The order-$n$ minors of certain $(n+k) \times n$ matrices

TL;DR

This paper establishes conditions for the non-vanishing of order-$n$ minors in certain rectangular matrices and constructs high-dimensional subspaces with specific Schmidt rank properties.

Contribution

It provides new criteria for minors' non-vanishing and explicit formulas for special matrix classes, advancing understanding of matrix minors and tensor subspace structures.

Findings

All order-$n$ minors are nonzero under certain conditions.

Explicit formulas for minors of special $(n+1) imes n$ matrices.

Constructed subspaces of $ ext{C}^m imes ext{C}^n$ with maximal dimension and prescribed Schmidt ranks.

Abstract

We determine sufficient conditions for certain classes of $(n+k) \times n$ matrices $E$ to have all order-$n$ minors to be nonzero. For a special class of $(n+1) \times n$ matrices $E,$ we give the formula for the order-$n$ minors. As an application we construct subspaces of $\C^m \otimes \C^n$ of maximal dimension, which does not contain any vector of Schmidt rank less than $k$ and which has a basis of Schmidt rank $k$ for $k=2,3,4$.

Equations

$$E_n(a,b)=b\sum _{j=2}^n|j-1⟩⟨j|-a\sum _{j=1}^n|j⟩⟨j|+a\sum _{j=1}^n|j+1⟩⟨j|-b\sum _{j=1}^{n-1}|j+2⟩⟨j|,$$

$$|\psi ⟩=\sum _{j=1}^k{\alpha }_j|u_j⟩\otimes |v_j⟩$$

$$\rho =\sum _j{p}_j|{\psi }_j⟩⟨{\psi }_j|,$$

$$\rho =\sum _j{p}_j|v_j⟩⟨v_j|,$$

$$0=⟨\psi |\rho |\psi ⟩=\sum _j{p}_j|⟨v_j|\psi ⟩{|}^2.$$

$$D_n(a,b)=b\sum _{j=2}^n|j-1⟩⟨j|-a\sum _{j=1}^n|j⟩⟨j|+a\sum _{j=1}^{n-1}|j+1⟩⟨j|-b\sum _{j=1}^{n-2}|j+2⟩⟨j|\in {\mathbb{M}}_n(\mathbb{C}).$$

$$F_n(a,b)$$

$$=-b\sum _{j=2}^n|j⟩⟨j-1|+a\sum _{j=1}^n|j⟩⟨j|-a\sum _{j=1}^{n-1}|j⟩⟨j+1|+b\sum _{j=1}^{n-2}|j⟩⟨j+2|$$

$$=D_n(-a,-b{)}^t.$$

$$A^{\uparrow }=\sum _{i,j=1}^{n-1}{a}_{i,j}|i⟩⟨j|\text{ and }{A}^{\downarrow }=\sum _{i,j=2}^n{a}_{i,j}|i⟩⟨j|,$$

$$c_{ij}=\{\begin{array}{ll}{a}_{ij}& \text{ if }1\le i,j\le n,\\ b_{i-n,j-n}& \text{ if }n+1\le i,j\le n+m\\ a& \text{ if }(i,j)=(n+1,n)\\ b& \text{ if }(i,j)=(n,n+1)\\ 0& \text{ otherwise }\end{array}.$$

$$|A{}_a{\oplus }_{b}B|=|A||B|-ab|A^{\uparrow }||B^{\downarrow }|,$$

$$\sigma (i)={\sigma }_1(i)(1\le i\le n)\text{ and }\sigma (i)={\sigma }_2(i)(n+1\le i\le n+m).$$

$$\sigma (n)=n+1,\sigma (n+1)=n,\sigma (i)={\tau }_1(i)(1\le i\le n-1)\text{ and }\sigma (i)={\tau }_2(i)(n+2\le i\le n+m).$$

$$\mathrm{sgn}(\sigma )\prod _{i=1}^{n+m}{c}_{i,\sigma (i)}=\{\begin{array}{ll}\mathrm{sgn}({\sigma }_1)\mathrm{sgn}({\sigma }_2){\prod }_{i=1}^n{c}_{i,{\sigma }_1(i)}{\prod }_{i=n+1}^{n+m}{c}_{i,{\sigma }_2(i)}& \text{ case}(1)\\ -ab\mathrm{sgn}({\tau }_1)\mathrm{sgn}({\tau }_2){\prod }_{i=1}^{n-1}{c}_{i,{\tau }_1(i)}{\prod }_{i=n+2}^{n+m}{c}_{i,{\tau }_2(i)}& \text{ case}(2)\\ 0& \text{ case}(3)\end{array}.$$

$$|C|$$

$$=\sum _{{\sigma }_1\in S(1,n),{\sigma }_2\in S(n+1,n+m)}\mathrm{sgn}({\sigma }_1)\mathrm{sgn}({\sigma }_2)\prod _{i=1}^n{c}_{i,{\sigma }_1(i)}\prod _{i=n+1}^{n+m}{c}_{i,{\sigma }_2(i)}$$

$$-ab\sum _{{\tau }_1\in S(1,n-1),{\tau }_2\in S(n+2,n+m)}\mathrm{sgn}({\tau }_1)\mathrm{sgn}({\tau }_2)\prod _{i=1}^{n-1}{c}_{i,{\tau }_1(i)}\prod _{i=n+2}^{n+m}{c}_{i,{\tau }_2(i)}$$

$$=\left(\sum _{{\sigma }_1\in S(1,n)}\mathrm{sgn}({\sigma }_1)\prod _{i=1}^n{a}_{i,{\sigma }_1(i)}\right)\left(\sum _{{\sigma }_2\in S(1,m)}\mathrm{sgn}({\sigma }_2)\prod _{i=1}^m{b}_{i,{\sigma }_2(i)}\right)$$

$$-ab\left(\sum _{{\tau }_1\in S(1,n-1)}\mathrm{sgn}({\tau }_1)\prod _{i=1}^{n-1}{a}_{i,{\tau }_1(i)}\right)\left(\sum _{{\tau }_2\in S(2,m)}\mathrm{sgn}({\tau }_2)\prod _{i=2}^m{b}_{i,{\tau }_2(i)}\right)$$

$$=|A||B|-ab|A^{\uparrow }||B^{\downarrow }|.$$

$$|E_n^k(a,b)|$$

$$=(-1{)}^{n-k+1}(|D_{k-1}(a,b)||D_{n-k+1}(a,b)|-|D_{k-2}(a,b)||D_{n-k}(a,b)|b^2).$$

$$|E_n^1(a,b)|$$

$$|E_n^{n+1}(a,b)|$$

$$|E_n^k(a,b)|$$

$$=|D_{k-1}(a,b)||F_{n-k+1}(a,b)|+|D_{k-2}(a,b)||F_{n-k}(a,b)|b^2.$$

$$|E_n^k(a,b)|$$

$$=(-1{)}^{n-k+1}(-a{)}^{k-1}(-a{)}^{n-k+1}$$

$$=(-1{)}^{k-1}{a}^n.$$

$$|E_n^k(a)|$$

$$=(-1{)}^{k+1}|E_n^{n-k+2}(a)|$$

$$=(-1{)}^n|E_n^{n-k+2}(a)|$$

$$d_k=\frac{1}{k!}\frac{d^k}{d{x}^k}{\left(\frac{1}{1+ax+a{x}^2+x^3}\right)|}_{x=0}.$$

$$d_k=\frac{1}{{\alpha }^{k+1}(\alpha -\beta )(\gamma -\alpha )}+\frac{1}{{\beta }^{k+1}(\alpha -\beta )(\beta -\gamma )}+\frac{1}{{\gamma }^{k+1}(\gamma -\alpha )(\beta -\gamma )}.$$

$$|E_n^k(a)|$$

$$=(-1{)}^{n-k+1}(d_{k-1}{d}_{n-k+1}-d_{k-2}{d}_{n-k}).$$

$$d_k=\frac{1}{k!}\frac{d^k}{d{x}^k}{\left(\frac{1}{1+ax+a{x}^2+x^3}\right)|}_{x=0}.$$

Full text

1. Introduction

In this article we investigate conditions under which certain $(n+k)\times n$ matrices $E$ have all order-$n$ minors to be nonzero. Using this we conclude that for such $(n+k)\times n$ matrix $E,$ any linear combination of the columns of $E$ will have at least $k+1$ nonzero entries. This property has an application to the computation of Schmidt rank.

For any $n\in{\mathbb{N}}$ and numbers $a$ and $b$ in ${\mathbb{C}},$ set the $(n+1)\times n$ matrix $E_{n}(a,b)\in{\mathbb{M}}_{n+1,n}({\mathbb{C}})$ as follows:

[TABLE]

where $|1\rangle,\ldots,|n+1\rangle$ are the standard basis column vectors of ${\mathbb{C}}^{n+1}$, as column vectors, and $\langle 1|,\ldots,\langle n|$ are the transposed standard basis column vectors of ${\mathbb{C}}^{n}$.

In Section 2 we obtain an expression for order-$n$ minors of $E_{n}(a,b)$ in terms of determinants of certain submatrices of $E_{n}(a,b).$ By exploiting a recurrence relation and the characteristic polynomial of a certain matrix, we derive formulae for order-$n$ minors of $E_{n}(a,b).$ For some examples of $E_{n}(a,b),$ the order-$n$ minors are computed in Section 3. In Section 4 we take an approach based on system of equations, and conclude that for $|a|\geq 5,$ all the order-$n$ minors of $E_{n}(a,1)$ are nonzero. We also analyse another class of $(n+2)\times n$ matrices $G_{n}(a,1)$ using the same approach.

Let ${\mathcal{H}}$ denote the bipartite Hilbert space ${\mathbb{C}}^{m}\otimes{\mathbb{C}}^{n}$. By Schmidt decomposition theorem (cf. Ref.  [4]), any pure state $\left|\psi\right\rangle\in{\mathcal{H}}$ can be written as

[TABLE]

for some $k\leq\min\{m,n\}$, where $\{\left|u_{j}\right\rangle:1\leq j\leq k\}$ and $\{\left|v_{j}\right\rangle:1\leq j\leq k\}$ are orthonormal sets in ${\mathbb{C}}^{m}$ and ${\mathbb{C}}^{n}$ respectively, and $\alpha_{j}$’s are positive real numbers satisfying $\sum_{j}\alpha_{j}^{2}=1$.

Definition 1.1**.**

In the Schmidt decomposition (2) of a pure bipartite state $\left|\psi\right\rangle$ the minimum number of terms required in the summation is known as the Schmidt rank of $\left|\psi\right\rangle$, and it is denoted by $SR(\left|\psi\right\rangle)$.

As an application of the results of the first four sections, we construct subspaces $\mathcal{T}$ of dimension $(m-k+1)(n-k+1)$ of bipartite finite dimensional Hilbert space ${\mathbb{C}}^{m}\otimes{\mathbb{C}}^{n}$ in Section 5 such that any vector in $\mathcal{T}$ has Schmidt rank greater than or equal to $k$ where $k=2,3$ and $4.$ In Ref. [2], it was proved that for a bipartite system ${\mathbb{C}}^{m}\otimes{\mathbb{C}}^{n},$ the dimension of any subspace of Schmidt rank greater than or equal to $k$ is bounded above by $(m-k+1)(n-k+1).$ Unlike Ref. [2], the subspaces $\mathcal{T}$ of ${\mathbb{C}}^{m}\otimes{\mathbb{C}}^{n}$ that we construct also have bases consisting of elements of Schmidt rank $k.$ For the case when a subspace of ${\mathbb{C}}^{m}\otimes{\mathbb{C}}^{n}$ is of Schmidt rank greater than or equal to 2 (that is, the subspace does not contain any product vector), the maximum dimension of that subspace is $(m-1)(n-1),$ and this was first proved in Ref. [4] and Ref. [6] (cf. Ref. [1]).

In the bipartite Hilbert space ${\mathbb{C}}^{m}\otimes{\mathbb{C}}^{n},$ for any $1\leq r\leq\min\{m,n\},$ there is at least some state $\left|\psi\right\rangle$ with $SR(\left|\psi\right\rangle)=r.$ Any state $\rho$ on a finite dimensional Hilbert space ${\mathcal{H}}$ can be written as

[TABLE]

where $\left|\psi_{j}\right\rangle$’s are pure states in ${\mathcal{H}}$ and $\{p_{j}\}$ forms a probability distribution. The following notion was introduced in Ref. [5]:

Definition 1.2**.**

The Schmidt number of a state $\rho$ on a bipartite finite dimensional Hilbert space ${\mathcal{H}}$ is defined to be the least natural number $k$ such that $\rho$ has a decomposition of the form given in (3) with $SR(\left|\psi_{j}\right\rangle)\leq k$ for all $j$. The Schmidt number of $\rho$ is denoted by $SN(\rho)$.

Schmidt number of a state on a bipartite Hilbert space is a measure of entanglement. Entanglement is the key property of quantum systems which is responsible for the higher efficiency of quantum computation and tasks like teleportation, super-dense coding, etc (cf. Ref. [3]). The following proposition establishes an important relation between Schmidt number of a state and the lower bound of Schmidt rank of any non-zero vector in the supporting subspace of the state:

Proposition 1.3**.**

Let $\mathcal{S}$ be a subspace of ${\mathcal{H}}={\mathbb{C}}^{m}\otimes{\mathbb{C}}^{n}$ which does not contain any non-zero vector of Schmidt rank lesser or equal to $k$. Then any state $\rho$ supported on $\mathcal{S}$ has Schmidt number at least $k+1$.

Proof.

Let $\rho$ be a state with Schmidt number, $SN(\rho)=r\leq k$. So, $\rho$ can be written as

[TABLE]

where $SR(|v_{j}\rangle)\leq k$ for all $j$ and $\{p_{j}\}$ forms a probability distribution. Let $|\psi\rangle\in\mathrm{Ker}(\rho)$. Then we have

[TABLE]

This means $|v_{j}\rangle\in\mathrm{Ker}(\rho)^{\perp}=\mathrm{Range}(\rho)$ for all $j$.

If such a $\rho$ is supported on $\mathcal{S}$, then the above statement gives a contradiction to the fact that $\mathcal{S}$ does not contain any vector of Schmidt rank lesser or equal to $k$. This completes the proof.

2. Main Results

Lemma 2.1**.**

For any $n\in{\mathbb{N}}$ and numbers $a$ and $b$ in ${\mathbb{C}}$ set the $n\times n$ matrix

[TABLE]

Then we have $|D_{n}(a,b)|=-a|D_{n-1}(a,b)|-ab|D_{n-2}(a,b)|-b^{3}|D_{n-3}(a,b)|$  $(n\geq 2)$, where $|D_{-1}(a,b)|=0$ and $|D_{0}(a,b)|=1$. Note that $|D_{1}(a,b)|=-a$.

We also define the $n\times n$ matrix $F_{n}(a,b)\in{\mathbb{M}}_{n}({\mathbb{C}})$ as follows:

[TABLE]

Then we have $|F_{n}(a,b)|=(-1)^{n}|D_{n}(a,b)|$  $(n\geq-1)$.

For an $n\times n$ matrix $A=\sum_{i,j=1}^{n}a_{ij}|i\rangle\langle j|\in{\mathbb{M}}_{n}({\mathbb{C}})$, we define $A^{\uparrow}$ and $A^{\downarrow}$ in ${\mathbb{M}}_{n-1}({\mathbb{C}})$ as follows:

[TABLE]

if $n\geq 2$.

For numbers $a,b\in{\mathbb{C}}$, $A=\sum_{i,j=1}^{n}a_{i,j}|i\rangle\langle j|\in{\mathbb{M}}_{n}({\mathbb{C}})$ and $B=\sum_{i,j=1}^{m}b_{i,j}|i\rangle\langle j|\in{\mathbb{M}}_{m}({\mathbb{C}})$, we define the matrix $C=\sum_{i,j=1}^{n+m}c_{i,j}|i\rangle\langle j|\in{\mathbb{M}}_{n+m}({\mathbb{C}})$ as follows:

[TABLE]

We denote this matrix $C$ by $A{}_{a}\!\!\oplus_{b}B$.

Lemma 2.2**.**

In this setting, we have

[TABLE]

where we set $|A^{\uparrow}|=|A^{\downarrow}|=1$ if $A\in{\mathbb{M}}_{1}({\mathbb{C}})$.

Proof.

We denote $A{}_{a}\!\!\oplus_{b}B=C=\sum_{i,j=1}^{n+m}c_{i,j}|i\rangle\langle j|$. For $k\leq l$, we denote $S(k,l)$ the set of all permutations on $\{k,k+1,\ldots,l-1,l\}$. Let $\sigma\in S(1,n+m)$ and ${\rm sgn}(\sigma)$ denote the signature of permutation $\sigma$. Then $\sigma$ has the form which is one of the following type:

1. (1)

there exist $\sigma_{1}\in S(1,n)$ and $\sigma_{2}\in S(n+1,n+m)$ such that

[TABLE] 2. (2)

there exist $\tau_{1}\in S(1,n-1)$ and $\tau_{2}\in S(n+2,n+m)$ such that

[TABLE] 3. (3)

$\sigma$ has the form of neither type (1) nor type (2).

Then we have

[TABLE]

This implies that

[TABLE]

Proposition 2.3**.**

For $1\leq k\leq n+1$ and $a$ and $b$ in ${\mathbb{C}}$, let $E_{n}^{k}(a,b)$ be a matrix which is obtained by deleting the $k$-th row of $E_{n}(a,b)$ which is defined in equation (1). Then we have

[TABLE]

Proof.

Since $E_{n}^{1}(a,b)=F_{n}(a,b)$, $E_{n}^{n+1}(a,b)=D_{n}(a,b)$, $|D_{0}(a,b)|=|F_{0}(a,b)|=1$ and $|D_{-1}(a,b)|=|F_{-1}(a,b)|=0$, we have

[TABLE]

In the case $2\leq k\leq n$, we have $E_{n}^{k}(a,b)=D_{k-1}(a,b){}_{(-b)}\oplus_{b}F_{n-k+1}(a,b)$. It implies that, by Lemma 2.2,

[TABLE]

If $b=0$ and $1\leq k\leq n+1$,

[TABLE]

If $b\not=0$, since $|E_{n}^{k}(a,b)|=b^{n}\left|E_{n}^{k}\left(\dfrac{a}{b},1\right)\right|$, we may assume that $b=1$.

After this we write $E_{n}(a,1)=E_{n}(a)$ and $D_{n}(a,1)=D_{n}(a)$.

Theorem 2.4**.**

For $n\in{\mathbb{N}}$ and $a\in{\mathbb{C}}$, let $d_{-1}=0$, $d_{0}=1$ and $d_{n}=|D_{n}(a)|$. Then for $1\leq k\leq n+1$ we have

[TABLE]

and

[TABLE]

Moreover, if $x^{3}+ax^{2}+ax+1=0$ has $3$ different solutions $\alpha,\beta,\gamma,$ then we have for $1\leq k$

[TABLE]

Proof.

By Proposition 2.3 we have

[TABLE]

On the contrary, $(-1)^{n}|E_{n}^{n-k+2}(a)|=(-1)^{n}(-1)^{k-1}(d_{n-k+1}d_{k-1}-d_{n-k}d_{k-2})=(-1)^{n-k+1}(d_{k-1}d_{n-k+1}-d_{k-2}d_{n-k})$.

Define the formal power series $f(x)=\sum\limits_{k=0}^{\infty}d_{k}x^{k}$, where the sequence $\{d_{n}\}_{n=0}^{\infty}$ satisfies the relation $d_{n}+ad_{n-1}+ad_{n-2}+d_{n-3}=0\ (n\geq 2)$. Since $(1+ax+ax^{2}+x^{3})f(x)=d_{0}+(d_{1}+ad_{0})x+(d_{2}+ad_{1}+ad_{0})x^{2}+\sum\limits_{k=3}^{\infty}(d_{k}+ad_{k-1}+ad_{k-2}+d_{k-3})x^{k}=1$, $f(x)=\dfrac{1}{1+ax+ax^{2}+x^{3}}$ is analytic at a neighborhood of [math]. By Maclaurin’s expansion of $f$, we have

[TABLE]

Suppose that $1+ax+ax^{2}+x^{3}=0$ has $3$ different solutions $\alpha,\beta,\gamma$. Then,

[TABLE]

Hence, for $1\leq k$

[TABLE]

3. Examples

Example 3.1**.**

Set $a=3$ and $b=1$. Then we have

[TABLE]

and

[TABLE]

Note that if $|E_{n}^{k}(3)|=0$, then $k=n+2$. Therefore, for any $1\leq k\leq n+1$ we have $|E_{n}^{k}(3)|\not=0,$ that is, all order-$n$ minors of $E_{n}(3)$ are nonzero.

Example 3.2**.**

Set $n=10$ and $a=2$. Then $|E_{10}^{k}(2)|=0$ for all $k\in\{1,2,\ldots,11\}$.

Indeed, the equation $x^{3}+2x^{2}+2x+1=(x+1)(x^{2}+x+1)=0$ has solutions $-1,\omega,\omega^{2}$, where $\omega=\dfrac{-1+\sqrt{3}\iota}{2}.$ Then we have

[TABLE]

since $\omega^{3}=1$. So the sequence $\{d_{k}\}_{k=-1}^{\infty}$ is periodic with the period 6 and

[TABLE]

By Theorem 2.4 it is easily computed that $|E_{10}^{1}(2)|=|E_{10}^{2}(2)|=\cdots=|E_{10}^{6}(2)|=0$, and we can get $|E_{10}^{k}(2)|=0$ for all $k\in\{1,2,\ldots,11\}$.

4. Invertibility of $E^{k}_{n}(a)$

In this section if $|a|\geq 5$, we show that for $1\leq k\leq n+1$ the matrix $E_{n}^{k}(a)$, which is obtained by deleting the $k$-th row of $E_{n}(a)$, is invertible.

The following lemma is useful.

Lemma 4.1**.**

Let $a(i)$, $b(i)$, $c(i)$, $d(i)\in{\mathbb{C}}$ with $|a(i)|=|d(i)|=1$ and $|b(i)|=|c(i)|=\alpha\geq 5$ $(i=1,2,\dots,k-1)$. If

[TABLE]

then $|x_{l+1}|\geq(\alpha-2)|x_{l}|$  $(l=1,2,\dots,k-1)$.

Proof.

Since $c(1)x_{1}+d(1)x_{2}=0$, we have $|x_{2}|=\alpha|x_{1}|\geq(\alpha-2)|x_{1}|$. Since $b(2)x_{1}+c(2)x_{2}+d(2)x_{3}=0$, we have

[TABLE]

We assume that $|x_{l+1}|\geq(\alpha-2)|x_{l}|$ and $|x_{l+2}|\geq(\alpha-2)|x_{l+1}|$. Then the relation $a(l+2)x_{l}+b(l+2)x_{l+1}+c(l+2)x_{l+2}+d(l+2)x_{l+3}=0$ implies that

[TABLE]

Hence, we have $|x_{l+1}|\geq(\alpha-2)|x_{l}|$  $(l=1,2,\dots,k-1)$ inductively.

Proposition 4.2**.**

For $a\in{\mathbb{C}}$ with $|a|\geq 5$ and $1\leq k\leq n+1$ the matrix $E_{n}^{k}(a)$ is invertible, that is, all order-$n$ minors of $E_{n}(a)$ are nonzero.

Proof.

For $1\leq k\leq n+1$ and $\boldsymbol{x}=(x_{1},x_{2},\ldots,x_{n})^{t}\in{\mathbb{C}}^{n}$, we will show that if $E_{n}^{k}(a)\boldsymbol{x}=\boldsymbol{0}$, then $\boldsymbol{x}=\boldsymbol{0}$.

(1) When $k=n+1$, $E_{n}^{n+1}(a)\boldsymbol{x}=\boldsymbol{0}$ means that

[TABLE]

Apply Lemma 4.1 for (*), we have

[TABLE]

By (**) we have

[TABLE]

So we can get $0=x_{n}=x_{n-1}=\cdots=x_{1}$.

(2) When $k=1$, we set $\boldsymbol{y}=(x_{n},x_{n-1},\ldots,x_{1})^{t}\in{\mathbb{C}}^{n}$. The equation $E_{n}^{1}(a)\boldsymbol{x}=\boldsymbol{0}$ is equivalent to $E_{n}^{n+1}(a)\boldsymbol{y}=\boldsymbol{0}$. So we have $\boldsymbol{y}=\boldsymbol{0}$, that is $\boldsymbol{x}=\boldsymbol{0}$.

(3) When $2\leq k\leq n$, $E_{n}^{k}(a)\boldsymbol{x}=\boldsymbol{0}$ means that

[TABLE]

Apply Lemma 4.1 for (*) and (**) like as (1) and (2), we have

[TABLE]

Since $|x_{k}|\geq(|a|-2)|x_{k}|$ and $|x_{k-1}|\geq(|a|-2)|x_{k}|$, we have $x_{k-1}=x_{k}=0$ and $\boldsymbol{x}=\boldsymbol{0}$.

Remark 4.3**.**

Proposition 4.2 is true even if we replace $E_{n}(a)$ by $\tilde{E}_{n}(a)$ with $|a|\geq 5$, where

[TABLE]

Lemma 4.4**.**

Let $a$ be a complex number with $|a|\geq 2$ and set

[TABLE]

Then $H_{n}(a)$ is invertible.

Proof.

We set $e_{n}=|H_{n}(a)|$. Then the sequence $\{e_{n}\}_{n=1}^{\infty}$ satisfies

[TABLE]

Let $x^{2}-ax+1=(x-\alpha)(x-\beta)$. Then we can get

[TABLE]

If $\alpha^{n+1}-\beta^{n+1}=0$ for some $n$, then we have $\alpha=\beta=1$ or $\alpha=\beta=-1$ because $|\alpha+\beta|=|a|\geq 2$ and $\alpha\beta=1$. This implies, for $n=0,1,2,\ldots$,

[TABLE]

So we have $e_{n}\neq 0$, that is, $H_{n}(a)$ is invertible for $|a|\geq 2$.

For any $n\in{\mathbb{N}}$ and a number $a\in{\mathbb{C}}$, set $(n+2)\times n$ matrix

[TABLE]

From the same argument as in Proposition 4.2 we can get the following:

Proposition 4.5**.**

For $1\leq i<j\leq n+2$ and a number $a\in\mathbb{C}$ with $|a|\geq 2$, let $G_{n}^{i,j}(a)$ be a matrix which is obtained by deleting $i$-th and $j$-th rows of $G_{n}(a)$. Then $G_{n}^{i,j}(a)$ is invertible, that is, all order-$n$ minors of $G_{n}(a)$ are nonzero.

Proof.

For $1\leq i<j\leq n+2$ and $\boldsymbol{x}=(x_{1},x_{2},\ldots,x_{n})^{t}\in\mathbb{C}^{n}$, we will show that if $G_{n}^{i,j}(a)\boldsymbol{x}=\boldsymbol{0}$, then $\boldsymbol{x}=\boldsymbol{0}$.

When $i=1,j=n+2$, $G_{n}^{1,n+2}(a)\boldsymbol{x}=H_{n}(a)\boldsymbol{x}=\boldsymbol{0}$ implies $\boldsymbol{x}=\boldsymbol{0}$ by Lemma 4.4.

When $1<i<j=n+2$, $G_{n}^{i,n+2}(a)\boldsymbol{x}=\boldsymbol{0}$ is equivalent to

[TABLE]

where $H^{1}_{n-i+2}$ is obtained by deleting the 1st row of $H_{n-i+2}$.

[TABLE]

Since $J_{k}(a)$ is invertible, this condition is also equivalent to

[TABLE]

By Lemma 4.4, we have $\boldsymbol{x}=\boldsymbol{0}$.

When $1=i<j<n+2$, $G_{n}^{1,j}(a)\boldsymbol{x}=\boldsymbol{0}$ is equivalent to

[TABLE]

So we have $\boldsymbol{x}=\boldsymbol{0}$.

When $1<i<j<n+2$, $G_{n}^{i,j}(a)\boldsymbol{x}=\boldsymbol{0}$ is equivalent to

[TABLE]

and also equivalent to

[TABLE]

So we have $\boldsymbol{x}=\boldsymbol{0}$.

5. Subspaces of Maximal Dimension with Bounded Schmidt Rank

In this section we use the $(n+1)\times n$ matrix $E_{n}(a)$ ($$a=3 or |a|\geq 5$$) to the construction of subspaces of ${\mathbb{C}}^{m}\otimes{\mathbb{C}}^{n}$ of maximal dimension, which do not contain any nonzero vector of Schmidt rank less than 4. Moreover, using the $(n+2)\times n$ matrix $G_{n}(a)$ we can construct subspaces $\mathcal{T}\subset\mathcal{S}$ of ${\mathbb{C}}^{m}\otimes{\mathbb{C}}^{n}$ such that any vector in $\mathcal{T}\backslash\{\boldsymbol{0}\}$ has Schmidt rank greater than or equal 4, but there is at least one vector with Schmidt rank $3$ in $\mathcal{S}\backslash{\mathcal{T}}$.

For $n\in\mathbb{N}$ and a number $a\in\mathbb{C}$, we consider the $(n+3)\times n$ matrix $B_{n}(a)$ (resp. $\tilde{B}_{n}(a)$) as follows:

[TABLE]

[TABLE]

Proposition 5.1**.**

For $1\leq i<j<k\leq n+3$ and $|a|\geq 5$, let $B_{n}^{i,j,k}$ be a matrix which is obtained by deleting $i$-th, $j$-th and $k$-th rows of $B_{n}(a)$. Then $B_{n}^{i,j,k}(a)\in\mathbb{M}_{n}(\mathbb{C})$ is invertible.

Proof.

It suffices to show that $B_{n}^{i,j,k}(a)\boldsymbol{x}=\boldsymbol{0}$ implies $\boldsymbol{x}=\boldsymbol{0}$, where $\boldsymbol{x}=(x_{1},x_{2},\ldots,x_{n})^{t}$.

(1) When $1=i<j<k=n+3$, $B_{n}^{1,j,n+3}(a)=E_{n}^{j-1}(a)$ is invertible by Proposition 4.2. So $\boldsymbol{x}=\boldsymbol{0}$.

(2) When $1<i$, $i+1<j<k=n+3$, $B_{n}^{i,j,n+3}(a)\boldsymbol{x}=\boldsymbol{0}$ is equivalent to

[TABLE]

where $l\in\{1,\ldots,i-1\}\cup\{i+1,\ldots,j-1\}\cup\{j+1,\ldots,n+2\}$. This condition is also equivalent to

[TABLE]

where $l\in\{i+1,\ldots,j-1\}\cup\{j+1,\ldots,n+2\}$. Then we have, by Lemma 4.1,

[TABLE]

Since $|x_{j-1}|\geq(|a|-2)|x_{j-2}|$ and $|x_{j-2}|\geq(|a|-2)|x_{j-1}|$, this implies $x_{j-1}=x_{j-2}=0$ and $\boldsymbol{x}=\boldsymbol{0}$.

(3) When $1<i$, $j=i+1<k=n+3$, $B_{n}^{i,i+1,n+3}(a)\boldsymbol{x}=\boldsymbol{0}$ is equivalent to

[TABLE]

where $l\in\{1,\ldots,i-1\}\cup\{i+2,\ldots,n+2\}$. This condition is also equivalent to

[TABLE]

By Proposition 4.2, we have $x_{i}=x_{i+1}=\cdots=x_{n}=0$ and $\boldsymbol{x}=\boldsymbol{0}$.

(4) When $1<i<j=n+2,k=n+3$, $B_{n}^{i,n+2,n+3}(a)\boldsymbol{x}=\boldsymbol{0}$ is equivalent to

[TABLE]

where $l\in\{1,\ldots,i-1\}\cup\{i+1,\ldots,n+1\}$. This condition is also equivalent to

[TABLE]

By Proposition 4.2, we have $x_{i}=x_{i+1}=\cdots=x_{n}=0$ and $\boldsymbol{x}=\boldsymbol{0}$.

(5) When $i=1<j<k<n+3$, $B_{n}^{1,j,k}(a)\boldsymbol{x}=\boldsymbol{0}$ is equivalent to

[TABLE]

So $\boldsymbol{x}=\boldsymbol{0}$ by (2), (3) and (4).

(6) When $1<i,j=i+1,k=i+2<n+3$, $B_{n}^{i,i+1,i+2}(a)\boldsymbol{x}=\boldsymbol{0}$ is equivalent to

[TABLE]

where $l\in\{1,\ldots,i-1\}\cup\{i+3,\ldots,n+3\}$. So we have $x_{1}=x_{2}=\cdots=x_{i-1}=0$ and $x_{i}=x_{i+1}=\cdots=x_{n}=0$, that is, $\boldsymbol{x}=\boldsymbol{0}$.

(7) When $1<i,i+1<j,k=j+1<n+3$, $B_{n}^{i,j,j+1}(a)\boldsymbol{x}=\boldsymbol{0}$ is equivalent to

[TABLE]

where $l\in\{1,\ldots,i-1\}\cup\{i+1,\ldots,j-1\}\cup\{j+2,\ldots,n+3\}$. This condition is also equivalent to

[TABLE]

By Proposition 4.2, we have $x_{i}=x_{i+1}=\cdots=x_{j-2}=0$ and $\boldsymbol{x}=\boldsymbol{0}$.

(8) When $1<i,j=i+1,j+1<k<n+3$, $B_{n}^{i,i+1,k}(a)\boldsymbol{x}=\boldsymbol{0}$ is equivalent to

[TABLE]

So $\boldsymbol{x}=\boldsymbol{0}$ by (7).

(9) When $1<i$, $i+1<j$, $j+1<k<n+3$, $B_{n}^{i,j,k}(a)\boldsymbol{x}=\boldsymbol{0}$ is equivalent to

[TABLE]

where $l\in\{1,\ldots,i-1\}\cup\{i+1,\ldots,j-1\}\cup\{j+1,\ldots,n+3\}$. This condition implies

[TABLE]

Since $|x_{j-1}|\geq(|a|-2)|x_{j-2}|$ and $|x_{j-2}|\geq(|a|-2)|x_{j-1}|$, we have $x_{j-1}=x_{j-2}=0$ and $\boldsymbol{x}=\boldsymbol{0}$.

Remark 5.2**.**

We consider the following system of equations:

[TABLE]

Then we have

[TABLE]

This means that, if $x_{1}\neq 0$, then $|x_{1}|<|x_{2}|<\cdots<|x_{n}|$. Applying this fact for (1) and (9) in the proof of Proposition 5.1, we can show that $B_{n}^{i,j,k}(3)$ is invertible for $1\leq i<j<k\leq n+3$.

Proposition 5.3**.**

For $1\leq i<j<k\leq n+3$ and $|a|\geq 5$ let $\tilde{B}_{n}^{i,j,k}(a)$ be a matrix which is obtained by deleting the $i,j,k$-th rows of $\tilde{B}_{n}(a)$. Then $|\tilde{B}_{n}^{i,j,k}(a)|\not=0$.

Proof.

It follows from the same argument as in Proposition 5.1 using Remark 4.3.

Corollary 5.4**.**

For $a=3$ or $|a|\geq 5$, the columns of $B_{n}(a)$ are linearly independent such that any nonzero linear combination of these columns has at least $4$ nonzero entries.

Proof.

Suppose that there are $\lambda_{1},\lambda_{2},\dots,\lambda_{n}\in{\mathbb{C}}$ such that $\sum\limits_{i=1}^{n}\lambda_{i}e_{i}=(x_{1},x_{2},\dots,x_{n+3})^{t}$ has less than $4$ nonzero entries, where $e_{i}$  $(1\leq i\leq n)$ are columns in $B_{n}(a)$ and $x_{i}\in{\mathbb{C}}$  $(1\leq i\leq n+3)$. Assume that for distinct elements $l,j$ and $k$ in $\{1,2,\ldots,n+3\},$ we have $x_{i}=0$ for all $i\in\{1,2,\ldots,n+3\}\backslash\{l,j,k\}.$ Then we have

[TABLE]

Since $B_{n}^{i,j,k}(a)$ is invertible by Proposition 5.1 and Remark 5.2, we get each $\lambda_{i}=0$ $(1\leq i\leq n)$, that is, $\sum\limits_{i=1}^{n}\lambda_{i}e_{i}=(x_{1},x_{2},\dots,x_{n+3})^{t}=\boldsymbol{0}$. This is a contradiction.

Corollary 5.5**.**

For $a=3$ or $|a|\geq 5$ the columns of $\tilde{B}_{n}(a)$ are linearly independent such that any nonzero linear combination of these columns has at least $4$ nonzero entries.

Proof.

When $|a|\geq 5$, it follows from Proposition 5.3 and the same argument as in Corollary 5.4.

When $a=3$, the matrix $\tilde{B}_{n}(3)$ implies the following system of equations:

[TABLE]

Then we have

[TABLE]

and $|x_{1}|<|x_{2}|<\cdots<|x_{n}|$ if $x_{1}\neq 0$. By the similar reason as Remark 5.2, we can show that $\tilde{B}_{n}^{i,j,k}(3)$ is invertible for $1\leq i<j<k\leq n+3$.

Theorem 5.6**.**

Let $m$ and $n$ be natural numbers such that $4\leq\min\{m,n\}$. Let $N=n+m-2$, and $\{|e_{i}\rangle\}_{i=1}^{m}$ (resp. $\{|f_{j}\rangle\}_{j=1}^{n}$) be the canonical basis for ${\mathbb{C}}^{m}$ (resp. ${\mathbb{C}}^{n}$). For $3\leq d\leq N-3$ define

[TABLE]

and $\mathcal{S}=\bigoplus\limits_{d=0}^{N}\mathcal{S}^{(d)}$. If $a=3$ or $|a|\geq 5$, then $\mathcal{S}$ does not contain any nonzero vector of Schmidt rank $\leq 3$ and $\dim\mathcal{S}=(m-3)(n-3)$.

Proof.

Let $\phi\colon{\mathbb{C}}^{m}\otimes{\mathbb{C}}^{n}\rightarrow M_{m\times n}({\mathbb{C}})$ be the linear isomorphism defined by $\phi(|\eta\rangle)=[c_{ij}]$ for each $|\eta\rangle=\sum_{i,j}c_{ij}|e_{i}\rangle\otimes|f_{j}\rangle\in{\mathbb{C}}^{m}\otimes{\mathbb{C}}^{n}$. Then $|\eta\rangle$ has Schmidt rank at least $r$ if and only if the corresponding matrix $[c_{ij}]$ is of rank at least $r$. Also, it is known that a matrix has rank at least $r$ if and only if it has a nonzero minor of order $r$. Thus, it is enough to construct a set of $(m-3)(n-3)$ linearly independent matrices, which are image under $\phi$ of a basis of $\mathcal{S},$ such that any nonzero linear combination of these matrices has a nonzero minor of order 4.

Label the anti-diagonals of any $m\times n$ matrix by non-negative integers $k,$ such that the first anti-diagonal from upper left (of length one) is labeled $k=0$ and value of $k$ increases from upper left to lower right. Let the length of the $k$-th anti-diagonal be denoted by

$[k]$. Note that if $m\leq n$, then

[TABLE]

Recall that for $3\leq k\leq N-3$, $\mathcal{S}^{(k)}$ is generated by the set

[TABLE]

Note that any element of $\phi(B_{k})$ is the matrix obtained from the $m\times n$ zero matrix by replacing the $k$-th anti-diagonal by $(0,\dots,0,-1,a,-a,1,0,\dots,0).$ For $3\leq k\leq N-3$, from Corollary 5.4, $\phi(B_{k})$ is a set of $[k]-3$ linearly independent matrices. Also, by Corollary 5.4 it follows that if $M$ is a matrix obtained by taking any nonzero linear combination of matrices from $\phi(B_{k}),$ then $M$ has at least $4$ nonzero entries in the $k$-th anti-diagonal and the entries of $M$, other than those on $k$-th anti-diagonal, are zeros. Since the determinant of the $4\times 4$ submatrix with these $4$ nonzero elements in its principal anti-diagonal is clearly nonzero, any linear combination of the matrices in $\phi(B_{k})$ has at least one nonzero order-$4$ minor, thus has rank at least $4.$

Let $B=\displaystyle\bigcup_{k=3}^{N-3}B_{k}$. Since elements from different $\phi(B_{k})$ have different nonzero anti-diagonal, $\phi(B)$ is linearly independent. Thus $B$ is a basis for $\mathcal{S}.$ Let $C$ be a matrix obtained by an arbitrary linear combination from the elements of $\phi(B)$. Let $\kappa$ be the largest $k$ for which the linear combination involves an element from $\phi(B_{k})$. The $\kappa$-th anti-diagonal of $C$ has at least $4$ nonzero elements. Because $\kappa$ labels the bottom-rightmost anti-diagonal of $C$ that contains nonzero elements, the $4\times 4$ submatrix of $C,$ with these 4 nonzero elements in the principal anti-diagonal, has only zero entries in all its anti-diagonals which are below the principal anti-diagonal. Hence the $4\times 4$ submatrix has nonzero determinant. Thus, the rank of $C$ is at least 4. We conclude that $\mathcal{S}$ does not contain any nonzero vector of Schmidt rank $\leq 3.$

The dimension of $\mathcal{S}$ is equal to the cardinality of $B$. Hence, the dimension of $\mathcal{S}$ is given by

[TABLE]

A similar argument holds in the case $m>n$.

Remark 5.7**.**

From the general fact on Schmidt rank it follows that all the elements of the basis

[TABLE]

of $\mathcal{S}$ have Schmidt rank $4$ if and only if $a\neq 0$.

Remark 5.8**.**

Theorem 5.6 is true even if we replace $\mathcal{S}$ by $\tilde{\mathcal{S}}$ with $a=3$ or $|a|\geq 5$ as follows:

[TABLE]

and $\mathcal{\tilde{S}}=\bigoplus\limits_{d=0}^{N}\mathcal{\tilde{S}}^{(d)}$. Note that it follows from Corollary 5.5.

Theorem 5.9**.**

Let $m$ and $n$ be natural numbers such that $4\leq\min\{m,n\}$. Let $N=n+m-2$, and $\{|e_{i}\rangle\}_{i=1}^{m}$ (resp. $\{|f_{j}\rangle\}_{j=1}^{n}$) be the canonical basis for ${\mathbb{C}}^{m}$ (resp. ${\mathbb{C}}^{n}$). For $3\leq d\leq N-3$ define

[TABLE]

and $\mathcal{T}=\bigoplus\limits_{d=0}^{N}\mathcal{T}^{(d)}$. Similarly, for $2\leq d\leq N-2$ define

[TABLE]

and $\mathcal{S}=\bigoplus\limits_{d=0}^{N}\mathcal{S}^{(d)}$. Then,

1. (1)

$\mathcal{T}\subset\mathcal{S}$. 2. (2)

If $|a|\geq 2$, then $\mathcal{S}$ does not contain any nonzero vector of Schmidt rank $\leq 2$ and $\dim\mathcal{S}=(m-2)(n-2)$. 3. (3)

If $a>4$, then $\mathcal{T}$ does not contain any nonzero vector of Schmidt rank $\leq 3$ and $\dim\mathcal{T}=(m-3)(n-3)$. 4. (4)

If $a>4$, then there is at least one vector with Schmidt rank $3$ in $\mathcal{S}\backslash{\mathcal{T}}$.

Proof.

(1) For any $|e_{i-1}\rangle\otimes|f_{j+2}\rangle+(a+1)|e_{i}\rangle\otimes|f_{i}\rangle+(a+1)|e_{i+1}\rangle\otimes|f_{j}\rangle+|e_{i+2}\rangle\otimes|f_{j-1}\rangle\in\mathcal{T}^{(d)}$ we take $|e_{i}\rangle\otimes|f_{j+1}\rangle+a|e_{i+1}\rangle\otimes|f_{j}\rangle+|e_{i+2}\rangle\otimes|f_{j-1}\rangle=|e_{i}\rangle\otimes|f_{(j-1)+2}\rangle+a|e_{i+1}\rangle\otimes|f_{(j-1)+1}\rangle+|e_{i+2}\rangle\otimes|f_{j-1}\rangle$ $|e_{i-1}\rangle\otimes|f_{j+2}\rangle+a|e_{i}\rangle\otimes|f_{j+1}\rangle+|e_{i+1}\rangle\otimes|f_{j}\rangle=|e_{i-1}\rangle\otimes|f_{j+2}\rangle+a|e_{(i-1)+1}\rangle\otimes|f_{j+1}\rangle+|e_{(i-1)+2}\rangle\otimes|f_{j}\rangle$ $\in\mathcal{S}^{(d)}$. Then

[TABLE]

Hence, $\mathcal{T}^{(d)}\subset\mathcal{S}^{(d)}$, and $\mathcal{T}\subset\mathcal{S}$.

(2) Using Proposition 4.5

we know that the columns of $G_{n}(a)$ are linearly independent such that any nonzero linear combination of these columns has at least 3 nonzero entries. Let

[TABLE]

Then any element of $\phi(C_{k})$ is the matrix by replacing the $k$-th anti-diagonal by $(0,\dots,0,1,a,1,0,\dots,0)$. For $2\leq k\leq N-2$, from the above observation, $\phi(C_{k})$ is the set of $[k]-2$ linearly independent, where as in the proof of Theorem 5.6, if $m\leq n$, then

[TABLE]

Also, it follows that if $M$ is a matrix obtained by taking any nonzero linear combinations of matrices from $\phi(C_{k})$, then $M$ has at least 3 nonzero entries in the $k$-th anti-diagonal, and the entries of $M$, other than those on $k$-th anti-diagonal are zeros. Since the determinant of the $3\times 3$ submatrix with these $3$ nonzero entries in its principal ant-diagonal is clearly nonzero, any linear combination of the matrices in $\phi(C_{k})$ has at least one nonzero order-$3$ minor, thus has rank at least $3$.

Hence the dimension of $\mathcal{S}$ is given by

[TABLE]

A similar argument holds in the case $m>n$.

(3) Since $a>4$, $a+1>5$. Hence, it follows from Remark 5.8.

(4) Each generator $|e_{i}\rangle\otimes|f_{j+2}\rangle+a|e_{i+1}\rangle\otimes|f_{j+1}\rangle+|e_{i+2}\rangle\otimes|f_{j}\rangle$ in $\mathcal{S}^{(d)}$  $(2\leq d\leq N-2)$ has Schmidt rank $3$.

Remark 5.10**.**

In Theorem 5.9 (3) if we replace the condition $a>4$ by the condition $|a+1|>5$, then it would give more examples.

Theorem 5.11**.**

Let $m$ and $n$ be natural numbers such that $4\leq\min\{m,n\}$. Let $N=n+m-2$, and $\{|e_{i}\rangle\}_{i=1}^{m}$ (resp. $\{|f_{j}\rangle\}_{j=1}^{n}$) be the canonical basis for ${\mathbb{C}}^{m}$ (resp. ${\mathbb{C}}^{n}$) and set $g(i,j)=|e_{i+1}\rangle\otimes|f_{j+1}\rangle+a|e_{i}\rangle\otimes|f_{j+2}\rangle$. Consider

[TABLE]

When $a>0$ and $a+\dfrac{1}{a}>4$, we have the similar observation as in Theorem 5.9 as follows:

1. (1)

$\mathcal{U}\subset\mathcal{T}\subset\mathcal{S}$, 2. (2)

Any element in $\mathcal{S}$ has Schmidt rank $\geq 2$, any generator in $\mathcal{S}$ has Schmidt rank $2$, and $\dim\mathcal{S}=(m-1)(n-1)$, 3. (3)

Any element in $\mathcal{T}$ has Schmidt rank $\geq 3$, any generator in $\mathcal{T}$ has Schmidt rank $3$, and $\dim\mathcal{T}=(m-2)(n-2)$, 4. (4)

Any element in $\mathcal{U}$ has Schmidt rank $\geq 4$, any generator in $\mathcal{U}$ has Schmidt rank $4$, and $\dim\mathcal{U}=(m-3)(n-3)$.

Proof.

(1) It is obvious.

(2) As in Theorem 5.9 we consider the $(n+1)\times n$ matrix $K_{n}(a)$

[TABLE]

Then we know that the columns of $K_{n}(a)$ are linearly independent such that any nonzero linear combination of these columns has at least 2 nonzero entries. Let

[TABLE]

Then any element of $\phi(D_{k})$ is the matrix by replacing the $k$-th anti-diagonal by $(0,\dots,0,1,a,0,\dots,0)$. For $1\leq k\leq N-1$, from the above observation, $\phi(D_{k})$ is the set of $[k]-1$ linearly independent, where as in the proof of Theorem 5.6, if $m\leq n$, then

[TABLE]

Also, it follows that if $M$ is a matrix obtained by taking any nonzero linear combinations of matrices from $\phi(D_{k})$, then $M$ has at least 2 nonzero entries in the $k$-th anti-diagonal, and the entries of $M$, other than those on $k$-th anti-diagonal are zeros. Since the determinant of the $2\times 2$ submatrix with these $2$ nonzero entries in its principal anti-diagonal is clearly nonzero, any linear combination of the matrices in $\phi(D_{k})$ has at least one nonzero order-$2$ minor, thus has rank at least $2$.

Hence the dimension of $\mathcal{S}$ is given by

[TABLE]

A similar argument holds in the case $m>n$.

(3) and (4) follow from the same argument in Theorem 5.6.

6. Acknowledgement

Part of this research was carried out during the fourth author’s stay at Department of Mathematics and Informatics, Chiba University. He would like to appreciate this institute for their hospitality. Research of the fourth author is partially supported by JSPS KAKENHI Grant Number JP17K05285.