$, right?

     – Minato Oct 18 '18 at 13:16
  • You are actually right. – Sumanta Oct 18 '18 at 13:19