1 Introduction
Let \(C^{m\times n}\)
\((R^{m\times n})\) be the set of all complex (real) matrices and let \(\mathbb{M}_{n}^{+}\) be the positive definite Hermitian matrices. Let \(Z^{n\times n}=\{A=(a_{ij})\in R^{n\times n}:a_{ij} \leq0, i\neq j, i, j \in\{1,2,\ldots,n\}\}\). For any \(A=(a_{ij}) \in C^{n\times n}\), its associated matrix is defined by \(A^{\prime}=( \alpha_{ij})\), where \(\alpha_{ii}=\vert a_{ii}\vert \), \(\alpha_{ij}=-\vert a_{ij}\vert \) (\(i\neq j\)). For \(A=(a_{ij})\), \(B=(b_{ij})\)
\(\in C^{m\times n}\), the Hadamard product of A and B is \(A \circ B =(a_{ij}b_{ij})\in C ^{m\times n}\) while their Fan product \(A*B=(c_{ij})\) is defined by \(c_{ii}=a_{ii}b_{ii}\) and \(c_{ij}=-a_{ij}b_{ij}\) for \(i\neq j\).
If \(A=(a_{ij}) \in C^{n\times n}\), then the \(k \times k\) leading principal submatrix of A is denoted by \(A_{k}\) (\(k\in\{1,2,\ldots,n\}\)). \(A_{\alpha}\) denotes the principal submatrix of A, with indices in \(\alpha\subseteq\{1,2,\ldots,n\}\). \(A\in R^{n\times n}\) is called an M-matrix if \(A\in Z^{n\times n}\) and \(\det A_{k}>0\) (\(\forall k\in\{1,2,\ldots,n\}\)), and we denote it by \(A\in M_{n}\). A matrix \(A\in C^{n\times n}\) is called an H-matrix if \(A^{\prime}\) is an M-matrix, and we denote it by \(A\in H_{n}\).
Anzeige
Lynn [1], Theorem 3.1, proved the following determinantal inequality for H-matrices: if \(A, B\in H_{n}\), then
i.e.
$$\det(A\circ B)^{\prime}+\det A^{\prime}\det B^{\prime} \geq \prod_{i=1}^{n}\vert b _{ii} \vert \det A^{\prime}+\prod_{i=1}^{n} \vert a_{ii}\vert \det B^{\prime}, $$
$$\begin{aligned} \det(A\circ B)^{\prime} \geq& \det A^{\prime}\det B^{\prime} \biggl( \frac{\prod_{i=1} ^{n}\vert a_{ii}\vert }{\det A^{\prime}}+\frac{\prod_{i=1}^{n}\vert b_{ii}\vert }{\det B^{\prime}}-1 \biggr). \end{aligned}$$
(1.1)
Chen [2], Theorem 2.7, obtained a determinantal inequality for positive definite matrices: if \(A=(a_{ij})\), \(B=(b_{ij})\in \mathbb{M}_{n}^{+}\), then
Lin [3] recently proved that a similar result to the block positive definite matrices holds for the block Hadamard product.
$$\begin{aligned} \det(A\circ B) \geq&\det A\det B \prod _{k=2}^{n} \biggl( \frac{a_{kk} \det A_{k-1}}{\det A_{k}}+ \frac{b_{kk}\det B_{k-1}}{\det B_{k}}-1 \biggr). \end{aligned}$$
(1.2)
Ando [4], Theorem 5.3, has given the following result: if \(A=(a_{ij})\), \(B=(b_{ij})\) are M-matrices, then
i.e.
$$\begin{aligned}& \det(A*B)+\det A\cdot\det B \\& \quad \geq \Biggl( \prod_{i=1}^{n}a_{ii} \Biggr) \cdot\det B+\det A \cdot \Biggl( \prod_{i=1}^{n}b_{ii} \Biggr), \end{aligned}$$
$$\begin{aligned} \det(A*B) \geq& \det A \det B \biggl( \frac{\prod_{i=1}^{n}a_{ii}}{ \det A} + \frac{\prod_{i=1}^{n}b_{ii}}{\det B}-1 \biggr). \end{aligned}$$
(1.3)
In this paper, we will present some determinantal inequalities for matrices which are generalizations of (1.1), (1.2), and (1.3).
Anzeige
2 Main results and some remarks
We give some lemmas before we present the main theorems of this paper.
Lemma 1
[4], Corollary 4.1.2
Let
\(A=(a_{ij})\in R^{n\times n}\)
be an M-matrix. If
\(\alpha_{i}\subseteq\{1,2,\ldots,n\}\) (\(i=1,2,3, \ldots,N\)) satisfies
\(\alpha_{i}\cap\alpha_{j}=\phi\)
for
\(i\neq j\)
and
\(\bigcup_{j=1}\alpha_{j}=\{1,2,\ldots,n\}\), then
$$\begin{aligned} \det A \leq& \prod_{i=1}^{N} \det(A_{\alpha_{i}}). \end{aligned}$$
In particular,
$$\begin{aligned} \det A \leq& \prod_{i=1}^{n}a_{ii}. \end{aligned}$$
(2.1)
Lemma 3
[5], Theorem 5.2.1
If
A, B
are positive definite matrices and
\(C=A\circ B\), then
C
is positive definite matrix.
Now we present the main results.
First of all, we give a determinantal inequality for the Hadamard product of finite number of H-matrices as follows:
Theorem 5
If
\(A_{1}=(a_{1}^{kl}), A_{2}=(a_{2}^{kl}),\ldots, A_{m}=(a_{m}^{kl})\) (\(k,l=1,\ldots,n\)) are
H-matrices, then
$$\begin{aligned} \det(A_{1}\circ\cdots\circ A_{m})^{\prime} \ge&\det(A_{1}\cdots A_{m}) \\ &{}\times \biggl( \frac{\prod_{i=1}^{n}\vert a_{1}^{ii}\vert }{\det A_{1}^{\prime}}+ \cdots+\frac{\prod_{i=1}^{n}\vert a_{m}^{ii}\vert }{\det A_{m}^{\prime }}-(m-1) \biggr). \end{aligned}$$
(2.2)
Proof
By Lemma 2, it is straightforward to observe that the Hadamard product \(A_{1}\circ\cdots\circ A_{m}\) is an H-matrix. Use induction on k. When \(k=2\), the result is (1.1). Suppose that (2.2) holds when \(k=m-1\)
When \(k=m\), we need to show
By (1.1), we have
By the inductive assumption, the above inequality is
Let
By (2.1), we have
and so
Thus by \(ab\ge a+b-1\) for \(a, b\ge1\), the above inequality (2.3) is
This completes the proof. □
$$\begin{aligned} \det(A_{1}\circ\cdots\circ A_{m-1})^{\prime} \ge&\det\bigl(A_{1}^{\prime} \cdots A _{m-1}^{\prime} \bigr) \\ &{} \times \biggl( \frac{\prod_{i=1}^{n}\vert a_{1}^{ii}\vert }{ \det A_{1}^{\prime}}+\cdots+\frac{\prod_{i=1}^{n}\vert a_{m-1}^{ii}\vert }{\det A _{m-1}^{\prime}}-(m-2) \biggr). \end{aligned}$$
$$\begin{aligned} \det(A_{1}\circ\cdots\circ A_{m})^{\prime} \ge&\det\bigl(A_{1}^{\prime} \cdots A _{m}^{\prime}\bigr) \\ &{} \times \biggl( \frac{\prod_{i=1}^{n}\vert a_{1}^{ii}\vert }{\det A _{1}^{\prime}}+\cdots+\frac{\prod_{i=1}^{n}\vert a_{m}^{ii}\vert }{\det A_{m}^{\prime}}-(m-1) \biggr). \end{aligned}$$
$$\begin{aligned} \det\bigl((A_{1}\circ\cdots\circ A_{m-1}) \circ A_{m}\bigr)^{\prime} \ge&\det\bigl( \bigl(A _{1}^{\prime}\circ\cdots\circ A_{m-1}^{\prime} \bigr)A_{m}^{\prime}\bigr) \\ &{}\times \biggl( \frac{ \prod_{i=1}^{n}( \vert a_{1}^{ii}\vert \cdots \vert a_{m-1}^{ii}\vert )}{\det (A_{1}^{\prime} \circ\cdots\circ A_{m-1}^{\prime})}+\frac{\prod_{i=1}^{n} \vert a_{m}^{ii}\vert }{ \det A_{m}^{\prime}} -1 \biggr). \end{aligned}$$
$$\begin{aligned} \det(A_{1}\circ\cdots\circ A_{m})^{\prime} \ge&\det\bigl(A_{1}^{\prime} \cdots A _{m-1}^{\prime}\bigr)\det A_{m}^{\prime} \\ &{}\times \biggl( \frac{\prod_{i=1}^{n}\vert a _{1}^{ii}\vert }{\det A_{1}^{\prime}}+\cdots+\frac{\prod_{i=1}^{n}\vert a_{m-1}^{ii}\vert }{ \det A_{m-1}^{\prime}}-(m-2) \biggr) \\ &{}\times \biggl( \frac{\prod_{i=1}^{n}( \vert a _{1}^{ii}\vert \cdots \vert a_{m-1}^{ii}\vert )}{\det(A_{1}^{\prime}\circ\cdots \circ A_{m-1}^{\prime})}+\frac{\prod_{i=1}^{n} \vert a_{m}^{ii}\vert }{\det A_{m}^{\prime}} -1 \biggr). \end{aligned}$$
(2.3)
$$\begin{aligned}& a = \biggl( \frac{\prod_{i=1}^{n}\vert a_{1}^{ii}\vert }{\det A_{1}^{\prime }}+\cdots +\frac{\prod_{i=1}^{n}\vert a_{m-1}^{ii}\vert }{\det A_{m-1}^{\prime }}-(m-2) \biggr), \\& b = \biggl( \frac{\prod_{i=1}^{n}( \vert a_{1}^{ii}\vert \cdots \vert a_{m-1}^{ii}\vert )}{ \det(A_{1}^{\prime}\circ\cdots\circ A_{m-1}^{\prime})}+\frac {\prod_{i=1}^{n} \vert a_{m}^{ii}\vert }{\det A_{m}^{\prime}} -1 \biggr). \end{aligned}$$
$$\begin{aligned}& \frac{\prod_{i=1}^{n}\vert a_{j}^{ii}\vert }{\det A_{j}^{\prime}} \geq 1,\quad j=1, \ldots, m, \\& \frac{\prod_{i=1}^{n}( \vert a_{1}^{ii}\vert \cdots \vert a_{m-1}^{ii}\vert )}{\det(A _{1}^{\prime}\circ\cdots\circ A_{m-1}^{\prime})} \geq 1, \end{aligned}$$
$$a,b\geq1. $$
$$\begin{aligned} \det(A_{1}\circ\cdots\circ A_{m})^{\prime} \ge&\det\bigl(A_{1}^{\prime} \cdots A _{m-1}^{\prime}\bigr)\det A_{m}^{\prime} \times a\times b \\ \ge&\det \bigl(A_{1}^{\prime}\cdots A_{m}^{\prime} \bigr)\times(a+b-1) \\ \geq&\det\bigl(A_{1}^{\prime}\cdots A_{m}^{\prime} \bigr) \\ &{}\times \biggl( \frac{\prod_{i=1}^{n}\vert a_{1}^{ii}\vert }{\det A_{1}^{\prime}}+ \cdots+\frac{\prod_{i=1}^{n}\vert a_{m}^{ii}\vert }{\det A_{m}^{\prime }}-(m-1) \biggr). \end{aligned}$$
Second, we achieve a determinantal inequality for the Hadamard product of positive definite matrices as follows:
Theorem 7
If
\(A_{i}\) (\(i=1,\ldots,m\)) (\(m\ge2\)) are
\(n\times n\)
positive definite matrices, the Hadamard product of
\(A_{i}=(a_{i}^{lt})\)
and
\(A_{j}=(a_{j}^{lt}) \) (\(i\neq j\)) is denoted by
\(A_{i}\circ A_{j}\), and
\(A_{i}^{(k)}\)
is the
\(k\times k\) (\(k=1,2,\ldots,n\)) leading principal submatrix of
\(A_{i}\), then
$$\begin{aligned} \det(A_{1}\circ\cdots\circ A_{m}) \ge&\det(A_{1}\cdots A_{m}) \\ & {} \times\prod_{\mu=2}^{n} \biggl( \frac{ a_{1}^{\mu\mu}\det A^{(\mu-1)} _{1}}{\det A^{(\mu)}_{1}} +\cdots+\frac{ a_{m}^{\mu\mu}\det A^{( \mu-1)}_{m}}{\det A_{m}^{(\mu)}}-(m-1) \biggr). \end{aligned}$$
(2.4)
Proof
By Lemma 3, it is straightforward to see that the Hadamard product \(A_{1}\circ\cdots\circ A_{m}\) is a positive definite matrix. Use induction on m. When \(k=2\), the result is (1.2). Suppose that (2.4) holds when \(k=m-1\). We have
When \(k=m\), we need to show
By (1.2), we have
By the inductive assumption, the above inequality is such that
Let
By Fischer’s inequality [5], p.506, we have
and so
Thus by \(a_{\mu}b_{\mu}\ge a_{\mu}+b_{\mu}-1\) for \(a_{\mu}, b _{\mu}\ge1\), the above inequality (2.5) is
This completes the proof. □
$$\begin{aligned} \det(A_{1}\circ\cdots\circ A_{m-1}) \ge&\det(A_{1} \cdots A_{m-1}) \\ &{}\times\prod_{\mu=2}^{n} \biggl( \frac{ a_{1}^{\mu\mu}\det A^{(\mu-1)} _{1}}{\det A^{(\mu)}_{1}} +\cdots+\frac{ a_{m-1}^{\mu\mu}\det A ^{(\mu-1)}_{m-1}}{\det A_{m-1}^{(\mu)}}-(m-2) \biggr). \end{aligned}$$
$$\begin{aligned} \det(A_{1}\circ\cdots\circ A_{m}) \ge&\det(A_{1}\cdots A_{m}) \\ &{} \times\prod_{\mu=2}^{n} \biggl( \frac{ a_{1}^{\mu\mu}\det A^{(\mu-1)} _{1}}{\det A^{(\mu)}_{1}} +\cdots+\frac{ a_{m}^{\mu\mu}\det A^{( \mu-1)}_{m}}{\det A_{m}^{(\mu)}}-(m-1) \biggr). \end{aligned}$$
$$\begin{aligned}& \det\bigl((A_{1}\circ\cdots\circ A_{m-1}) \circ A_{m}\bigr) \\& \quad \ge\det\bigl((A _{1}\circ\cdots\circ A_{m-1})A_{m} \bigr) \\& \qquad {}\times\prod_{\mu=2}^{n} \biggl( \frac{(a_{1}^{\mu\mu} \cdots a_{m-1}^{\mu\mu})\det(A_{1} \circ\cdots\circ A_{m-1})^{(\mu-1)}}{\det(A_{1}\circ\cdots \circ A_{m-1})^{(\mu)}}+\frac{a_{m}^{\mu\mu}\det A^{(\mu-1)}_{m}}{ \det A^{(\mu)}_{m}} -1 \biggr). \end{aligned}$$
$$\begin{aligned}& \det(A_{1}\circ\cdots\circ A_{m-1})\det A_{m} \\& \qquad {}\times\prod_{ \mu=2}^{n} \biggl( \frac{(a_{1}^{\mu\mu} \cdots a_{m-1}^{\mu\mu}) \det(A_{1}\circ\cdots\circ A_{m-1})^{(\mu-1)}}{\det(A_{1}\circ \cdots\circ A_{m-1})^{(\mu)}}+\frac{a_{m}^{\mu\mu}\det A^{(\mu-1)} _{m}}{\det A^{(\mu)}_{m}} -1 \biggr) \\& \quad \ge\det(A_{1}\cdots A_{m-1}) \det A_{m} \\& \qquad {} \times\prod_{\mu=2}^{n} \biggl( \frac{ a_{1}^{\mu \mu}\det A^{(\mu-1)}_{1}}{\det A^{(\mu)}_{1}} +\frac{ a_{2}^{ \mu\mu}\det A^{(\mu-1)}_{2}}{\det A^{(\mu)}_{2}}+\cdots+\frac{ a _{m}^{\mu\mu}\det A^{(\mu-1)}_{m-1}}{\det A_{m-1}^{(\mu)}}-(m-2) \biggr) \\& \qquad {}\times\prod_{\mu=2}^{n} \biggl( \frac{(a_{1}^{\mu\mu} \cdots a _{m-1}^{\mu\mu})\det(A_{1}\circ\cdots\circ A_{m-1})^{(\mu-1)}}{ \det(A_{1}\circ\cdots\circ A_{m-1})^{(\mu)}}+\frac{a_{m}^{\mu \mu}\det A^{(\mu-1)}_{m}}{\det A^{(\mu)}_{m}} -1 \biggr). \end{aligned}$$
(2.5)
$$\begin{aligned}& a_{\mu}=\frac{ a_{1}^{\mu\mu}\det A^{(\mu-1)}_{1}}{\det A^{( \mu)}_{1}} +\frac{ a_{2}^{\mu\mu}\det A^{(\mu-1)}_{2}}{\det A^{( \mu)}_{2}}+\cdots+ \frac{ a_{n}^{\mu\mu}\det A^{(\mu-1)}_{m-1}}{ \det A_{m-1}^{(\mu)}}-(m-2), \\& b_{\mu}=\frac{(a_{1}^{\mu\mu} \cdots a_{m-1}^{\mu\mu})\det(A _{1}\circ\cdots\circ A_{m-1})^{(\mu-1)}}{\det(A_{1}\circ\cdots \circ A_{m-1})^{(\mu)}}+\frac{a_{m}^{\mu\mu}\det A^{(\mu-1)}_{m}}{ \det A^{(\mu)}_{m}} -1. \end{aligned}$$
$$\begin{aligned}& \frac{a_{i}^{\mu\mu}\det A^{(\mu-1)}_{i}}{\det A^{(\mu)}_{i}} \geq1,\quad i=1,\ldots,m, \\& \frac{(a_{1}^{\mu\mu} \cdots a_{m-1}^{\mu\mu})\det(A_{1}\circ \cdots\circ A_{m-1})^{(\mu-1)}}{\det(A_{1}\circ\cdots\circ A_{m-1})^{( \mu)}}-1 \geq 0, \end{aligned}$$
$$a_{\mu}, b_{\mu}\geq1. $$
$$\begin{aligned}& \det(A_{1}\circ\cdots\circ A_{m}) \\& \quad \geq\det(A_{1}\cdots A_{m-1}) \det A_{m}\times\prod_{\mu=2}^{n} a_{\mu}b_{\mu} \\& \quad \geq\det(A _{1}\cdots A_{m})\times\prod _{\mu=2}^{n}( a_{\mu}+b_{\mu}-1) \\& \quad \geq\det(A_{1}\cdots A_{m}) \prod _{\mu=2}^{n} \biggl( \frac{ a_{1} ^{\mu\mu}\det A^{(\mu-1)}_{1}}{\det A^{(\mu)}_{1}} +\cdots+ \frac{a _{m}^{\mu\mu}\det A^{(\mu-1)}_{m}}{\det A^{(\mu)}_{m}} -(m-1) \biggr). \end{aligned}$$
Finally, a result on Fan product of M-matrices is obtained in the following theorem.
Theorem 9
If
\(A_{1}=(a_{1}^{kl}), A_{2}=(a_{2}^{kl}),\ldots, A_{m}=(a_{m}^{kl})\) (\(k,l=1,\ldots,n\)) are
M-matrices, then
$$\begin{aligned} \det(A_{1}* \cdots * A_{m}) \ge &\det(A_{1}\cdots A_{m}) \\ &{}\times \biggl( \frac{\prod_{i=1}^{n}a_{1}^{ii}}{\det A_{1}}+\cdots+\frac{ \prod_{i=1}^{n}a_{m}^{ii}}{\det A_{m}}-(m-1) \biggr). \end{aligned}$$
(2.6)
Proof
By Lemma 4, it is straightforward to see that the Hadamard product \(A_{1}\ast\cdots\ast A_{m}\) is an M-matrix. Use induction on k. When \(k=2\), the result is (1.3). Let \(k=m-1\), (2.6) holds:
When \(k=m\), we need to show
By (1.3), we have
By the inductive assumption, the above inequality is
Let
By (2.1), we have
and so
So by \(ab\ge a+b-1\) for \(a, b\ge1\), the above inequality (2.7) is
This completes the proof. □
$$\begin{aligned} \det(A_{1}\circ\cdots \circ A_{m-1}) \ge&\det(A_{1} \cdots A_{m-1}) \\ &{} \times \biggl( \frac{\prod_{i=1}^{n}a_{1}^{ii}}{\det A_{1}}+\cdots +\frac{ \prod_{i=1}^{n}a_{m-1}^{ii}}{\det A_{m-1}}-(m-2) \biggr). \end{aligned}$$
$$\begin{aligned} \det(A_{1}* \cdots * A_{m}) \ge& \det(A_{1}\cdots A_{m}) \\ &{} \times \biggl( \frac{\prod_{i=1}^{n}a_{1}^{ii}}{\det A_{1}}+\cdots+\frac{ \prod_{i=1}^{n}a_{m}^{ii}}{\det A_{m}}-(m-1) \biggr). \end{aligned}$$
$$\begin{aligned} \det\bigl((A_{1}* \cdots * A_{m-1}) \ast A_{m}\bigr) \ge&\det\bigl((A_{1}* \cdots * A_{m-1})A_{m}\bigr) \\ &{}\times \biggl( \frac{\prod_{i=1}^{n}( a_{1} ^{ii} \cdots a_{m-1}^{ii})}{\det(A_{1}* \cdots * A_{m-1})}+\frac{ \prod_{i=1}^{n} a_{m}^{ii}}{\det A_{m}} -1 \biggr). \end{aligned}$$
$$\begin{aligned} \det(A_{1}* \cdots * A_{m}) \ge&\det(A_{1}\cdots A_{m-1})\det A _{m} \\ &{} \times \biggl( \frac{\prod_{i=1}^{n}a_{1}^{ii}}{\det A_{1}}+ \cdots+\frac{\prod_{i=1}^{n}a_{m-1}^{ii}}{\det A_{m-1}}-(m-2) \biggr) \\ &{}\times \biggl( \frac{\prod_{i=1}^{n}( a_{1}^{ii} \cdots a_{m-1}^{ii})}{ \det(A_{1}* \cdots * A_{m-1})}+\frac{\prod_{i=1}^{n} a_{m}^{ii}}{ \det A_{m}} -1 \biggr). \end{aligned}$$
(2.7)
$$\begin{aligned}& a = \biggl( \frac{\prod_{i=1}^{n}a_{1}^{ii}}{\det A_{1}}+\cdots+\frac{ \prod_{i=1}^{n}a_{m-1}^{ii}}{\det A_{m-1}}-(m-2) \biggr), \\& b = \biggl( \frac{\prod_{i=1}^{n}( a_{1}^{ii} \cdots a_{m-1}^{ii})}{ \det(A_{1}\circ\cdots\circ A_{m-1})}+\frac{\prod_{i=1}^{n} a_{m} ^{ii}}{\det A_{m}} -1 \biggr). \end{aligned}$$
$$\begin{aligned}& \frac{\prod_{i=1}^{n}a_{j}^{ii}}{\det A_{j}}\geq1,\quad j=1,\ldots, m, \\& \frac{\prod_{i=1}^{n}( a_{1}^{ii} \cdots a_{m-1}^{ii})}{\det(A_{1} \circ\cdots\circ A_{m-1})} \geq 1, \end{aligned}$$
$$a,b\geq1. $$
$$\begin{aligned} \det(A_{1}*\cdots * A_{m}) \ge& \det(A_{1}\cdots A_{m}) \\ &{}\times \biggl( \frac{\prod_{i=1}^{n}a_{1}^{ii}}{\det A_{1}}+\cdots+\frac{ \prod_{i=1}^{n}a_{m}^{ii}}{\det A_{m}}-(m-1) \biggr). \end{aligned}$$
Acknowledgements
We are grateful to Dr. Limin Zou for fruitful discussions. This research was supported by the key project of the applied mathematics of Hainan Normal University, the natural science foundation of Hainan Province (No. 20161005), the Chongqing Graduate Student Research Innovation Project (No. CYS14020) and the Doctoral scientific research foundation of Hainan Normal University.
Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Competing interests
The authors declare to have no competing interests.
Authors’ contributions
Xiaohui Fu carried out all the proofs of the results and gave the generalizations of Fan product. Yang Liu participated in the design of the study and drafted the manuscript. All authors read and approved the final manuscript.