A Solution to a Permanent Conjecture of Pate

1. Introduction

Let $A=\left(a_{ij}\right)$ be an $n\times n$ complex matrix. The permanent of A is defined by replacing the alternating signs in the determinant expansion with positive signs. Explicitly, $perA={\sum }_{\sigma \in S_n}{\prod }_{k=1}^n{a}_{k\sigma \left(k\right)}$, where $S_n$ denotes the symmetric group on $\{1,2,\dots ,n\}$.

Permanent theory is an active area of research in matrix theory and combinatorics. Despite extensive study, many intriguing questions and long-standing conjectures remain open; see, e.g., [2,3,7,8,9,12].

We denote by $M_n\left(\mathbb{C}\right)$ the set of all $n\times n$ complex matrices. For $A\in M_n\left(\mathbb{C}\right)$ with $n\ge 2$, let $A\left(i\right|j)$ be the submatrix obtained by deleting row i and column j of A, where $1\le i,j\le n$. Our main result is the following statement.

Theorem 1. 

Let $A=\left(a_{ij}\right)$ be an $n\times n$ positive semidefinite matrix. Then,

$$\begin{array}{cc}\hfill & a_{ii}perA\left(i\right|i)-a_{ij}perA\left(i\right|j)\hfill \\ \hfill & -a_{ji}perA\left(j\right|i)+a_{jj}perA\left(j\right|j)\le 2perA.\hfill \end{array}$$

(1)

Inequality (1) appears as Conjecture 11 in Wanless’s article [11]. It originates from Pate’s article [10] as Inequality (3) (with $i=1,j=2$), where Pate asks whether the inequality holds and remarks that it is “not likely to be easy.” We answer in the affirmative the question of Pate and Wanless.

2. A Lemma

We begin with the following lemma.

Lemma. 

Let $J_x,J_y,$ and $J_z$ be self-adjoint operators on a finite-dimensional complex Hilbert space. Suppose that $J_x,J_y,$ and $J_z$ satisfy, for some $j\ge 0$,

$$[J_x,J_y]=-i{J}_z,J_x^2+J_y^2+J_z^2=j(j+1)I.$$

(2)

(Here, $\left[\right]$ denotes commutator.) Let $M=j+\frac{3}{2}$. Then, for any $\theta \in [0,1]$,

$${(J_z+\theta MI)}^2+(1-{\theta }^2)J_y^2⪰{\displaystyle \frac{{\theta }^2}{4}}I.$$

(3)

(Here, ⪰ denotes the Loewner partial order.)

Proof. 

For an arbitrarily fixed unit vector $\psi$, we write

$$z=\parallel J_z{\psi \parallel }^2,\eta =\parallel J_y{\psi \parallel }^2,\xi ={\parallel J_x\psi \parallel }^2,\mu =\langle J_z\psi ,\psi \rangle .$$

The Cauchy–Schwarz inequality immediately yields two inequalities:

$${\mu }^2={\left|\mu \right|}^2=|\langle J_z\psi ,\psi \rangle {|}^2\le {\parallel J_z\psi \parallel }^2=z$$

(4)

and

$$\left|\mu \right|=\left|\langle [J_x,J_y]\psi ,\psi \rangle \right|=2\left|Im\langle J_x\psi ,J_y\psi \rangle \right|\le 2\sqrt{\xi \eta }.$$

(5)

By (2), we obtain

$$\xi +z+\eta =j(j+1).$$

(6)

Set $C=(j+1)(j+2)=M^2-\frac{1}{4}$. Then, $C-\eta \ge 2(j+1)>0$ as $\eta \le j(j+1)$.

To show (3), we evaluate the quadratic form of the left-hand side of (3) minus ${\displaystyle \frac{{\theta }^2}{4}}I$ at $\psi$ and express it as a quadratic function in $\theta$:

$$q\left(\theta \right)=z+\eta +2\theta M\mu +{\theta }^2(C-\eta ).$$

(7)

Thus, to show (3), or equivalently, $q\left(\theta \right)\ge 0$, it suffices to show that

$$M^2{\mu }^2\le (z+\eta )(C-\eta ).$$

(8)

Let $R=(z+\eta )(C-\eta )$. By (4), we see that (8) holds if $R\ge M^{2}z$.

We now show that (8) also holds for the case where $R<M^{2}z$.

Note that $R<M^{2}z$ is $(z+\eta )(C-\eta )<(C+{\displaystyle \frac{1}{4}})z$, which is equivalent to

$$z>{\displaystyle \frac{\eta (C-\eta )}{\eta +\frac{1}{4}}}:=z_0.$$

(9)

Define a linear function

$$\Phi \left(t\right)=(t+\eta )(C-\eta )-4M^2\eta \left(j(j+1)-t-\eta \right).$$

Using (6), we have

$$\Phi \left(z\right)=R-4M^2\eta \xi .$$

(10)

Observe that ${\Phi }^{\prime }\left(t\right)=C(1+4\eta )>0$ and that $\Phi \left(t\right)$ has the root

$$r={\displaystyle \frac{4M^2\eta (j(j+1)-\eta )-\eta (C-\eta )}{C(1+4\eta )}}.$$

One checks by computation that $z_0-r={\displaystyle \frac{2\eta {(2j+3)}^2}{(4\eta +1)(j+2)}}\ge 0$. Thus, $\Phi \left(z_0\right)\ge \Phi \left(r\right)=0$. (One may also show directly that $\Phi \left(z_0\right)=2\eta (j+1){(2j+3)}^2\ge 0$.)

By (9), $z>z_0$. So, $\Phi \left(z\right)>\Phi \left(z_0\right)\ge 0$. By (5) and (10), we have

$$R-M^2{\mu }^2\ge R-4M^2\eta \xi =\Phi \left(z\right)>0,$$

(11)

which proves (8) in the second case. This completes the proof. □

3. Symmetric Tensors and Creation Operators.

In this section, we establish a norm inequality for tensors with creation operators that will be used to prove our main result (Theorem 1). The inequality is also of interest in its own right.

We begin with some basic concepts in tensor theory. Let $\mathcal{K}$ be a finite-dimensional complex Hilbert space and let d be a nonnegative integer. The dth tensor power ${\mathcal{K}}^{\otimes d}$, see, e.g., ([6] p. 141), is equipped with the inner product

$$\langle u_1\otimes \dots \otimes u_d,w_1\otimes \dots \otimes w_d\rangle ={\displaystyle \prod _{\ell =1}^d}\langle u_{\ell },w_{\ell }\rangle$$

and the symmetrization operator

$$P_d(u_1\otimes \dots \otimes u_d)={\displaystyle \frac{1}{d!}}\sum _{\sigma \in S_d}{u}_{\sigma \left(1\right)}\otimes \dots \otimes u_{\sigma \left(d\right)}$$

is the orthogonal projection onto ${\mathcal{K}}^{\odot d}$; see ([6] p. 153). We write $u_1\odot \dots \odot u_d$ for the displayed symmetrized tensor and use the inner product inherited from ${\mathcal{K}}^{\otimes d}$. We set ${\mathcal{K}}^{\otimes 0}={\mathcal{K}}^{\odot 0}=\mathbb{C}$ and $P_0=I_{\mathbb{C}}$; the empty symmetric product is 1. For $w\in \mathcal{K}$ and $g\in {\mathcal{K}}^{\odot d}$, define

$$w\odot g=P_{d+1}(w\otimes g),c_w^{*}g=\sqrt{d+1}w\odot g.$$

For $d\ge 1$, define $c_w:{\mathcal{K}}^{\odot d}\to {\mathcal{K}}^{\odot (d-1)}$ as the adjoint of $c_w^{*}:{\mathcal{K}}^{\odot (d-1)}\to {\mathcal{K}}^{\odot d}$, and set $c_{w}1=0$. Thus, $c_w^{*}$ raises the degree by one, whereas $c_w$ lowers it by one. We compose these operators on the algebraic direct sum

$${\mathcal{F}}_{\mathrm{alg}}\left(\mathcal{K}\right)={\displaystyle \underset{d\ge 0}{\overset{\mathrm{alg}}{⨁}}}{\mathcal{K}}^{\odot d},$$

whose elements are finite sums of homogeneous symmetric tensors. Different degrees are orthogonal. Adjoints are taken between the corresponding homogeneous spaces.

To compute these operators, choose an orthonormal basis $e_1,\dots ,e_q$ of $\mathcal{K}$. For a multi-index $\mathbf{n}=(n_1,\dots ,n_q)$ with nonnegative integer entries, write $\left|\mathbf{n}\right|=n_1+\dots +n_q$ and $\mathbf{n}!=n_1!\dots n_q!$. When $\left|\mathbf{n}\right|=d$, the symmetrization of $e_1^{\otimes n_1}\otimes \dots \otimes e_q^{\otimes n_q}$ has $d!/\mathbf{n}!$ distinct, mutually orthogonal elementary tensors, each with coefficient $\mathbf{n}!/d!$. Consequently,

$${∥e_1^{\odot n_1}\odot \dots \odot e_q^{\odot n_q}∥}^2={\displaystyle \frac{\mathbf{n}!}{d!}}.$$

Different multi-indices give orthogonal tensors, and their span is ${\mathcal{K}}^{\odot d}$. An orthonormal basis is therefore given by

$$E_{\mathbf{n}}=\sqrt{{\displaystyle \frac{d!}{\mathbf{n}!}}}{e}_1^{\odot n_1}\odot \dots \odot e_q^{\odot n_q},\left|\mathbf{n}\right|=d.$$

Here $E_{\mathbf{0}}=1$. If ${\epsilon }_{\ell }$ is the multi-index with a single 1 in position ℓ, the definition of $c_{e_{\ell }}^{*}$ gives

$$c_{e_{\ell }}^{*}{E}_{\mathbf{n}}=\sqrt{d+1}\sqrt{{\displaystyle \frac{d!}{\mathbf{n}!}}}\sqrt{{\displaystyle \frac{(\mathbf{n}+{\epsilon }_{\ell })!}{(d+1)!}}}{E}_{\mathbf{n}+{\epsilon }_{\ell }}=\sqrt{n_{\ell }+1}{E}_{\mathbf{n}+{\epsilon }_{\ell }}.$$

Taking adjoints on successive degrees gives

$$c_{e_{\ell }}{E}_{\mathbf{n}}=\left\{\begin{array}{cc}\sqrt{n_{\ell }}{E}_{\mathbf{n}-{\epsilon }_{\ell }},\hfill & n_{\ell }\ge 1,\hfill \\ 0,\hfill & n_{\ell }=0.\hfill \end{array}\right.$$

For an arbitrary symmetric product of degree $d\ge 1$, they give

$$c_w(w_1\odot \dots \odot w_d)={\displaystyle \frac{1}{\sqrt{d}}}{\displaystyle \sum _{\ell =1}^d}\langle w_{\ell },w\rangle w_1\odot \dots \odot \widehat{w_{\ell }}\odot \dots \odot w_d,$$

where the hat denotes omission. Both sides are multilinear in $w_1,\dots ,w_d$ and conjugate-linear in w. It is therefore enough to check the formula when these vectors are basis vectors, in which case it follows from the preceding formulas.

Theorem 2. 

Let $\mathcal{K}$ be a finite-dimensional complex Hilbert space, let m be a nonnegative integer, let $x,y\in \mathcal{K}$, and let $f\in {\mathcal{K}}^{\odot m}$. Then,

$${∥x\otimes c_y^{*}f-y\otimes c_x^{*}f∥}^2\le 2{∥c_x^{*}{c}_y^{*}f∥}^2.$$

(12)

The norm on the left is taken in $\mathcal{K}\otimes {\mathcal{K}}^{\odot (m+1)}$ and the norm on the right in ${\mathcal{K}}^{\odot (m+2)}$.

Proof. 

Step 1. Reduction to an orthonormal pair. If $x=0$, $y=0$, or $f=0$, both sides of (12) vanish. Since $w\mapsto c_w^{*}$ is linear, replacing $x,y$ by $\lambda x,\nu y$ multiplies both vectors in (12) by $\lambda \nu$. We may therefore assume $\parallel x\parallel =\parallel y\parallel =1$. Multiplying y by a scalar of modulus one leaves both norms unchanged, so we may also assume

$$c=\langle x,y\rangle \in [0,1].$$

If $c=1$, then ${\parallel x-y\parallel }^2=2-2c=0$, and the left side of (12) is zero. Suppose henceforth that $c<1$, and set

$$u={\displaystyle \frac{x+y}{\sqrt{2(1+c)}}},v={\displaystyle \frac{x-y}{\sqrt{2(1-c)}}},\alpha ={\displaystyle \frac{1+c}{2}},\beta ={\displaystyle \frac{1-c}{2}}.$$

The identities ${\parallel x+y\parallel }^2=2(1+c)$, ${\parallel x-y\parallel }^2=2(1-c)$, and

$$\langle x+y,x-y\rangle ={\parallel x\parallel }^2-\langle x,y\rangle +\langle y,x\rangle -{\parallel y\parallel }^2=0$$

show that $u,v$ are orthonormal. Solving for $x,y$ gives

$$x=\sqrt{\alpha }u+\sqrt{\beta }v,y=\sqrt{\alpha }u-\sqrt{\beta }v.$$

(13)

Write

$$\begin{array}{c}{a}^{*}=c_u^{*},b^{*}=c_v^{*},a=c_u,b=c_v,\\ N_1=a^{*}a,N_2=b^{*}b,N=N_1+N_2.\end{array}$$

Extend $u,v$ to an orthonormal basis $e_1=u,e_2=v,e_3,\dots ,e_q$ of $\mathcal{K}$. The preceding basis formulas give

$$\begin{array}{cccc}\hfill a^{*}a{E}_{\mathbf{n}}& =n_1{E}_{\mathbf{n}},\hfill & \hfill a{a}^{*}{E}_{\mathbf{n}}& =(n_1+1)E_{\mathbf{n}},\hfill \\ \hfill b^{*}b{E}_{\mathbf{n}}& =n_2{E}_{\mathbf{n}},\hfill & \hfill b{b}^{*}{E}_{\mathbf{n}}& =(n_2+1)E_{\mathbf{n}}.\hfill \end{array}$$

Operators acting on different coordinates commute. For example, $a{b}^{*}$ and $b^{*}a$ both send $E_{\mathbf{n}}$ to $\sqrt{n_1(n_2+1)}{E}_{\mathbf{n}-{\epsilon }_1+{\epsilon }_2}$ when $n_1\ge 1$, and both vanish when $n_1=0$. The other mixed commutators are checked in the same way. Thus, with $[A,B]=AB-BA$,

$$[a,a^{*}]=[b,b^{*}]=I,[a,b]=[a,b^{*}]=[a^{*},b]=[a^{*},b^{*}]=0$$

(14)

on ${\mathcal{F}}_{\mathrm{alg}}\left(\mathcal{K}\right)$.

Step 2. Decomposition into eigenspaces of N. For each degree d, the operator N is diagonal in the basis above:

$$N{E}_{\mathbf{n}}=(n_1+n_2)E_{\mathbf{n}}.$$

For $d\ge 0$ and $0\le k\le d$, let

$${\mathcal{H}}_{d,k}=span\{E_{\mathbf{n}}:\left|\mathbf{n}\right|=d,n_1+n_2=k\}.$$

This is the k-eigenspace of N on ${\mathcal{K}}^{\odot d}$. In particular,

$${\mathcal{K}}^{\odot m}={\displaystyle \underset{k=0}{\overset{m}{⨁}}}{\mathcal{H}}_{m,k}$$

is an orthogonal sum. We put ${\mathcal{H}}_{d,k}=\left\{0\right\}$ when $d<0$, $k<0$, or $k>d$. The basis formulas show that

$$a^{*},b^{*}:{\mathcal{H}}_{d,k}⟶{\mathcal{H}}_{d+1,k+1},a,b:{\mathcal{H}}_{d,k}⟶{\mathcal{H}}_{d-1,k-1},$$

with annihilation equal to zero on degree 0. Since $x,y$ belong to $span\{u,v\}$, (13) implies

$$c_x^{*}{c}_y^{*}{\mathcal{H}}_{m,k}\subseteq {\mathcal{H}}_{m+2,k+2},$$

and, for $f_k\in {\mathcal{H}}_{m,k}$,

$$x\otimes c_y^{*}{f}_k-y\otimes c_x^{*}{f}_k\in \mathcal{K}\otimes {\mathcal{H}}_{m+1,k+1}.$$

For distinct k, these target spaces are orthogonal on each side. Writing $f={\sum }_{k=0}^m{f}_k$ therefore gives

$$\begin{array}{cc}\hfill \parallel c_x^{*}{c}_y^{*}{f\parallel }^2& ={\displaystyle \sum _{k=0}^m}{\parallel c_x^{*}{c}_y^{*}{f}_k\parallel }^2,\hfill \\ \hfill \parallel x\otimes c_y^{*}f-y\otimes c_x^{*}{f\parallel }^2& ={\displaystyle \sum _{k=0}^m}{\parallel x\otimes c_y^{*}{f}_k-y\otimes c_x^{*}{f}_k\parallel }^2.\hfill \end{array}$$

It therefore suffices to prove (12) for $f\in {\mathcal{H}}_{m,k}$. We fix a nonzero space ${\mathcal{H}}_{m,k}$ and work with a vector f satisfying

$$Nf=kf,0\le k\le m.$$

(15)

Step 3. The two expressions in (12). By linearity and (13),

$$c_x^{*}=\sqrt{\alpha }{a}^{*}+\sqrt{\beta }{b}^{*},c_y^{*}=\sqrt{\alpha }{a}^{*}-\sqrt{\beta }{b}^{*}.$$

The mixed terms cancel because $a^{*}{b}^{*}=b^{*}{a}^{*}$, and so

$$c_x^{*}{c}_y^{*}=\alpha a^{*2}-\beta b^{*2}.$$

(16)

Expanding the tensor expression gives

$$x\otimes c_y^{*}f-y\otimes c_x^{*}f=2\sqrt{\alpha \beta }\left(v\otimes a^{*}f-u\otimes b^{*}f\right).$$

The two tensors on the right are orthogonal because $u\perp v$. Moreover, (14) and (15) give

$$\begin{array}{cc}\hfill \parallel a^{*}{f\parallel }^2+{\parallel b^{*}f\parallel }^2& =\langle (a{a}^{*}+b{b}^{*})f,f\rangle \hfill \\ & =\langle (N_1+N_2+2I)f,f\rangle =(k+2){\parallel f\parallel }^2.\hfill \end{array}$$

It follows that

$${∥x\otimes c_y^{*}f-y\otimes c_x^{*}f∥}^2=4\alpha \beta (k+2){\parallel f\parallel }^2.$$

(17)

Since $\alpha >0$, we may put $r=\beta /\alpha \in [0,1]$. Equation (16) reads $c_x^{*}{c}_y^{*}=\alpha (a^{*2}-r{b}^{*2})$. Thus, (12) for this f is equivalent to

$${∥(a^{*2}-r{b}^{*2})f∥}^2\ge 2r(k+2){\parallel f\parallel }^2.$$

(18)

Step 4. Calculation of $Q^{*}Q$ and $Q{Q}^{*}$. Set

$$\begin{array}{c}Q=a^2-r{b}^2,Q^{*}=a^{*2}-r{b}^{*2},\\ L=N_1-r{N}_2,\delta =1-r,Y=i(a^{*}b-b^{*}a).\end{array}$$

The operator Q lowers the degree by two, and $Q^{*}$ raises it by two. Both $Q{Q}^{*}$ and $Q^{*}Q$ preserve each homogeneous space and its N-eigenspaces. In particular, the maps

$$Q^{*}:{\mathcal{H}}_{m,k}⟶{\mathcal{H}}_{m+2,k+2},Q:{\mathcal{H}}_{m+2,k+2}⟶{\mathcal{H}}_{m,k}$$

are adjoints of one another. Hence,

$$\parallel Q^{*}{f\parallel }^2=\langle Q{Q}^{*}f,f\rangle .$$

The basis formulas, including the cases $n_1=0,1$, give

$$a^{*2}{a}^2=N_1(N_1-I),a^2{a}^{*2}=(N_1+I)(N_1+2I).$$

The corresponding identities with $b,N_2$ in place of $a,N_1$ also hold. Thus,

$$[a^2,a^{*2}]=4N_1+2I,[b^2,b^{*2}]=4N_2+2I.$$

To calculate $Y^2$, write $U=a^{*}b$ and $V=b^{*}a$. The mixed commutation relations give

$$\begin{array}{cccc}\hfill U^2& =a^{*2}{b}^2,\hfill & \hfill V^2& =b^{*2}{a}^2,\hfill \\ \hfill UV& =a^{*}b{b}^{*}a=N_1(N_2+I),\hfill & \hfill VU& =b^{*}a{a}^{*}b=N_2(N_1+I).\hfill \end{array}$$

Therefore

$$\begin{array}{cc}\hfill Y^2& =-{(U-V)}^2=-U^2-V^2+UV+VU\hfill \\ & =-a^{*2}{b}^2-b^{*2}{a}^2+2N_1{N}_2+N_1+N_2.\hfill \end{array}$$

Expanding $Q^{*}Q$ and substituting the preceding formula yields

$$\begin{array}{cc}\hfill Q^{*}Q& =N_1(N_1-I)+r^2{N}_2(N_2-I)-r(a^{*2}{b}^2+b^{*2}{a}^2)\hfill \\ & =N_1^2+r^2{N}_2^2-2r{N}_1{N}_2+r{Y}^2-(1+r)N_1-r(1+r)N_2.\hfill \end{array}$$

Since $N_1{N}_2=N_2{N}_1$, this is

$$Q^{*}Q=L^2+r{Y}^2-(1+r)(N_1+r{N}_2).$$

(19)

Also, $[a^2,b^{*2}]=[b^2,a^{*2}]=0$, so

$$[Q,Q^{*}]=[a^2,a^{*2}]+r^2[b^2,b^{*2}].$$

Consequently,

$$Q{Q}^{*}=Q^{*}Q+4N_1+4r^2{N}_2+2(1+r^2)I.$$

(20)

On ${\mathcal{H}}_{m,k}$, the equality $N_1+N_2=kI$ and (19)–(20) give

$$\begin{array}{cc}\hfill Q{Q}^{*}-2r(k+2)I& =L^2+r{Y}^2+3(1-r)N_1-3r(1-r)N_2+2{(1-r)}^{2}I\hfill \\ & =L^2+3\delta L+2{\delta }^{2}I+r{Y}^2.\hfill \end{array}$$

Factoring the polynomial in L gives

$$Q{Q}^{*}-2r(k+2)I=(L+\delta I)(L+2\delta I)+r{Y}^2$$

(21)

on ${\mathcal{H}}_{m,k}$.

Step 5. The operators $J_x,J_y,J_z$. On the finite-dimensional space ${\mathcal{H}}_{m,k}$, define

$$j={\displaystyle \frac{k}{2}},J_z={\displaystyle \frac{N_1-N_2}{2}},J_x={\displaystyle \frac{U+V}{2}},J_y={\displaystyle \frac{i(U-V)}{2}}={\displaystyle \frac{Y}{2}}.$$

The operators $N_1,N_2$ preserve ${\mathcal{H}}_{m,k}$. So do $U=a^{*}b$ and $V=b^{*}a$: they transfer one factor from the v-direction to the u-direction, or conversely, without changing either the total degree or $n_1+n_2$. Their restrictions satisfy $U^{*}=V$. Since $N_1,N_2$ are also self-adjoint on ${\mathcal{H}}_{m,k}$, so are $J_x,J_y,J_z$.

The formulas for $UV$ and $VU$ above give

$$[U,V]=N_1-N_2.$$

It follows that

$$[J_x,J_y]={\displaystyle \frac{i}{4}}[U+V,U-V]=-{\displaystyle \frac{i}{2}}[U,V]=-i{J}_z.$$

Furthermore,

$$\begin{array}{cc}\hfill J_x^2+J_y^2& ={\displaystyle \frac{1}{4}}\left({(U+V)}^2-{(U-V)}^2\right)\hfill \\ & ={\displaystyle \frac{1}{2}}(UV+VU)=N_1{N}_2+{\displaystyle \frac{N_1+N_2}{2}}.\hfill \end{array}$$

Adding $J_z^2$ yields

$$\begin{array}{cc}\hfill J_x^2+J_y^2+J_z^2& =N_1{N}_2+{\displaystyle \frac{N_1+N_2}{2}}+{\displaystyle \frac{{(N_1-N_2)}^2}{4}}\hfill \\ & ={\displaystyle \frac{{(N_1+N_2)}^2}{4}}+{\displaystyle \frac{N_1+N_2}{2}}={\displaystyle \frac{k(k+2)}{4}}I.\hfill \end{array}$$

Since $j=k/2$, we have proved

$$J_x^2+J_y^2+J_z^2=j(j+1)I.$$

(22)

Thus, $J_x,J_y,J_z$ satisfy the hypotheses of the lemma in Section 2.

Step 6. Application of the lemma. Put

$$s=1+r>0,\theta ={\displaystyle \frac{1-r}{1+r}}={\displaystyle \frac{\delta }{s}}.$$

Since $0\le r\le 1$, we have $0\le \theta \le 1$, and

$$s^2(1-{\theta }^2)={(1+r)}^2-{(1-r)}^2=4r.$$

On ${\mathcal{H}}_{m,k}$, the relations $N_1+N_2=2jI$ and $N_1-N_2=2J_z$ give

$$N_1=jI+J_z,N_2=jI-J_z,L=s{J}_z+\delta jI.$$

Consequently,

$$\begin{array}{cc}\hfill (L+\delta I)(L+2\delta I)& =s^2\left(J_z+\theta (j+1)I\right)\left(J_z+\theta (j+2)I\right)\hfill \\ & =s^2\left({\left(J_z+\theta \left(j+{\displaystyle \frac{3}{2}}\right)I\right)}^2-{\displaystyle \frac{{\theta }^2}{4}}I\right).\hfill \end{array}$$

Also $r{Y}^2=4r{J}_y^2=s^2(1-{\theta }^2)J_y^2$. Substituting both identities into (21) gives

$$\begin{array}{cc}\hfill Q{Q}^{*}-2r(k+2)I=s^2(& {\left(J_z+\theta \left(j+{\displaystyle \frac{3}{2}}\right)I\right)}^2\hfill \\ \hfill & +(1-{\theta }^2)J_y^2-{\displaystyle \frac{{\theta }^2}{4}}I).\hfill \end{array}$$

(23)

By our lemma, the operator in parentheses is positive semidefinite. Therefore

$$\parallel (a^{*2}-r{b}^{*2}){f\parallel }^2=\parallel Q^{*}{f\parallel }^2=\langle Q{Q}^{*}f,f\rangle \ge 2r(k+2){\parallel f\parallel }^2,$$

which proves (18). Returning to (16) and (17), we obtain

$$\begin{array}{cc}\hfill 2\parallel c_x^{*}{c}_y^{*}{f\parallel }^2& =2{\alpha }^2{\parallel Q^{*}f\parallel }^2\hfill \\ & \ge 4{\alpha }^2{r(k+2)\parallel f\parallel }^2=4\alpha \beta (k+2){\parallel f\parallel }^2\hfill \\ & =\parallel x\otimes c_y^{*}f-y\otimes c_x^{*}{f\parallel }^2.\hfill \end{array}$$

This proves (12) on each ${\mathcal{H}}_{m,k}$. Summing over the orthogonal components from Step 2 proves it for arbitrary $f\in {\mathcal{K}}^{\odot m}$. The argument includes $m=0$, for which ${\mathcal{K}}^{\odot 0}={\mathcal{H}}_{0,0}=\mathbb{C}$. □

4. Proof of Theorem 1

For $A=\left(a_{ij}\right)\in M_n\left(\mathbb{C}\right)$, set

$${\mathcal{C}}_1\left(A\right)={\left(a_{ij}perA(i\mid j)\right)}_{i,j=1}^n.$$

We are now ready to prove our main result.

Proof of Theorem 1. 

If $i=j$, the left-hand side of (1) is zero, and $perA\ge 0$, as also follows from (27) below. We may therefore assume $i\ne j$. A simultaneous permutation of rows and columns reduces the proof to the case with $i=1$ and $j=2$. Choose vectors $v_1,\dots ,v_n$ in a finite-dimensional complex Hilbert space $\mathcal{K}$ such that $a_{pq}=\langle v_p,v_q\rangle$. The relation between symmetric tensors and permanents [5], with the normalization used here, is

$$〈u_1\odot \dots \odot u_d,w_1\odot \dots \odot w_d〉={\displaystyle \frac{1}{d!}}per{\left(\langle u_p,w_q\rangle \right)}_{p,q=1}^d.$$

(24)

For $1\le p\le n$, put $q_p={⨀}_{\ell \ne p}{v}_{\ell }$ and $z_p=v_p\otimes q_p$. It follows that

$$\langle z_p,z_q\rangle ={\displaystyle \frac{a_{pq}perA\left(p\right|q)}{(n-1)!}}.$$

(25)

Thus, ${\mathcal{C}}_1\left(A\right)$ is $(n-1)!$ times the Gram matrix of $z_1,\dots ,z_n$.

Set $m=n-2$, $x=v_1$, $y=v_2$, and $f=v_3\odot \dots \odot v_n\in {\mathcal{K}}^{\odot m}$, with $f=1$ when $n=2$. Since

$$q_1={\displaystyle \frac{c_y^{*}f}{\sqrt{m+1}}},q_2={\displaystyle \frac{c_x^{*}f}{\sqrt{m+1}}},$$

equation (25) gives

$$\begin{array}{c}\hfill {(e_1-e_2)}^{*}{\mathcal{C}}_1\left(A\right)(e_1-e_2)=m!{∥x\otimes c_y^{*}f-y\otimes c_x^{*}f∥}^2,\end{array}$$

(26)

where $e_1-e_2={(1,-1,0,\dots ,0)}^T\in {\mathbb{R}}^n$.

Also, (24) and the definition of the creation operators give

$$perA=n!\parallel v_1\odot \dots \odot v_n{\parallel }^2=m!{\parallel c_x^{*}{c}_y^{*}f\parallel }^2.$$

(27)

Combining (26) and (27) with Theorem 2 proves (1) in Theorem 1. □

5. Further Discussions

A closely related statement, stronger than inequality (1), is Bapat and Sunder’s Conjecture 3 in [1] (1986). In his study of Bapat and Sunder’s conjecture by means of permanent functions and compound matrices, Pate [10] (2008) showed that

$$\langle {\mathcal{C}}_k\left(A\right)x,x\rangle \le {\parallel x\parallel }^{2}perA$$

for every positive semidefinite Hermitian matrix A and every $(0,1)$-vector x, where

$${\mathcal{C}}_k\left(A\right)={\left(perA[\alpha \mid \beta ]perA(\alpha \mid \beta )\right)}_{\alpha ,\beta }.$$

Here $\alpha$ and $\beta$ range over the k-element subsets of $\{1,\dots ,n\}$, listed in lexicographic order. The matrix $A[\alpha \mid \beta ]$ has rows indexed by $\alpha$ and columns indexed by $\beta$, and $A(\alpha \mid \beta )=A[{\alpha }^c\mid {\beta }^c]$. Pate also indicated that Bapat and Sunder’s conjecture would imply inequality (1).

Drury [4] (2018) gave a negative answer to Bapat and Sunder’s conjecture by providing a counterexample. However, Pate’s question, namely whether inequality (1) holds, was left open in [10]. It is answered here by Theorem 1.

Question. 

Pate showed that

$$\langle {\mathcal{C}}_k\left(A\right)x,x\rangle \le {\parallel x\parallel }^{2}perA$$

if x is a $(0,1)$-vector. Is it true that

$$\langle {\mathcal{C}}_k\left(A\right)x,x\rangle \le {\parallel x\parallel }^{2}perA$$

for $x=e_1-e_2$ and every k with $2\le k\le n-1$? Here $e_1,e_2$ are the first two standard basis vectors of ${\mathbb{C}}^{\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)}$.

The answer is negative. We provide a family of counterexamples for $k=5$.

For $t>0$, let

$$A_t=X_t^{*}{X}_t,where{X}_t=\left(\begin{array}{cccccccccc}0& 0& i& i& 1& 1& 1& 1& t& t\\ 1& 1& t& t& 1& -1& 0& 0& i& i\end{array}\right).$$

The columns come from the three orthogonal pairs

$$\left\{\left({\displaystyle \genfrac{}{}{0pt}{}{0}{1}}\right),\left({\displaystyle \genfrac{}{}{0pt}{}{1}{0}}\right)\right\},\left\{\left({\displaystyle \genfrac{}{}{0pt}{}{i}{t}}\right),\left({\displaystyle \genfrac{}{}{0pt}{}{t}{i}}\right)\right\},\left\{\left({\displaystyle \genfrac{}{}{0pt}{}{1}{1}}\right),\left({\displaystyle \genfrac{}{}{0pt}{}{1}{-1}}\right)\right\}.$$

Each vector in the first two pairs occurs twice, and each vector in the third pair occurs once. Thus, $A_t$ is positive semidefinite of rank two; indeed, $X_t{X}_t^{*}=(6+2t^2)I_2$.

The first two 5-subsets in lexicographic order are

$$\alpha =\{1,2,3,4,5\},\beta =\{1,2,3,4,6\}.$$

Put $q=t^2$ and

$$\begin{array}{cc}\hfill a& =6q^2+6q+1,\hfill \\ \hfill b& =4q(q+1){(2q+1)}^2,\hfill \\ \hfill c& =20(3q^4+3q^3+5q^2+3q+3).\hfill \end{array}$$

A direct calculation gives

$$per{A}_t={(4!)}^{2}c,{\mathcal{C}}_5\left(A_t\right)[\{\alpha ,\beta \}\mid \{\alpha ,\beta \}]={(4!)}^2\left(\begin{array}{cc}{a}^2& -b\\ -b& a^2\end{array}\right).$$

Therefore, for $x=e_1-e_2\in {\mathbb{C}}^{\left({\displaystyle \genfrac{}{}{0pt}{}{10}{5}}\right)}$,

$$x^{*}{\mathcal{C}}_5\left(A_t\right)x-2per{A}_t=2{(4!)}^2(a^2+b-c).$$

The sign follows from the identity

$$\begin{array}{cc}\hfill a^2+b-c& =-8q^4+44q^3-32q^2-44q-59\hfill \\ & =21+40(4-q)+4(q-3)(4-q)(2q^2+3q+5).\hfill \end{array}$$

All terms in the last line are nonnegative when $3\le q\le 4$, and the first is 21. Hence, every $t\in [\sqrt{3},2]$ gives a counterexample. For instance, $t=2$ gives

$${\displaystyle \frac{x^{*}{\mathcal{C}}_5\left(A_2\right)x}{{\parallel x\parallel }^{2}per{A}_2}}={\displaystyle \frac{21121}{21100}}>1.$$