Bordered minor and rank of a matrix

Question

Let $M\in\mathbf{R}^{n\times n}$ be a matrix. Suppose that there is a $k\times k$ minor $M_k$ of rank k. Now this reference (Algebra For Iit Jee 7.65) here states that if all the $k+1$th minors bordering the minor $M_k$ vanish this implies that the rank of $M$ is in fact $k$. Is this obvious? And if so, what exactly is a bordering minor?

Answer 1

I don't know what a "bordering minor" is. This does not seem to be a word with a standard meaning. However, here is a theorem that is true:

Theorem 1. Let $\mathbf{k}$ be a field. Let $A\in\mathbf{k}^{u\times v}$ be a matrix. In the following, whenever $i_{1},i_{2},\ldots,i_{k}\in\left\{ 1,2,\ldots,u\right\} $ and $j_{1},j_{2},\ldots,j_{\ell}\in\left\{ 1,2,\ldots,v\right\} $, we shall denote by $A\left[ \dfrac{j_{1} ,j_{2},\ldots,j_{\ell}}{i_{1},i_{2},\ldots,i_{k}}\right] $ the $k\times\ell $-matrix whose $\left( p,q\right) $-th entry is the $\left( i_{p} ,j_{q}\right) $-th entry of $A$ for every $\left( p,q\right) \in\left\{ 1,2,\ldots,k\right\} \times\left\{ 1,2,\ldots,\ell\right\} $.

Let $k\in\mathbb{N}$. Let $i_{1},i_{2},\ldots,i_{k}\in\left\{ 1,2,\ldots ,u\right\} $ and $j_{1},j_{2},\ldots,j_{k}\in\left\{ 1,2,\ldots,v\right\} $ be such that $\det\left( A\left[ \dfrac{j_{1},j_{2},\ldots,j_{k}} {i_{1},i_{2},\ldots,i_{k}}\right] \right) \neq0$. Assume that every $i^{\prime}\in\left\{ 1,2,\ldots,u\right\} \setminus\left\{ i_{1} ,i_{2},\ldots,i_{k}\right\} $ and $j^{\prime}\in\left\{ 1,2,\ldots ,v\right\} \setminus\left\{ j_{1},j_{2},\ldots,j_{k}\right\} $ satisfy

(1) $\det\left( A\left[ \dfrac{j_{1},j_{2},\ldots,j_{k},j^{\prime} }{i_{1},i_{2},\ldots,i_{k},i^{\prime}}\right] \right) =0$.

Then, $\operatorname*{rank}A=k$.

Proof. Assume the contrary; i.e., assume that $\operatorname*{rank}A\neq k$.

First, we observe that the numbers $i_{1},i_{2},\ldots,i_{k}$ are distinct (because otherwise, the matrix $A\left[ \dfrac{j_{1},j_{2},\ldots,j_{k} }{i_{1},i_{2},\ldots,i_{k}}\right] $ would have two equal rows, which would yield $\det\left( A\left[ \dfrac{j_{1},j_{2},\ldots,j_{k}}{i_{1} ,i_{2},\ldots,i_{k}}\right] \right) =0$, contradicting $\det\left( A\left[ \dfrac{j_{1},j_{2},\ldots,j_{k}}{i_{1},i_{2},\ldots,i_{k}}\right] \right) \neq0$). Similarly, the numbers $j_{1},j_{2},\ldots,j_{k}$ are distinct.

The rows of the matrix $A\left[ \dfrac{j_{1},j_{2},\ldots,j_{k}}{i_{1} ,i_{2},\ldots,i_{k}}\right] $ are linearly independent (since $\det\left( A\left[ \dfrac{j_{1},j_{2},\ldots,j_{k}}{i_{1},i_{2},\ldots,i_{k}}\right] \right) \neq0$). Hence, the rows of the matrix $A\left[ \dfrac{1,2,\ldots ,v}{i_{1},i_{2},\ldots,i_{k}}\right] $ are also linearly independent (since the rows of the matrix $A\left[ \dfrac{j_{1},j_{2},\ldots,j_{k}}{i_{1} ,i_{2},\ldots,i_{k}}\right] $ are fragments of the rows of the matrix $A\left[ \dfrac{1,2,\ldots,v}{i_{1},i_{2},\ldots,i_{k}}\right] $, and therefore any linear dependence relation between the latter would yield a linear dependence relation between the former). In other words, the $i_{1}$-th, the $i_{2}$-th, etc., the $i_{k}$-th rows of the matrix $A$ are linearly independent. Hence, the matrix $A$ has $k$ linearly independent rows; thus, $\operatorname*{rank}A\geq k$. Combined with $\operatorname*{rank}A\neq k$, this yields $\operatorname*{rank}A>k$. Thus, the row space of $A$ cannot be spanned by just $k$ of its rows (because if it could, then its dimension would be $\leq k$, which would contradict the fact that its dimension is $\operatorname*{rank}A>k$). Therefore, there exists at least one $i^{\prime}\in\left\{ 1,2,\ldots,u\right\} \setminus\left\{ i_{1},i_{2},\ldots,i_{k}\right\} $ such that the $i^{\prime}$-th row of $A$ does not belong to the span of the $i_{1}$-th, the $i_{2}$-th, etc., the $i_{k}$-th rows of $A$. Fix such an $i^{\prime}$. We know that:

• The $i^{\prime}$-th row of $A$ does not belong to the span of the $i_{1} $-th, the $i_{2}$-th, etc., the $i_{k}$-th rows of $A$.

• The $i_{1}$-th, the $i_{2}$-th, etc., the $i_{k}$-th rows of $A$ are linearly independent.

Combining these two facts, we conclude that $i_{1}$-th, the $i_{2}$-th, etc., the $i_{k}$-th, and the $i^{\prime}$-th rows of $A$ are linearly independent. In other words, the rows of the matrix $A\left[ \dfrac{1,2,\ldots,v} {i_{1},i_{2},\ldots,i_{k},i^{\prime}}\right] $ are linearly independent. Since this matrix has $k+1$ rows, we thus conclude that $\operatorname*{rank} \left( A\left[ \dfrac{1,2,\ldots,v}{i_{1},i_{2},\ldots,i_{k},i^{\prime} }\right] \right) =k+1>k$. Therefore, the column space of $A\left[ \dfrac{1,2,\ldots,v}{i_{1},i_{2},\ldots,i_{k},i^{\prime}}\right] $ cannot be spanned by just $k$ of its columns (since if it could, then its dimension would be $\leq k$, which would contradict the observation that its dimension is $\operatorname*{rank}\left( A\left[ \dfrac{1,2,\ldots,v}{i_{1} ,i_{2},\ldots,i_{k},i^{\prime}}\right] \right) >k$). Thus, there exists at least one $j^{\prime}\in\left\{ 1,2,\ldots,v\right\} \setminus\left\{ j_{1},j_{2},\ldots,j_{k}\right\} $ such that the $j^{\prime}$-th column of $A\left[ \dfrac{1,2,\ldots,v}{i_{1},i_{2},\ldots,i_{k},i^{\prime}}\right] $ does not belong to the span of the $j_{1}$-th, the $j_{2}$-th, etc., the $j_{k}$-th columns of $A\left[ \dfrac{1,2,\ldots,v}{i_{1},i_{2},\ldots ,i_{k},i^{\prime}}\right] $. Fix such a $j^{\prime}$.

On the other hand, the columns of the matrix $A\left[ \dfrac{j_{1} ,j_{2},\ldots,j_{k}}{i_{1},i_{2},\ldots,i_{k}}\right] $ are linearly independent (since $\det\left( A\left[ \dfrac{j_{1},j_{2},\ldots,j_{k} }{i_{1},i_{2},\ldots,i_{k}}\right] \right) \neq0$). In other words, the $j_{1}$-th, the $j_{2}$-th, etc., the $j_{k}$-th columns of the matrix $A\left[ \dfrac{1,2,\ldots,v}{i_{1},i_{2},\ldots,i_{k}}\right] $ are linearly independent. Therefore, the $j_{1}$-th, the $j_{2}$-th, etc., the $j_{k}$-th columns of $A\left[ \dfrac{1,2,\ldots,v}{i_{1},i_{2},\ldots ,i_{k},i^{\prime}}\right] $ are linearly independent (because the $j_{1}$-th, the $j_{2}$-th, etc., the $j_{k}$-th columns of the matrix $A\left[ \dfrac{1,2,\ldots,v}{i_{1},i_{2},\ldots,i_{k}}\right] $ are fragments of the $j_{1}$-th, the $j_{2}$-th, etc., the $j_{k}$-th columns of $A\left[ \dfrac{1,2,\ldots,v}{i_{1},i_{2},\ldots,i_{k},i^{\prime}}\right] $, and therefore any linear dependence relation between the latter would yield a linear dependence relation between the former). Now we know that:

• The $j^{\prime}$-th column of $A\left[ \dfrac{1,2,\ldots,v}{i_{1} ,i_{2},\ldots,i_{k},i^{\prime}}\right] $ does not belong to the span of the $j_{1}$-th, the $j_{2}$-th, etc., the $j_{k}$-th columns of $A\left[ \dfrac{1,2,\ldots,v}{i_{1},i_{2},\ldots,i_{k},i^{\prime}}\right] $.

• The $j_{1}$-th, the $j_{2}$-th, etc., the $j_{k}$-th columns of $A\left[ \dfrac{1,2,\ldots,v}{i_{1},i_{2},\ldots,i_{k},i^{\prime}}\right] $ are linearly independent.

Combining these two facts, we conclude that the $j_{1}$-th, the $j_{2}$-th, etc., the $j_{k}$-th, and the $j^{\prime}$-th columns of $A\left[ \dfrac{1,2,\ldots,v}{i_{1},i_{2},\ldots,i_{k},i^{\prime}}\right] $ are linearly independent. In other words, the columns of the matrix $A\left[ \dfrac{j_{1},j_{2},\ldots,j_{k},j^{\prime}}{i_{1},i_{2},\ldots,i_{k} ,i^{\prime}}\right] $ are linearly independent. Hence, $\det\left( A\left[ \dfrac{j_{1},j_{2},\ldots,j_{k},j^{\prime}}{i_{1},i_{2},\ldots,i_{k} ,i^{\prime}}\right] \right) \neq0$. But this contradicts (1). Thus, we have found a contradiction, and Theorem 1 is proven.