1 Introduction and preliminaries

Inequality is an active research topic in recent years, the classical convexity has been generalized and extended in a diverse manner.One of them is the pre-invexity, introduced by WEIR et al^[1]as a significant generalization of convex functions.Many researchers have studied the basic properties of the pre-invex functions and their role in optimization, variational inequalities and equilibrium problems^[2-4].

Hadamard and Fischer inequalities are prima-ry inequalities for the real symmetric positive def-inite matrix, and there are many inequalities can be proved by using these two inequalities.There are many methods to prove these inequalities^[5-7].Some results have been established inspired by the Hadamard inequality^[8-9].

The real symmetric positive definite matrix has many properties and has been used in many ar-eas^[10-12].Some properties can be used to prove the Hadamard inequality.In this note, inspired by the results established in [13-14], a new eigenvalue inequality related to the real symmetric positive definite matrix is proposed, and the Hadamard and Fischer inequality are proved by using this new inequality.

Firstly, let us introduce some notations and lemmas.I_n is the identity matrix with order n×n.For a square matrix A, we use λ[A], det(A) and A^T represent the set of the eigenvalues, the deter-minant and the transpose of A, respectively.

Next, we introduce two lemmas.The following result about the block matrix determinant is well known^[10].

Lemma 1   If matrix A is invertible, then for any block matrix, we have

$ \det \left( {\left[ {\begin{array}{*{20}{c}} \mathit{\boldsymbol{A}}&\mathit{\boldsymbol{B}}\\ \mathit{\boldsymbol{C}}&\mathit{\boldsymbol{D}} \end{array}} \right]} \right) = \det \left( \mathit{\boldsymbol{A}} \right)\det \left( {\mathit{\boldsymbol{D}} - \mathit{\boldsymbol{C}}{\mathit{\boldsymbol{A}}^{ - 1}}\mathit{\boldsymbol{B}}} \right), $

(1)

or if marix D is invertible, then we have

$ \det \left( {\left[ {\begin{array}{*{20}{c}} \mathit{\boldsymbol{A}}&\mathit{\boldsymbol{B}}\\ \mathit{\boldsymbol{C}}&\mathit{\boldsymbol{D}} \end{array}} \right]} \right) = \det \left( \mathit{\boldsymbol{D}} \right)\det \left( {\mathit{\boldsymbol{A}} - \mathit{\boldsymbol{B}}{\mathit{\boldsymbol{D}}^{ - 1}}\mathit{\boldsymbol{C}}} \right). $

(2)

Lemma 2   If A ∈ R^m×n is a full column-rank matrix, the A(A^TA)^-1A^T is idempotent and the eigenvalues of A(A^TA)^-1A^T are 1 or 0, there exists an orthogonal matrix Q such that

$ \mathit{\boldsymbol{Q}}\left[ {\mathit{\boldsymbol{A}}{{\left( {{\mathit{\boldsymbol{A}}^{\rm{T}}}\mathit{\boldsymbol{A}}} \right)}^{ - 1}}{\mathit{\boldsymbol{A}}^{\rm{T}}}} \right]\mathit{\boldsymbol{Q}} = {\rm{diag}}\left[ {1, \cdots ,1,0, \cdots ,0} \right] = :\Lambda . $

Furthermore, we have rank [A]=n.

This lemma was suggested in [14], for convenience, we give the proof here.

Proof   If σ∈λ[A(A^TA)^-1A^T], then there exists a nonzero vector x∈R^m, satisfying

$ \mathit{\boldsymbol{A}}{\left( {{\mathit{\boldsymbol{A}}^{\rm{T}}}\mathit{\boldsymbol{A}}} \right)^1}{\mathit{\boldsymbol{A}}^{\rm{T}}}\mathit{\boldsymbol{x = }}\sigma \mathit{\boldsymbol{x}}. $

Thus, we have

$ \begin{array}{*{20}{c}} {{{\left[ {\mathit{\boldsymbol{A}}{{\left( {{\mathit{\boldsymbol{A}}^{\rm{T}}}\mathit{\boldsymbol{A}}} \right)}^{ - 1}}{\mathit{\boldsymbol{A}}^{\rm{T}}}\mathit{\boldsymbol{x}}} \right]}^{\rm{T}}}\left[ {\mathit{\boldsymbol{A}}{{\left( {{\mathit{\boldsymbol{A}}^{\rm{T}}}\mathit{\boldsymbol{A}}} \right)}^{ - 1}}{\mathit{\boldsymbol{A}}^{\rm{T}}}\mathit{\boldsymbol{x}}} \right] = {{\left( {\sigma \mathit{\boldsymbol{x}}} \right)}^{\rm{T}}}\left( {\sigma \mathit{\boldsymbol{x}}} \right),}\\ {{\mathit{\boldsymbol{x}}^{\rm{T}}}\left[ {\mathit{\boldsymbol{A}}{{\left( {{\mathit{\boldsymbol{A}}^{\rm{T}}}\mathit{\boldsymbol{A}}} \right)}^{ - 1}}{\mathit{\boldsymbol{A}}^{\rm{T}}}} \right]\left[ {\mathit{\boldsymbol{A}}{{\left( {{\mathit{\boldsymbol{A}}^{\rm{T}}}\mathit{\boldsymbol{A}}} \right)}^1}{\mathit{\boldsymbol{A}}^{\rm{T}}}} \right]\mathit{\boldsymbol{x = }}{\sigma ^2}{{\left\| \mathit{\boldsymbol{x}} \right\|}^2},}\\ {{\mathit{\boldsymbol{x}}^{\rm{T}}}\left[ {\mathit{\boldsymbol{A}}{{\left( {{\mathit{\boldsymbol{A}}^{\rm{T}}}\mathit{\boldsymbol{A}}} \right)}^{ - 1}}{\mathit{\boldsymbol{A}}^{\rm{T}}}\mathit{\boldsymbol{A}}{{\left( {{\mathit{\boldsymbol{A}}^{\rm{T}}}\mathit{\boldsymbol{A}}} \right)}^1}{\mathit{\boldsymbol{A}}^{\rm{T}}}} \right]\mathit{\boldsymbol{x = }}{\sigma ^2}{{\left\| \mathit{\boldsymbol{x}} \right\|}^2},}\\ {{\mathit{\boldsymbol{x}}^{\rm{T}}}\left[ {\mathit{\boldsymbol{A}}{{\left( {{\mathit{\boldsymbol{A}}^{\rm{T}}}\mathit{\boldsymbol{A}}} \right)}^1}{\mathit{\boldsymbol{A}}^{\rm{T}}}} \right]\mathit{\boldsymbol{x = }}{\sigma ^{ - 2}}{{\left\| \mathit{\boldsymbol{x}} \right\|}^2},}\\ {{\mathit{\boldsymbol{x}}^{\rm{T}}}\sigma \mathit{\boldsymbol{x}}{ = ^2}{{\left\| \mathit{\boldsymbol{x}} \right\|}^2},}\\ {\sigma {{\left\| \mathit{\boldsymbol{x}} \right\|}^2} = {\sigma ^2}{{\left\| \mathit{\boldsymbol{x}} \right\|}^2}.} \end{array} $

Since‖x‖²≠0, A(A^TA)^-1A^T has eigenvalues σ=0 or σ=1.Because of the symmetry of A(A^TA)^-1A^T, there exists a real orthogonal matrix Q:=[q₁, q₂, …, q_m]∈R^m×m such that

$ {\mathit{\boldsymbol{Q}}^{\rm{T}}}\left[ {\mathit{\boldsymbol{A}}{{\left( {{\mathit{\boldsymbol{A}}^{\rm{T}}}\mathit{\boldsymbol{A}}} \right)}^{ - 1}}{\mathit{\boldsymbol{A}}^{\rm{T}}}} \right]\mathit{\boldsymbol{Q}} = {\rm{diag}}\left[ {1, \cdots ,1,0, \cdots ,0} \right] = :\Lambda . $

On the other hand, since (A^TA)^-1A^T is the left pseudo-inverse of A, we have

$ \begin{array}{*{20}{c}} {{\rm{rank}}\left[ \mathit{\boldsymbol{A}} \right] = {\rm{rank}}\left[ {{\mathit{\boldsymbol{Q}}^{\rm{T}}}\left[ {\mathit{\boldsymbol{A}}{{\left( {{\mathit{\boldsymbol{A}}^{\rm{T}}}\mathit{\boldsymbol{A}}} \right)}^{ - 1}}{\mathit{\boldsymbol{A}}^{\rm{T}}}} \right]\mathit{\boldsymbol{Q}}} \right] = }\\ {{\rm{rank}}\left[ {\mathit{\boldsymbol{A}}{{\left( {{\mathit{\boldsymbol{A}}^{\rm{T}}}\mathit{\boldsymbol{A}}} \right)}^{ - 1}}{\mathit{\boldsymbol{A}}^{\rm{T}}}} \right] = {\rm{rank}}\left[ \mathit{\boldsymbol{A}} \right] = n.} \end{array} $

This proves lemma 2.

2 An inequality of a class of matrices

In this section, we will establish a new property about the eigenvalues related to the symmetric positive definite matrix.If A ∈ R^n×n is a symmetric positive definite matrix, then there exists an invertible matrix B such that A=BB^T.Suppose that B can be expressed as a block matrix

$ \mathit{\boldsymbol{B}}: = \left[ {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{B}}_1}}\\ {{\mathit{\boldsymbol{B}}_2}} \end{array}} \right]: = \left[ {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{B}}_{11}}}&{{\mathit{\boldsymbol{B}}_{12}}}\\ {{\mathit{\boldsymbol{B}}_{21}}}&{{\mathit{\boldsymbol{B}}_{22}}} \end{array}} \right], $

where B₁₁ ∈ R^m×m, B₁₂ ∈ R^m×p, B₂₁ ∈ R^p×m, and B₂₂ ∈ R^p×p.Using these symbols, the matrix A can be expressed as

$ \mathit{\boldsymbol{B}}{\mathit{\boldsymbol{B}}^{\rm{T}}} = \left[ {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{B}}_1}}\\ {{\mathit{\boldsymbol{B}}_2}} \end{array}} \right]\left[ {\mathit{\boldsymbol{B}}_1^{\rm{T}},\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right] = \left[ {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{B}}_1}\;\mathit{\boldsymbol{B}}_1^{\rm{T}}\;{\mathit{\boldsymbol{B}}_1}\;\mathit{\boldsymbol{B}}_2^{\rm{T}}}\\ {{\mathit{\boldsymbol{B}}_2}\;\mathit{\boldsymbol{B}}_1^{\rm{T}}\;{\mathit{\boldsymbol{B}}_2}\;\mathit{\boldsymbol{B}}_2^{\rm{T}}} \end{array}} \right] = :\left[ {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{A}}_{11}}\;{\mathit{\boldsymbol{A}}_{12}}}\\ {{\mathit{\boldsymbol{A}}_{21}}\;{\mathit{\boldsymbol{A}}_{22}}} \end{array}} \right]. $

Set

$ \mathit{\boldsymbol{N}}: = \left[ {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_1^{\rm{T}}}&0\\ 0&{{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \end{array}} \right] \in {{\bf{R}}^{m \times p}}. $

With these symbols, the following result holds.

Theorem 1   If the eigenvalues of the matrix N^-1A are δ₁, δ₂, …, δ_n, then 0 < δ₁δ₂…δ_n≤1.

Proof   Let f(λ): =det(λI_n-N^-1A)be the characteristic polynomial of matrix N^-1A, we have

$ \begin{array}{l} f\left( \lambda \right) = \\ \det \left( {\lambda {\mathit{\boldsymbol{I}}_n} - \left[ {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{I}}_m}}&{{{\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_1^{\rm{T}}} \right)}^{ - 1}}\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right)}\\ {{{\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right)}^{ - 1}}\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right)}&{{\mathit{\boldsymbol{I}}_p}} \end{array}} \right]} \right). \end{array} $

(3)

We verify that 2 not belongs to the eigenvalues of the matrix N^-1 A.If 2 is the eigenvalue of the matrix N^-1A, then

$ \begin{array}{l} f\left( 2 \right) = \\ \det \left( {\left[ {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{I}}_m}}&{ - {{\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_1^{\rm{T}}} \right)}^{ - 1}}\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right)}\\ { - {{\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right)}^{ - 1}}\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_1^{\rm{T}}} \right)}&{{\mathit{\boldsymbol{I}}_p}} \end{array}} \right]} \right) = 0. \end{array} $

On the other hand,

$ \begin{array}{l} f\left( 2 \right) = \\ \det \left( {\left[ {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{I}}_m}}&{ - {{\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_1^{\rm{T}}} \right)}^{ - 1}}\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right)}\\ { - {{\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right)}^{ - 1}}\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_1^{\rm{T}}} \right)}&{{\mathit{\boldsymbol{I}}_p}} \end{array}} \right]} \right) = \\ \det \left( {\left[ {\begin{array}{*{20}{c}} {{{\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_1^{\rm{T}}} \right)}^{ - 1}}}&{\bf{0}}\\ {\bf{0}}&{{{\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right)}^{ - 1}}} \end{array}} \right] \times \left[ {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_1^{\rm{T}}}&{ - {\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_2^{\rm{T}}}\\ { - {\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^{\rm{T}}}&{{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \end{array}} \right]} \right) = \\ \det \left( {\left[ {\begin{array}{*{20}{c}} {{{\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_1^{\rm{T}}} \right)}^{ - 1}}}&{\bf{0}}\\ {\bf{0}}&{{{\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right)}^{ - 1}}} \end{array}} \right]} \right) \times \\ \det \left( {\left[ {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{B}}_1}}\\ { - {\mathit{\boldsymbol{B}}_2}} \end{array}} \right]\left[ {\mathit{\boldsymbol{B}}_1^{\rm{T}} - \mathit{\boldsymbol{B}}_2^{\rm{T}}} \right]} \right) \ne 0. \end{array} $

This is a contradiction, so 2 is not a eigenvalue of the matrix N^-1A.

According to lemma 1, suppose that m≥p, equation (3) can be manipulated as

$ \begin{array}{l} f\left( \lambda \right) = \det \left( {\left[ {\begin{array}{*{20}{c}} {\left( {\lambda - 1} \right){\mathit{\boldsymbol{I}}_m}}&{ - {{\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_1^{\rm{T}}} \right)}^{ - 1}}\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right)}\\ { - {{\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right)}^{ - 1}}\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right)}&{\left( {\lambda - 1} \right){\mathit{\boldsymbol{I}}_p}} \end{array}} \right]} \right) = \\ \;\;\;\;\;\;\;\;\;\det \left( {\left( {\lambda - 1} \right){\mathit{\boldsymbol{I}}_m}} \right)\det \left( {\left( {\lambda - 1} \right){\mathit{\boldsymbol{I}}_p} - {{\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right)}^{ - 1}} \times } \right.\\ \;\;\;\;\;\;\;\;\;\left. {\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right){{\left( {\lambda - 1} \right)}^{ - 1}}{\mathit{\boldsymbol{I}}_m}{{\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_1^{\rm{T}}} \right)}^{ - 1}}\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right)} \right) = \\ \;\;\;\;\;\;\;\;\;\det \left( {\left( {\lambda - 1} \right){\mathit{\boldsymbol{I}}_m}} \right)\det \left( {\left( {\lambda - 1} \right){\mathit{\boldsymbol{I}}_p} - {{\left( {\lambda - 1} \right)}^{ - 1}}{\mathit{\boldsymbol{I}}_p} \times } \right.\\ \;\;\;\;\;\;\;\;\;\left. {{{\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right)}^{ - 1}}\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_1^{\rm{T}}} \right){{\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_1^{\rm{T}}} \right)}^{ - 1}}\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right)} \right) = \\ \;\;\;\;\;\;\;\;\;{\left( {\lambda - 1} \right)^{m - p}}\det \left( {{{\left( {\lambda - 1} \right)}^2}{\mathit{\boldsymbol{I}}_p} - {{\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right)}^{ - 1}} \times } \right.\\ \;\;\;\;\;\;\;\;\;\left. {{\mathit{\boldsymbol{B}}_2}\left[ {\mathit{\boldsymbol{B}}_1^{\rm{T}}{{\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_1^{\rm{T}}} \right)}^{ - 1}}{\mathit{\boldsymbol{B}}_1}} \right]\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right). \end{array} $

(4)

Since B₁^T(B₁B₁^T)^-1B₁ is a real symmetric matrix, according to lemma 2, there exists an orthogonal matrix Q: = [q₁, q₂, …, q_n] such that

$ \begin{array}{l} \left[ {\mathit{\boldsymbol{B}}_1^{\rm{T}}{{\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_1^{\rm{T}}} \right)}^{ - 1}}{\mathit{\boldsymbol{B}}_1}} \right]\left[ {{\mathit{\boldsymbol{q}}_1},{\mathit{\boldsymbol{q}}_2}, \cdots ,{\mathit{\boldsymbol{q}}_n}} \right] = \\ \;\;\;\;\;\;\;\left[ {{\mathit{\boldsymbol{q}}_1},{\mathit{\boldsymbol{q}}_2}, \cdots ,{\mathit{\boldsymbol{q}}_n}} \right]{\rm{diag}}\left[ {1, \cdots ,1,0, \cdots ,0} \right] = \\ \;\;\;\;\;\;\;\left[ {{\mathit{\boldsymbol{q}}_1},{\mathit{\boldsymbol{q}}_2}, \cdots ,{\mathit{\boldsymbol{q}}_m},0,0, \cdots ,0} \right], \end{array} $

(5)

where m=rank[B₁^T(B₁B₁^T)¹B₁].

Let C: =(B₂B₂^T)^-1B₂[B₁^T(B₁B₁^T)^-1B₁]B₂^T∈ R^p×p, and suppose that ρ∈λ[C].There exists a nonzero vector y∈R^p such that Cy=ρy.Manipulating this equation gives

$ \left\{ \begin{array}{l} {\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right)^{ - 1}}{\mathit{\boldsymbol{B}}_2}\left[ {\mathit{\boldsymbol{B}}_1^{\rm{T}}{{\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_1^{\rm{T}}} \right)}^{ - 1}}{\mathit{\boldsymbol{B}}_1}} \right]\mathit{\boldsymbol{B}}_2^{\rm{T}}y = \rho y,\\ {\mathit{\boldsymbol{B}}_2}\left[ {\mathit{\boldsymbol{B}}_1^{\rm{T}}{{\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_1^{\rm{T}}} \right)}^{ - 1}}{\mathit{\boldsymbol{B}}_1}} \right]\mathit{\boldsymbol{B}}_2^{\rm{T}}y = \rho \left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^{\rm{T}}} \right)y. \end{array} \right. $

(6)

It is clear that B₂^T$\mathcal{y}$ can be expressed as a linear combination of q₁, q₂, …, q_n, i.e.,

$ \mathit{\boldsymbol{B}}_2^{\rm{T}}y = {k_1}{\mathit{\boldsymbol{q}}_1} + {k_2}{\mathit{\boldsymbol{q}}_2} + \cdots + {k_n}{\mathit{\boldsymbol{q}}_n}, $

(7)

where k_i∈R, i=1, 2, …, n and k₁²+k₂²+…+k_n²≠0.Using equations(5) and(7), we have

$ \begin{array}{l} {\left( {\mathit{\boldsymbol{B}}_2^{\rm{T}}y} \right)^{\rm{T}}}\left( {\mathit{\boldsymbol{B}}_2^{\rm{T}}y} \right) = {\left( {{k_1}{\mathit{\boldsymbol{q}}_1} + {k_2}{\mathit{\boldsymbol{q}}_2} + \cdots + {k_n}{\mathit{\boldsymbol{q}}_n}} \right)^{\rm{T}}} \times \\ \;\;\;\;\;\;\;\left( {{k_1}{\mathit{\boldsymbol{q}}_2} + {k_2}{\mathit{\boldsymbol{q}}_2} + \cdots + {k_n}{\mathit{\boldsymbol{q}}_n}} \right) = \\ \;\;\;\;\;\;\;k_1^2 + k_2^2 + \cdots + k_n^2, \end{array} $

(8)

$ \begin{array}{l} \left[ {\mathit{\boldsymbol{B}}_1^T\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_1^T} \right)^{-1}{\mathit{\boldsymbol{B}}_1}} \right]\mathit{\boldsymbol{B}}_2^Ty = \left[ {\mathit{\boldsymbol{B}}_1^T\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_1^T} \right)^{-1}{\mathit{\boldsymbol{B}}_1}} \right] \times \\ \;\;\;\;\;\;\;\left( {{k_1}{\mathit{\boldsymbol{q}}_1} + {k_2}{\mathit{\boldsymbol{q}}_2} + \cdots + {k_n}{\mathit{\boldsymbol{q}}_n}} \right) = \\ \;\;\;\;\;\;\;{k_1}{\mathit{\boldsymbol{q}}_1} + {k_2}{\mathit{\boldsymbol{q}}_2} + \cdots + {k_m}{\mathit{\boldsymbol{q}}_m}, \end{array} $

(9)

The both side of equation (6) multiply by $\mathcal{y}$^T gives

$ {y^{\rm{T}}}{\mathit{\boldsymbol{B}}_2}\left[ {\mathit{\boldsymbol{B}}_1^T\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_1^T} \right){\mathit{\boldsymbol{B}}_1}} \right]\mathit{\boldsymbol{B}}_2^Ty = {y^{\rm{T}}}{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^Ty. $

(10)

According to equations(7) and (9), the left-side of equation (10) can be rewritten as

$ \begin{array}{l} {\left( {{k_1}{\mathit{\boldsymbol{q}}_1} + {k_2}{\mathit{\boldsymbol{q}}_2} + \cdots + {k_n}{\mathit{\boldsymbol{q}}_n}} \right)^{\rm{T}}} \times \\ \;\;\;\;\;\;\;\;\;\left( {{k_1}{\mathit{\boldsymbol{q}}_1} + {k_2}{\mathit{\boldsymbol{q}}_2} + \cdots + {k_m}{\mathit{\boldsymbol{q}}_m}} \right) = \\ \;\;\;\;\;\;\;\;\;k_1^2 + k_2^2 + \cdots + k_m^2, \end{array} $

(11)

Combining equations(10)(8) and(11) gives

$ \left( {k_1^2 + k_2^2 + \cdots + k_m^2} \right) = \rho \left( {k_1^2 + k_2^2 + \cdots + k_n^2} \right). $

Hence, we have

$ 0 \le \rho = \frac{{k_1^2 + k_2^2 + \cdots + k_m^2}}{{k_1^2 + k_2^2 + \cdots + k_n^2}} \le 1. $

(12)

Since 0≤ρ≤1, we suppose that λ[C]={ρ₁², ρ₁², …, ρ_p²}with 0≤ρ_p≤…≤ρ₁≤1.According to the Schur decomposition theorem, there exists an orthogonal matrix U such that

$ {\mathit{\boldsymbol{U}}^{\rm{T}}}\mathit{\boldsymbol{CU}} = {\rm{diag}}\left[ {\rho _1^2,\rho _2^2, \cdots ,\rho _p^2} \right] + \mathit{\boldsymbol{R}}, $

(13)

where R is strictly upper triangular.Substituting equation(13) into equation(4) and simplifying it, give

$ \begin{array}{l} f\left( \lambda \right) = {\left( {\lambda - 1} \right)^{m - p}}\det \left( {{{\left( {\lambda - 1} \right)}^2}{\mathit{\boldsymbol{I}}_p} - \mathit{\boldsymbol{C}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;{\left( {\lambda - 1} \right)^{m - p}}\det \left( {{\mathit{\boldsymbol{U}}^{\rm{T}}}} \right)\det \left( {{{\left( {\lambda - 1} \right)}^2}\mathit{\boldsymbol{I}}p - \mathit{\boldsymbol{C}}} \right)\det \left( \mathit{\boldsymbol{U}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;{\left( {\lambda - 1} \right)^{m - p}}\det \left( {{\mathit{\boldsymbol{U}}^{\rm{T}}}{{\left( {\lambda - 1} \right)}^2}{\mathit{\boldsymbol{I}}_p}\mathit{\boldsymbol{U}} - {\mathit{\boldsymbol{U}}^{\rm{T}}}\mathit{\boldsymbol{CU}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;{\left( {\lambda - 1} \right)^{m - p}}\det \left( {{{\left( {\lambda - 1} \right)}^2}{\mathit{\boldsymbol{I}}_p} - {\rm{diag}}\left[ {\rho _1^2,\rho _2^2, \cdots ,\rho _p^2} \right] - \mathit{\boldsymbol{R}}} \right) = \\ \;\;\;\;\;\;\;\;\;\;\;\;{\left( {\lambda - 1} \right)^{m - p}}\left[ {{{\left( {\lambda - 1} \right)}^2} - \rho _1^2} \right] \cdots \left[ {{{\left( {\lambda - 1} \right)}^2} - \rho _p^2} \right] = \\ \;\;\;\;\;\;\;\;\;\;\;\;{\left( {\lambda - 1} \right)^{m - p}}\left( {\lambda - 1 - {\rho _1}} \right)\left( {\lambda - 1 + {\rho _1}} \right) \times \cdots \times \\ \;\;\;\;\;\;\;\;\;\;\;\;\left( {\lambda - 1 - {\rho _p}} \right)\left( {\lambda - 1 + {\rho _p}} \right). \end{array} $

From this equation, we can see that the eigenvalues of N^-1A are

$ 1 + {\rho _1},1 - {\rho _1}, \cdots ,1 + {\rho _p},1 - {\rho _p},1, \cdots ,1. $

(14)

From this equation, we have

$ \begin{array}{l} 0 \le {{\rm{\delta }}_1}{{\rm{\delta }}_2} \cdots {{\rm{\delta }}_n} = \\ \;\;\;\;\;\;\left( {1 + {\rho _1}} \right)\left( {1 - {\rho _1}} \right) \cdots \left( {1 + {\rho _p}} \right)\left( {1 - {\rho _p}} \right) = \\ \;\;\;\;\;\;\left( {1 - \rho _1^2} \right)\left( {1 - \rho _2^2} \right) \cdots \left( {1 - \rho _p^2} \right) \le 1. \end{array} $

(15)

Since 2 is not the eigenvalue of the matrix N^-1A, an improvement of inequality (12) is 0≤ρ < 1.

Correspondingly, inequality (15) can be im-proved as

$ \begin{array}{l} 0 \le {{\rm{\delta }}_1}{{\rm{\delta }}_2} \cdots {{\rm{\delta }}_n} = \\ \;\;\;\;\;\;\left( {1 + {\rho _1}} \right)\left( {1 - {\rho _1}} \right) \cdots \left( {1 + {\rho _p}} \right)\left( {1 - {\rho _p}} \right) = \\ \;\;\;\;\;\;\left( {1 - \rho _1^2} \right)\left( {1 - \rho _2^2} \right) \cdots \left( {1 - \rho _p^2} \right) \le 1. \end{array} $

The proof is completed.

Remark 1   The above proof shows that the sup-position m≥p is not essential.In fact, if m < p, according to equation (2) of lemma 1, we have

$ \begin{array}{l} f\left( \lambda \right) = \\ \;\;\;\;\;\det \left( {\left[ {\begin{array}{*{20}{c}} {\left( {\lambda - 1} \right){\mathit{\boldsymbol{I}}_m}}&{ - {{\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_1^T} \right)}^{ - 1}}\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_2^T} \right)}\\ { - {{\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^T} \right)}^{ - 1}}\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_1^T} \right)}&{\left( {\lambda - 1} \right){\mathit{\boldsymbol{I}}_p}} \end{array}} \right]} \right) = \\ \;\;\;\;\;\det \left( {\left( {\lambda - 1} \right){\mathit{\boldsymbol{I}}_m}} \right)\det \left( {\left( {\lambda - 1} \right){\mathit{\boldsymbol{I}}_m} - {{\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_1^T} \right)}^{ - 1}} \times } \right.\\ \;\;\;\;\;\left. {\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_2^T} \right){{\left( {\lambda - 1} \right)}^{ - 1}}{\mathit{\boldsymbol{I}}_{\rm{p}}}{{\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^T} \right)}^{ - 1}}\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_1^T} \right)} \right) = \\ \;\;\;\;\;\det \left( {\left( {\lambda - 1} \right){\mathit{\boldsymbol{I}}_p}} \right)\det \left( {\left( {\lambda - 1} \right){\mathit{\boldsymbol{I}}_m} - {{\left( {\lambda - 1} \right)}^{ - 1}} \times } \right.\\ \;\;\;\;\;\left. {{\mathit{\boldsymbol{I}}_m}{{\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_1^T} \right)}^{ - 1}}{\mathit{\boldsymbol{B}}_1}\left[ {\mathit{\boldsymbol{B}}_2^T{{\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^T} \right)}^{ - 1}}{\mathit{\boldsymbol{B}}_2}} \right]\mathit{\boldsymbol{B}}_1^T} \right) = \\ \;\;\;\;\;{\left( {\lambda - 1} \right)^{p - m}}\det \left( {{{\left( {\lambda - 1} \right)}^2}{\mathit{\boldsymbol{I}}_m} - {{\left( {{\mathit{\boldsymbol{B}}_1}\mathit{\boldsymbol{B}}_1^T} \right)}^{ - 1}}{\mathit{\boldsymbol{B}}_1} \times } \right.\\ \;\;\;\;\;\left. {\left[ {\mathit{\boldsymbol{B}}_2^T{{\left( {{\mathit{\boldsymbol{B}}_2}\mathit{\boldsymbol{B}}_2^T} \right)}^{ - 1}}{\mathit{\boldsymbol{B}}_2}} \right]\mathit{\boldsymbol{B}}_1^T} \right). \end{array} $

This manipulation does not change the subsequent proof.

3 New proof of the Fischer inequality

In this section, we will use the results in theo-rem 1 to prove two determinant inequalities related to the symmetric positive definite matrix, that is, the Fischer inequality and the Hadamard inequali-ty.

Theorem 2   Considering the following symmetric positive definite block matrix

$ \mathit{\boldsymbol{M}}: = \left( {\begin{array}{*{20}{c}} {{\mathit{\boldsymbol{M}}_{11}}}&{{\mathit{\boldsymbol{M}}_{12}}}& \cdots &{{\mathit{\boldsymbol{M}}_{1k}}}\\ {{\mathit{\boldsymbol{M}}_{21}}}&{{\mathit{\boldsymbol{M}}_{22}}}& \cdots &{{\mathit{\boldsymbol{M}}_{2k}}}\\ \vdots & \vdots &{}& \vdots \\ {{\mathit{\boldsymbol{M}}_{k1}}}&{{\mathit{\boldsymbol{M}}_{k2}}}& \cdots &{{\mathit{\boldsymbol{M}}_{kk}}} \end{array}} \right), $

here M_ii, i =1, 2, …, k, are the definite submatri-ces, then

$ \det\left( \mathit{\boldsymbol{M}} \right) \le \det\left( {{\mathit{\boldsymbol{M}}_{11}}} \right)\det\left( {{\mathit{\boldsymbol{M}}_{22}}} \right) \cdots \det \left( {{\mathit{\boldsymbol{M}}_{kk}}} \right). $

Proof   For k=2.set $\boldsymbol{N} = \left[{\begin{array}{*{20}{c}} {{\boldsymbol{M}_{11}}}&0\\ 0&{{\boldsymbol{M}_{22}}} \end{array}} \right]$, according to theorem 1, we have

$ \det\left( {{\mathit{\boldsymbol{N}}^{ - 1}}\mathit{\boldsymbol{M}}} \right) \le \det\left( {{\mathit{\boldsymbol{N}}^{ - 1}}} \right)\det\left( \mathit{\boldsymbol{M}} \right) = {{\rm{ \mathsf{ δ} }}_1}{{\rm{ \mathsf{ δ} }}_2} \cdots {{\rm{ \mathsf{ δ} }}_n} \le 1. $

That is,

$ \det\left( \mathit{\boldsymbol{M}} \right) \le \det\left( \mathit{\boldsymbol{N}} \right) = \det\left( {{\mathit{\boldsymbol{M}}_{11}}} \right)\det\left( {{\mathit{\boldsymbol{M}}_{22}}} \right). $

For k>2, using this manipulation successively gives

$ \det\left( \mathit{\boldsymbol{M}} \right) \le \det\left( {{\mathit{\boldsymbol{M}}_{11}}} \right)\det\left( {{\mathit{\boldsymbol{M}}_{22}}} \right) \cdots \det \left( {{\mathit{\boldsymbol{M}}_{kk}}} \right). $

The proof is completed.

It is clear that Hadamard inequality is the spe-cial case of Fischer inequality when k=n, so the following inequality holds.

Theorem 3   If M=(m_ij) ∈ R^n×nis a symmetric positive definite matrix, then

$ \det\left( \mathit{\boldsymbol{M}} \right) \le {m_{11}}{m_{22}} \cdots {m_{nn}}. $

4 Conclusions and future work

The eigenvalues of a class of matrices related to the real symmetric positive definite matrix are discussed in this paper, and an inequality about the eigenvalues is established.Using this result, the Fischer inequality and the Hadamard inequality of the positive definite matrix are proved.

S and T denote the subsets of set $\mathscr{W}$:= {1, 2, …, n} and $\mathscr{S}$ and $\mathscr{T}$ satisfy $\mathscr{S}$ ∪$\mathscr{T}$=$\mathscr{W}$.cd($\mathscr{S}$) denotes the cardinality of set $\mathscr{S}$.$\mathscr{S}$^c represents the complementary set of $\mathscr{S}$.$\mathscr{M}$_S denotes the principal submatrix determined by set $\mathscr{S}$.

Consider the following of the Koteljanskii, Fan and Szasz inequalities^[15],

$ \begin{array}{*{20}{c}} {\det \left( \mathit{\boldsymbol{M}} \right) \le \frac{{\det \left( {{\mathit{\boldsymbol{M}}_S}} \right)\det \left( {{\mathit{\boldsymbol{M}}_T}} \right)}}{{\det \left( {{\mathit{\boldsymbol{M}}_{S \cup T}}} \right)}},}\\ {\det \left( \mathit{\boldsymbol{M}} \right) \le \det \left( {{\mathit{\boldsymbol{M}}_S}} \right)\prod\limits_{j \in {S^c}} {\frac{{\det \left( {{\mathit{\boldsymbol{M}}_{S \cup \left\{ j \right\}}}} \right)}}{{\det \left( {{\mathit{\boldsymbol{M}}_S}} \right)}}} ,}\\ {\prod\limits_{{\rm{cd}}\left( S \right) = j + 1} {\det {{\left( {{\mathit{\boldsymbol{M}}_S}} \right)}^{\left( {\begin{array}{*{20}{c}} {n - 1}\\ {j - 1} \end{array}} \right)}}} \le \prod\limits_{{\rm{cd}}\left( T \right) = j} {\det {{\left( {{\mathit{\boldsymbol{M}}_T}} \right)}^{\left( {\begin{array}{*{20}{c}} {n - 1}\\ j \end{array}} \right)}}} ,}\\ {j = 1,2, \cdots ,n - 1,} \end{array} $

here, M∈R^n×nand $\left( \begin{array}{l} n\\ j \end{array} \right)$ denotes the combinatorial number.Using this method, these inequalities can be proved and it will be discussed in the future.