# Left invertible non-square matrices and their sub-matrices over a commutative ring

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!

• 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. Oct 3 at 12:15
• @user376343. thank you! I have changed the $1$ in the second row to $-1$. Oct 3 at 12:16
• @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. Oct 3 at 12:17
• @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 $R=\mathbb{C}[t]$. Oct 3 at 12:32

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 $$C\in R^{m\times n}$$.
$$1 = \det\big(I_n\big) = \det\big(BC\big) = \sum_{S} \det\big(B_{[n],S}\big)\det\big(C_{S,[n]}\big) = \sum_{S} \det\big(B_{[n],S}\big)\cdot \alpha_k$$
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 $$A$$.
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: $$M_{1,1}= \begin{pmatrix} t+1 \end{pmatrix}$$, $$M_{2,1}= \begin{pmatrix} t \end{pmatrix}$$.