On special matrices related to Cauchy and Toeplitz matrices

TL;DR

This paper investigates determinants of matrices related to Cauchy and Toeplitz types and applies these results to find the rank of specific non-square matrices.

Contribution

It provides new formulas for determinants of certain matrices linked to Cauchy and Toeplitz matrices and uses these to determine ranks of related non-square matrices.

Findings

Derived explicit determinant formulas for related matrices

Established rank criteria for certain non-square matrices

Connected determinant properties to matrix rank calculations

Abstract

In this paper, we are going to calculate the determinant of a certain type of square matrices, which are related to the well-known Cauchy and Toeplitz matrices. Then, we will use the results to determine the rank of special non-square matrices.

Equations

$$C_{(i-r,0)}=(-1{)}^r\prod _{e<\ell \in I}{d}_{(i,\ell )}{d}_{(\ell ,e)},C_{(i-r,j)}=(-1{)}^{r+j}{a}_i\prod _{e<\ell \in I_j}{a}_{\ell }{d}_{(i,\ell )}{d}_{(\ell ,e)},$$

$$D_r=(-1{)}^r\prod _{e<\ell \in I}{a}_{\ell }(a_{\ell }-a_e)=(-1{)}^r\prod _{e<\ell \in I}{a}_{\ell }{d}_{(\ell ,e)}\ne 0,$$

$$x_{ij}=\frac{1}{x_i-y_j},1\le i,j\le n.$$

$$|X_n|=\frac{{\prod }_{i<j\in I}(x_i-x_j)(y_i-y_j)}{{\prod }_{i\in I}{\prod }_{j\in I}(x_i-y_j)},I=\{1,2,\cdots ,n\}.$$

$$x_{ij}=\frac{(y_1-y_j)}{(x_i-y_1)}\cdot \frac{1}{(x_i-y_j)}(1\le i,j\le n).$$

$$|U|=\frac{1}{(x_1-y_1)}\cdot \frac{{\prod }_{j=2}^n(y_1-y_j)}{{\prod }_{1=2}^n(x_i-y_1)}\cdot \left|\begin{array}{cccc}1& \frac{1}{(x_1-y_2)}& \cdots & \frac{1}{(x_1-y_n)}\\ 1& \frac{1}{(x_2-y_2)}& \cdots & \frac{1}{(x_2-y_n)}\\ ⋮& \cdots & ⋮& ⋮\\ 1& \frac{1}{(x_n-y_2)}& \cdots & \frac{1}{(x_n-y_n)}\end{array}\right|.$$

$$x_{i1}^{\prime }=0,x_{ij}^{\prime }=\frac{(x_1-x_i)}{(x_1-y_j)}\cdot \frac{1}{(x_i-y_j)}2\le i,j\le n.$$

$$|X_n|=\frac{1}{(x_1-y_1)}\prod _{i,j=2}^n\frac{(y_1-y_j)(x_1-x_i)}{(x_i-y_1)(x_1-y_i)}\left|\begin{array}{cccc}\frac{1}{(x_2-y_2)}& \frac{1}{(x_2-y_3)}& \cdots & \frac{1}{(x_2-y_n)}\\ \frac{1}{(x_3-y_2)}& \frac{1}{(x_3-y_3)}& \cdots & \frac{1}{(x_3-y_n)}\\ ⋮& \cdots & ⋮& ⋮\\ \frac{1}{(x_n-y_2)}& \frac{1}{(x_n-y_3)}& \cdots & \frac{1}{(x_n-y_n)}\end{array}\right|.$$

$$|X_n|=\frac{1}{{\prod }_{i\in I}(x_i-y_i)}\cdot \frac{{\prod }_{i<j\in I}(y_i-y_j)(x_j-x_i)}{{\prod }_{i<j\in I}(x_i-y_j)(x_j-y_i)}=\frac{{\prod }_{i<j\in I}(x_i-x_j)(y_i-y_j)}{{\prod }_{i\in I}{\prod }_{j\in I}(x_i-y_j)}.$$

$$|{\mathbf{H}}_n|=\frac{c_n^4}{c_{2n}},c_n=\prod _{i=1}^{n-1}i!.$$

$$\frac{1}{|{\mathbf{H}}_n|}=\frac{c_{2n}}{c_n^4}=n!\cdot \prod _i^{2n-1}\left(\genfrac{}{}{0ex}{}{i}{[i/2]}\right).$$

$$V_n:=\left[\begin{array}{ccccc}{v}_0& v_1& v_2& \cdots & v_{n-1}\\ v_{-1}& v_0& v_1& \cdots & v_{n-2}\\ v_{-2}& v_{-1}& v_0& \cdots & v_{n-3}\\ ⋮& ⋮& ⋮& \cdots & ⋮\\ v_{1-n}& v_{2-n}& v_{3-n}& \cdots & v_0\end{array}\right].$$

$$a_s{d}_{(\ell ,e)}-a_{\ell }{d}_{(s,e)}=-a_e{d}_{(s,\ell )},\frac{d_{(t,e)}}{d_{(t,\ell )}}-\frac{d_{(s,e)}}{d_{(s,\ell )}}=\frac{d_{(t,s)}{d}_{(\ell ,e)}}{d_{(t,\ell )}{d}_{(s,\ell )}}.$$

$$a_s{d}_{(\ell ,e)}-a_{\ell }{d}_{(s,e)}$$

$$=(a_s-a_{\ell })d_{(\ell ,e)}-a_{\ell }{d}_{(s,\ell )}$$

$$=d_{(s,\ell )}(d_{(\ell ,e)}-a_{\ell })=-a_e{d}_{(s,\ell )}$$

$$\frac{d_{(t,e)}}{d_{(t,\ell )}}-\frac{d_{(s,e)}}{d_{(s,\ell )}}$$

$$=\frac{1}{d_{(t,\ell )}{d}_{(s,\ell )}}\cdot \left|\begin{array}{cc}{d}_{(t,t)}& d_{(t,\ell )}\\ d_{(s,e)}& d_{(s,\ell )}\end{array}\right|$$

$$=\frac{1}{d_{(t,\ell )}{d}_{(s,\ell )}}\cdot \left|\begin{array}{cc}{d}_{(t,e)}-d_{(s,e)}& d_{(t,\ell )}-d_{(s,\ell )}\\ d_{(s,e)}& d_{(s,\ell )}\end{array}\right|$$

$$=\frac{1}{d_{(t,\ell )}{d}_{(s,\ell )}}\cdot \left|\begin{array}{cc}{d}_{(t,s)}& d_{(t,s)}\\ d_{(s,e)}& d_{(s,\ell )}\end{array}\right|$$

$$=\frac{d_{(t,s)}}{d_{(t,\ell )}{d}_{(s,\ell )}}\cdot \left|\begin{array}{cc}1& 1\\ d_{(s,e)}& d_{(s,\ell )}\end{array}\right|$$

$$=\frac{d_{(t,s)}(d_{(s,\ell )}-d_{(s,e)})}{d_{(t,\ell )}{d}_{(s,\ell )}}=\frac{d_{(t,s)}{d}_{(\ell ,e)}}{d_{(t,\ell )}{d}_{(s,\ell )}}.$$

$$A_n:=\left[\begin{array}{ccccc}1& \frac{a_{i_1}}{d_{(i_1,e_1)}}& \frac{a_{i_1}}{d_{(i_1,e_2)}}& \cdots & \frac{a_{i_1}}{d_{(i_1,e_n)}}\\ 1& \frac{a_{i_2}}{d_{(i_2,e_1)}}& \frac{a_{i_2}}{d_{(i_2,e_2)}}& \cdots & \frac{a_{i_2}}{d_{(i_2,e_n)}}\\ ⋮& ⋮& \cdots & ⋮& ⋮\\ 1& \frac{a_{i_n}}{d_{(i_n,e_1)}}& \frac{a_{i_n}}{d_{(i_n,e_2)}}& \cdots & \frac{a_{i_n}}{d_{(i_n,e_n)}}\\ 1& \frac{a_{i_{n+1}}}{d_{(i_{n+1},e_1)}}& \frac{a_{i_{n+1}}}{d_{(i_{n+1},e_2)}}& \cdots & \frac{a_{i_{n+1}}}{d_{(i_{n+1},e_n)}}\end{array}\right],$$

$$|A_n|=\frac{D_r\cdot {\prod }_{s^{\prime }<s\in J}{d}_{(i_s,i_{s^{\prime }})}}{{\prod }_{s\in J}{\prod }_{j\in I}{d}_{(i_s,e_j)}}.$$

$$|A_n|$$

$$=\left|\begin{array}{cccc}1& \frac{-a_{i_1}}{d_{(i_1,e_1)}}& \cdots & \frac{-a_{i_1}}{d_{(i_1,e_n)}}\\ 0& \frac{-a_{e_1}{d}_{(i_2,i_1)}}{d_{(i_1,e_1)}{d}_{(i_2,e_1)}}& \cdots & \frac{-a_{e_n}{d}_{(i_2,i_n)}}{d_{(i_1,e_n)}{d}_{(i_2,e_n)}}\\ ⋮& ⋮& \cdots & ⋮\\ 0& \frac{-a_{e_1}{d}_{(i_{n+1},i_1)}}{d_{(i_1,e_1)}{d}_{(i_{n+1},e_1)}}& \cdots & \frac{-a_{e_n}{d}_{(i_{n+1},i_n)}}{d_{(i_1,e_n)}{d}_{(i_{n+1},e_n)}}\end{array}\right|$$

$$|A_n|=(-1{)}^n\prod _{j=1}^n\frac{a_{e_j}}{d_{(i_1,e_j)}}\cdot \prod _{s=2}^{n+1}{d}_{(i_s,i_1)}\cdot |B_n|$$

$$B_n:=\left[\begin{array}{cccc}\frac{1}{d_{(i_2,e_1)}}& \frac{1}{d_{(i_2,e_2)}}& \cdots & \frac{1}{d_{(i_2,e_n)}}\\ \frac{1}{d_{(i_3,e_1)}}& \frac{1}{d_{(i_3,e_2)}}& \cdots & \frac{1}{d_{(i_3,e_n)}}\\ ⋮& \cdots & ⋮& ⋮\\ \frac{1}{d_{(i_{n+1},e_1)}}& \frac{1}{d_{(i_{n+1},e_2)}}& \cdots & \frac{1}{d_{(i_{n+1},e_n)}}.\end{array}\right]$$

$$x_1=a_{i_2},\cdots ,x_n=a_{i_{n+1}},y_1=a_{e_1},\cdots ,y_n=a_{e_n},$$

$$|B_n|=\frac{{\prod }_{s^{\prime }<s\in \{2,\cdots ,n+1\}}{d}_{(i_s,i_{s^{\prime }})}\cdot {\prod }_{i<j\in J}{d}_{(e_j,e_i)}}{{\prod }_{s=2}^{n+1}{\prod }_{j=1}^n{d}_{(i_s,e_j)}},$$

$$|A_n|=\frac{(-1{)}^n{\prod }_{j=1}^n\frac{a_{e_j}}{d_{(i_1,e_j)}}\cdot {\prod }_{s^{\prime }<s\in J}{d}_{(i_s,i_{s^{\prime }})}}{{\prod }_{s=1}^{n+1}{\prod }_{j=1}^n{d}_{(i_s,e_j)}}=\frac{D_r\cdot {\prod }_{s^{\prime }<s\in J}{d}_{(i_s,i_{s^{\prime }})}}{{\prod }_{s=1}^{n+1}{\prod }_{j=1}^n{d}_{(i_s,e_j)}}.$$

$$V_n=\left[\begin{array}{cccccc}\frac{1}{n}& \frac{1}{n-1}& \frac{1}{n-2}& \cdots & \frac{1}{2}& 1\\ \frac{1}{n+1}& \frac{1}{n}& \frac{1}{n-1}& \cdots & \frac{1}{3}& \frac{1}{2}\\ \frac{1}{n+2}& \frac{1}{n+1}& \frac{1}{n}& \cdots & \frac{1}{4}& \frac{1}{3}\\ ⋮& ⋮& ⋮& \cdots & ⋮& ⋮\\ \frac{1}{2n-2}& \frac{1}{2n-3}& \frac{1}{2n-4}& \cdots & \frac{1}{n}& \frac{1}{n-1}\\ \frac{1}{2n-1}& \frac{1}{2n-2}& \frac{1}{2n-3}& \cdots & \frac{1}{n+1}& \frac{1}{n}\end{array}\right]=(-1{)}^k{\mathbf{H}}_n,$$

$$|C^{\prime }|=(-1{)}^{r^2+3r}{D}_r^{r+1}\prod _{s^{\prime }<s\in J}{d}_{(i_s,i_{s^{\prime }})}.$$

$$|C^{\prime }|$$

Full text

1. Introduction and main result

Throughout this paper, we fix a field $F$ of characteristic zero, $2\leq r<n$ arbitrary integers, and $\{a_{\ell}\}_{\ell=1}^{\infty}\subset F^{*}$ an infinite sequence of distinct elements. For any pair of indexes $(\ell,e)$, we define $d_{(\ell,e)}:=a_{\ell}-a_{e}$ and $d_{\ell}:=d_{(\ell+1,\ell)}$. Then, we consider the $(n-r)\times(r+1)$ matrix $C:=[C_{(i-r,j)}]$, such that for $j=0,\cdots,r$ and $r+1\leq i\leq n$ we have

[TABLE]

[TABLE]

where $I=\{1,\cdots,r\}$ and $I_{j}:=I\backslash\{j\}$ for $j\in I$. Define ${\bf C}_{r}^{n}:=[C|D]$ as a $(n-r)\times(n+1)$ blocked matrix, where $D$ is a $(n-r)\times(n-r)$ diagonal matrix with entries $D_{r}$. A special case of ${\bf C}_{r}^{n}$ is related to the Hilbert and Toeplitz’s matrices [1, 2, 3]. For more details, see Section 2.

The main result of this paper concerns with calculating the rank of the matrix ${\bf C}_{r}^{n}$,

Theorem 1.1**.**

Let $F$ be a field of characteristic zero and $\{a_{\ell}\}_{\ell=1}^{\infty}\subset F^{*}$ be an infinite sequence of distinct elements of $F$. Then the matrix ${\bf C}_{r}^{n}$ has full rank $n-r$ for integers $2\leq r<n$.

In order to prove the above theorem, we will calculate the determinant of certain square matrices which are related to the well-known Cauchy’s matrices [4, 5]. We notice that it is used the author’s forthcoming paper [6] to show the non-singularity of a certain family of complete intersection varieties satisfying the Bombieri-Lang conjecture in the Diophantine geometry [7].

The organization of the present paper is as follows. In the section 2, we recall the definition and determinant of the square Cauchy’s and Hilbert’s matrix over a field of characteristic zero. In section 3, we calculate the determinant of a certain square matrix that are related to the Cauchy’s matrix. Finally, in Section 4, we use the result of Section 3 to prove Theorem 1.1.

2. Cauchy’s and Toeplitz matrices

In 1841, Augustin Louis Cauchy introduced a certain type of matrices with certain properties, see [4, 5]. We are going to recall the definition and determinant of these matrices in this section.

An $n\times n$ square Cauchy’s matrix defined by disjoint subsets of distinct nonzero elements $\{x_{1},\cdots,x_{n}\}$ and $\{y_{1},\cdots,y_{n}\}$ in a field of characteristic zero $F$, is the square matrix $X_{n}:=[x_{ij}]$ with

[TABLE]

Note that any submatrix of a Cauchy’s matrix is itself a Cauchy’s matrix. The determinant of a Cauchy’s matrix is known as Cauchy’s determinant in the literature, which is always nonzero because $x_{i}\not=y_{j}$. Following proposition shows that how one can calculate the determinant of Cauchy’s matrices.

Proposition 2.1**.**

Let $n\geq 1$ be an integer and $X_{n}$ a $n\times n$ Cauchy’s matrix defined as above over a field $F$ of characteristic zero. Then

[TABLE]

Proof.

See the lemma (11.3) in [9] for an analytic proof, when $F={\mathbb{C}}$. For an arbitrary field $F$, we will use the elementary column and row operations to get the desired result. Subtracting the first column of $U$ from others gives that

[TABLE]

Extracting the factor $1/(x_{i}-y_{1})$ from $i$-th row for $i=1,\cdots,n$, and $y_{1}-y_{j}$ from $j$-th column for $j=2,\cdots,n$ leads to

[TABLE]

Now, denoting the last determinant by $|x^{\prime}_{ij}|$ and subtracting its first row from others, we get

[TABLE]

Extracting the factor $(x_{1}-x_{i})$ from each rows, and $1/(x_{1}-y_{j})$ from each column, for $2\leq i,j\leq n$, gives that

[TABLE]

Repeating this procedure, we obtain that

[TABLE]

∎

In [1], Hilbert introduced a certain square matrix which is a special case of the Cauchy square matrix. The Hilbert’s matrix is an $n\times n$ matrix ${\bf H}_{n}=[h_{ij}]$ with entries $h_{ij}=1/(i+j-1),$ where $1\leq i,j\leq n.$ Using the proposition 2.1, one can calculate the determinant of a Hilbert’s matrix as

[TABLE]

He also mentioned that the determinant of ${\bf H}_{n}$ is the reciprocal of a well known integer which follows from the following identity

[TABLE]

For more information see the sequence A005249 in OEIS [8]. For a recent work related to the Cauchy’s and Hilbert’s matrices one can see [10].

The other type of matrices, which we are going to recall here, are the Toeplitz matrices. An $n\times n$ Toeplitz matrix with entries in a field $F$ is the square matrix

[TABLE]

These are one of the most well studied and understood classes of matrices that arise in most areas of the mathematics: algebra [11], algebraic geometry [12], and graph theory [13]. In [3], the author obtained a unique LU factorizations and an explicit formula for the determinant and also the inversion of Toeplitz matrices. And, the inverse, determinants, eigenvalues, and eigenvectors of symmetric Toeplitz matrices over real number field with linearly increasing entries have been studied in [14]. In [15], the author showed that every $n\times n$ square matrix is generically a product of $\left\lfloor n/2\right\rfloor+1$ and always a product of at most $2n+5$ Toeplitz matrices.

3. Determinant of certain square matrix

In this section, we calculate the determinant of certain square matrices with entries in a field $F$ of characteristic zero, which are related to the determinant of Cauchy’s matrix. In special case, the determinant of our matrix is related to the determinant of a certain Toeplitz matrix. First, let us to give the following elementary result for a given infinite sequence $\{a_{\ell}\}_{\ell=1}^{\infty}$ of distinct nonzero elements in a field $F$ of characteristic zero.

Lemma 3.1**.**

For indexes $e,\ell,s,$ and $t$, we have

[TABLE]

Proof.

For indexes $e,\ell,$ and $s$, by definition $d_{(s,e)}=d_{(s,\ell)}+d_{(\ell,e)}$, so

[TABLE]

For indexes $e,\ell,s,$ and $t$, one has

[TABLE]

∎

For any integer $n\geq 1$, define $(n+1)\times(n+1)$ matrix $A_{n}$ as:

[TABLE]

where $\{a_{i_{1}},\cdots,a_{i_{n+1}}\}$ and $\{a_{e_{1}},\cdots,a_{e_{n}}\}$ are disjoint subsets of the infinite sequence $\{a_{\ell}\}_{\ell=1}^{\infty}$. The following proposition gives the determinant of $A_{n}$. We will use Lemma 3.1 in its proof.

Proposition 3.2**.**

Let $I=\{1,2,\cdots,n\}$ and $J=\{1,2,\cdots,n+1\}$. Then, one has

[TABLE]

Proof.

Subtracting first row from others and using lemma (3.2), gives that

[TABLE]

By extracting the factor $-a_{e_{j}}/d_{(i_{1},e_{j})}$ from each columns $(1\leq j\leq n)$ and $d_{(i_{s},i_{1})}$ from each rows $(2\leq s\leq n+1)$, one gets that

[TABLE]

where

[TABLE]

Since the matrix $B_{n}$ is a Cauchy’s matrix defined by

[TABLE]

so using Proposition 2.1 we have

[TABLE]

and hence,

[TABLE]

∎

We note that the matrix $B_{n}$ in the proof of the above proposition is related to a certain $n\times n$ Toeplitz matrix. Indeed, if we consider the sequence $a_{\ell}=1/\ell$ for $\ell=1,2,\cdots$ and indexes $e_{j}=j$ and $i_{s}=n+s-1$ for $j=1,\cdots,n$ and $s=1,\cdots,n+1$, then a simple calculation shows that $B_{n}=(-1)^{n}(2n)!V_{n},$ where $V_{n}$ is the following $n\times n$ Toeplitz matrix

[TABLE]

where $k=n/2$ if $n$ is even and $k=(n-1)/2$ if $n$ is odd; and the last equality comes by changing $j$-th column with $(n-j+1)$-th column of $V_{n}$.

4. Proof of theorem 1.1

In order to prove Theorem 1.1, we need the following result.

Proposition 4.1**.**

Given integers $2\leq r<n$ satisfying $r+1\leq n-r$, consider the indexes $r+1\leq i_{1},\cdots,i_{r+1}\leq n$ and let $I=\{1,2,\cdots,r\}$ and $J=\{1,\cdots,r+1\}$. Then, the matrix $C^{\prime}:=[C_{i_{s}-r,j}]$, where $s\in J$ and $0\leq j\leq r$, has nonzero determinant as

[TABLE]

Proof.

Extracting the factor $(-1)^{r}\prod_{e<\ell\in I}d_{(\ell,e)}$ and $(-1)^{r+j}\prod_{e<\ell\in I_{j}}a_{\ell}d_{(\ell,e)}$, respectively, from first column and the $j$-th column for $s\in J$ and $2\leq j\leq r$, where $I_{j}=I\backslash\{j\}$, gives that

[TABLE]

where $C^{\prime\prime}$ is the following $(r+1)\times(r+1)$ matrix

[TABLE]

By extracting the factor $\prod_{\ell\in I}d_{(i_{s},\ell)}$ from $s$-th row $1\leq s\leq r+1$, we obtain

[TABLE]

Considering $t=r$ and $e_{j}=j$ for $j=1,\cdots,r$, and using Proposition 3.2 for calculating the last determinant, one can conclude that

[TABLE]

∎

We notice that above proposition is a special case of the next general one.

Proposition 4.2**.**

Given integers $2\leq r<n$, and $m\leq\min\{r+1,n-r\}$, any $m\times m$ sub-matrix of the matrix $C=[C_{(i-r,j)}]$ has non-zero determinant, where $r+1\leq i\leq n$ and $0\leq j\leq r$; therefore $C$ has maximal rank equal to $\min\{n-r,r+1\}$.

Proof.

We may assume that $r+1\leq n-r$, the other case is similar. For $m\leq r+1$, we denote by $C_{m}$ any $m\times m$ sub-matrix of $C$. By proposition 4.1, the determinant of $C_{m}$ is nonzero for $m=r+1$. Thus, we may suppose that $m<r+1$ and $C_{m}=[C_{(i_{s}-r,j_{t})}]$, where $r+1\leq i_{s}\leq n-r$, $0\leq j_{s^{\prime}}\leq r$ for $1\leq s,s^{\prime}\leq m$. If we suppose that $0=j_{1}<j_{2},\cdots,j_{m}$, then

[TABLE]

such that

[TABLE]

[TABLE]

where $I=\{1,2,\cdots,r\}$ and $I_{j_{s^{\prime}}}=I\backslash\{j_{s^{\prime}}l\}$. Extracting $(-1)^{r}\prod_{e<\ell\in I}d_{(\ell,e)}$ and $(-1)^{r+j_{s^{\prime}}}\prod_{e<\ell\in I_{j_{s^{\prime}}}}a_{\ell}d_{(\ell,e)}$ from first and $s^{\prime}$-th columns, respectively, and then $\prod_{\ell\in I}d_{(i_{s},\ell)}$ from $s$-th row for $1\leq s<s^{\prime}\leq m$, gives that

[TABLE]

where $r^{\prime}=mr+j_{2}+\cdots+j_{m}$ and the above is nonzer by Propositions 3.2. Otherwise, if suppose that $1\leq j_{1}<j_{2}<\cdots<j_{m}$, then extracting the factor $(-1)^{r+j_{s^{\prime}}}a_{i_{s}}\prod_{e<\ell\in I_{j_{s^{\prime}}}}a_{\ell}d_{(\ell,e)}$ from ${s^{\prime}}$-th column, and then $\prod_{\ell\in I}d_{(i_{s},\ell)}$ from $s$-th row of the matrix $C_{m}=[C_{i_{s}-r,j_{s^{\prime}}}]$, where $1\leq s,s^{\prime}\leq m$, gives that

[TABLE]

where $r^{\prime\prime}=mr+j_{1}+\cdots+j_{m}$ and the last determinant is nonzero by Propositions 2.1. This completes the proof of the proposition. ∎

Now we are ready to prove the main theorem 1.1, using the above results.

Proof.

For integers $2\leq r<n$, recall that ${\bf C}_{r}^{n}:=[C|D]$ is a $(n-r)\times(n+1)$ blocked matrix, where $C=[C_{(i-r,j)}]$ is $(n-r)\times(r+1)$ matrix defined as in the first section and $D$ is a $(n-r)\times(n-r)$ diagonal matrix with entries $D_{r}$. By Proposition 4.2, any $m\times m$ sub-matrix of the matrix $C$ has non-zero determinant and $C$ has maximal rank equal to $\min\{n-r,r+1\}$. It is clear that the matrix $D$ has full rank equal to $n-r$. By exchanging the columns, if it is necessary, one can see that any $(n-r)\times(n-r)$ submatrix of ${\bf C}_{r}^{n}$ is a diagonal blocked matrix with blocks equal to $D_{r}$ or $m\times m$ submatrices of $C$ with $1\leq m\leq\min\{n-r,r+1\}$, which have non-zero determinant. Therefore, any $(n-r)\times(n-r)$ submatrix of ${\bf C}_{r}^{n}$ has nonzero determinant, and hence it has maximal rank $n-r$, as desired. ∎