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Let $$ A= \begin{pmatrix} 1 & 1 \\ 2 & 1 \\ 1 & 4 \end{pmatrix} $$

The matrix $A$ has rank two (=number of independent columns), so it is full, hence has a left inverse which is $$ A_L= \begin{pmatrix} -1 & 1 & 0\\ 2 & -1 & 0\\ \end{pmatrix} $$ (It is easy to check that $A_LA=I_{2 \times 2}$).

$A$ has three invertible square sub-matrices: $ A_{1,2}= \begin{pmatrix} 1 & 1 \\ 2 & 1 \end{pmatrix} $ $ A_{1,3}= \begin{pmatrix} 1 & 1 \\ 1 & 4 \end{pmatrix} $ $ A_{2,3}= \begin{pmatrix} 2 & 1 \\ 1 & 4 \end{pmatrix} $. Each of the three sub-matrices is invertible, but generally it may happen that there are non-invertible sub-matrices, as the following example shows: Let $$ B= \begin{pmatrix} 1 & 0 \\ 1 & 0 \\ 0 & 1 \end{pmatrix} $$ $ B_{1,2}= \begin{pmatrix} 1 & 0 \\ 1 & 0 \end{pmatrix} $ $ B_{1,3}= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $ $ B_{2,3}= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} $. Here $B_{1,3}$ is not invertible.

Claim 1: Let $k$ be a field of characteristic zero and let $A$ be an $m \times n$ matrix over $k$, $m > n$. Assume that $A$ is left invertible (hence all $m$ columns of $A$ are $k$-linearly independent). Then there exists an $n \times n$ sub-matrix of $A$ which is invertible (equivalently, has an invertible determinant = a nonzero determinant, since we work over a field $k$).

Question 1: Is Claim 1 true? It seems true and follows from the assumption that $A$ has rank $m$.

Claim 2: Let $R$ be a commutative ring and let $A$ be an $m \times n$ matrix over $k$, $m > n$. Assume that $A$ is left invertible (hence all $m$ columns of $A$ are $R$-linearly independent, so they generate a free $R$-module of rank $m$). Then there exists an $n \times n$ sub-matrix of $A$ which is invertible (equivalently, has an invertible determinant = the determinant is an invertible element of $R$, see this).

Question 2: Is Claim 2 true? It seems true if Claim 1 is true or am I missing something? Maybe the result of left invertibility is not equivalent to having full column rank $m$? (I think this result stull holds. see this).

Thank you very much! I apologize if my question is trivial.

Edit: Maybe all we can say is that there is square $n \times n$ sub-matrix $S$ of $A$ with non-zero determinant $d_S \in R$, but this does not imply that $S$ itself is an invertible matrix, since $d_s$ may not be invertible in $R$. The following is not a counterexample: $A= \begin{pmatrix} t \\ t \end{pmatrix} $. $A$ has rank one, $S= \begin{pmatrix} t \end{pmatrix} $ is not invertible, but $A$ itselt is not left invertible!

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    $\begingroup$ The size of the largest non-singular matrix is known as the determinantal rank of the matrix. It is known, in the field case, that it is the same as the row/column rank. I don't know about commutative rings though. $\endgroup$ Oct 3, 2023 at 12:15
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    $\begingroup$ @user376343. thank you! I have changed the $1$ in the second row to $-1$. $\endgroup$
    – user237522
    Oct 3, 2023 at 12:16
  • $\begingroup$ @TheoBendit, thank you very much! You can write your comment as a partial answer. I think/hope that the same result holds over commutative rings, but I have not carefully checked all the details yet. $\endgroup$
    – user237522
    Oct 3, 2023 at 12:17
  • $\begingroup$ @TheoBendit, maybe there would be a problem if we work over a commutative ring which is not a field...Instead of nonzero minors we should consider invertible minors. I will try to find a counterexample over $R=\mathbb{C}[t]$. $\endgroup$
    – user237522
    Oct 3, 2023 at 12:32

2 Answers 2

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Over a commutative ring it is better to relate determinants to values taken in the image of your matrix [Cramer's Rule] and the minors do relate to surjectivity in the following way:

Since $A^T$ has a right inverse it is surjective and the proper claim is its $n\times n$ minors generate the unit ideal. [In a PID like $\mathbb Z$ you can also talk about a gcd of $1$.] With $R$ denoting the commutative ring, define $B:=A^T$ and we know $I_n = BC$ for some $C\in R^{m\times n}$.

Proof:
$1 = \det\big(I_n\big) = \det\big(BC\big) = \sum_{S} \det\big(B_{[n],S}\big)\det\big(C_{S,[n]}\big) = \sum_{S} \det\big(B_{[n],S}\big)\cdot \alpha_k$
by Cauchy-Binet [a polynomial identity that holds over $\mathbb C$ so it holds over every commutative ring. Ref e.g. https://en.wikipedia.org/wiki/Cauchy%E2%80%93Binet_formula for notation clarifications though the OP has flipped the roles of $m$ and $n$.] You can of course transpose this at the end to talk about minors of $A$.

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The following is a counterexample to Claim 2: $M= \begin{pmatrix} t+1 \\ t \end{pmatrix} $. $M$ is left invertible: $M_L= \begin{pmatrix} 1 & -1 \end{pmatrix} $. Each of the two sub-matrices is not invertible: $M_{1,1}= \begin{pmatrix} t+1 \end{pmatrix} $, $M_{2,1}= \begin{pmatrix} t \end{pmatrix} $.

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