Per-unit system: Difference between revisions

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{{Hatnote|Another inequality is called the [[Hermite–Hadamard inequality]].}}
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In [[mathematics]], '''Hadamard's inequality''', first published by [[Jacques Hadamard]] in 1893,<ref name=MandS>Maz'ya & Shaposhnikova</ref> is a bound on the [[determinant]] of a [[matrix (mathematics)|matrix]] whose entries are [[complex number]]s in terms of the lengths of its column vectors. In geometrical terms, when restricted to real numbers, it bounds the [[volume]] in [[Euclidean space]] of ''n'' dimensions marked out by ''n'' vectors ''v<sub>i</sub>'' for 1 ≤ ''i'' ≤ ''n'' in terms of the lengths of these vectors ||''v<sub>i</sub>''||.
 
Specifically, Hadamard's inequality states that if ''N'' is the matrix having columns<ref>The result is sometimes stated in terms of row vectors. That this is equivalent is seen by applying the transpose.</ref> ''v<sub>i</sub>'', then
:<math>|\det(N)|\le \prod_{i=1}^n \|v_i\|,</math>
and equality is achieved if and only if the vectors are [[orthogonal]] or at least one of the columns is 0.
 
==Alternate forms and corollaries==
A corollary is that if the entries of an ''n'' by ''n'' matrix ''N'' are bounded by B, so |''N<sub>ij</sub>''|≤''B'' for all ''i'' and ''j'', then
:<math>|\det(N)|\le B^nn^{n/2}.</math>
In particular, if the entries of N are +1 and &minus;1 only then<ref>Garling</ref>
:<math>|\det(N)|\le n^{n/2}.</math>
In [[combinatorics]], matrices ''N'' for which equality holds, i.e. those with orthogonal columns, are called [[Hadamard matrix|Hadamard matrices]].
 
A [[positive-semidefinite matrix]] ''P'' can be written as ''N''<sup>*</sup>''N'', where ''N''<sup>*</sup> denotes the [[conjugate transpose]] of ''N'' (see [[Cholesky decomposition]]). Then
:<math>\det(P)=\det(N)^2 \le \prod_{i=1}^n \|v_i\|^2 = \prod_{i=1}^n p_{ii}.</math>
So, the determinant of a [[positive definite matrix]] is less than or equal to the product of its diagonal entries. Sometimes this is also known as Hadamard's inequality.<ref name=MandS/>
 
==Proof==
The result is trivial if the matrix N is [[Invertible matrix|singular]], so assume the columns of N are linearly independent. By dividing each column by its length, it can be seen that the result is equivalent to the special case where each column has length 1, in other words if ''e<sub>i</sub>'' are unit vectors and ''M'' is the matrix having the ''e<sub>i</sub>'' as columns then
:<math>|\det M|\le 1,</math>
and equality is achieved if and only if the vectors are an [[orthogonal set]], that is when the matrix is [[Unitary matrix|unitary]]. The general result now follows:
:<math>|\det N| = \bigg (\prod_{i=1}^n \|v_i\| \bigg) |\det M| \leq \prod_{i=1}^n \|v_i\|.</math>
 
For the positive definite case, let ''P'' =''M<sup>*</sup>M'' and let the eigenvalues of ''P'' be λ<sub>1</sub>, λ<sub>2</sub>, … λ<sub>''n''</sub>. By assumption, each entry in the diagonal of ''P'' is 1, so the [[Trace (linear algebra)|trace]] of ''P'' is ''n''. Applying the [[inequality of arithmetic and geometric means]],
:<math>\det P=\prod_{i=1}^n \lambda_i \le \bigg({1 \over n}\sum_{i=1}^n \lambda_i\bigg)^n = \left({1 \over n} \mathrm{tr} P\right)^n = 1^n = 1,</math>
so
:<math>|\det M| = \sqrt{\det P} \le 1.</math>
 
If there is equality then each of the λ<sub>''i''</sub>'s must all be equal and their sum is ''n'', so they must all be 1. The matrix ''P'' is Hermitian, therefore diagonalizable, so it is the identity matrix—in other words the columns of ''M'' are an orthonormal set and the columns of ''N'' are an orthogonal set.<ref>Proof follows, with minor modifications, the second proof given in Maz'ya & Shaposhnikova. See also {{PlanetMath|urlname=7541|title=Proof of Hadamard's inequality}}.</ref>
 
Many other proofs can be found in the literature.
 
==References==
{{Reflist}}
*{{cite book |title=Jacques Hadamard: A Universal Mathematician
|first1=Vladimir|last1=Maz'ya|first2=T. O.|last2=Shaposhnikova
|publisher=AMS|year=1999|isbn=0-8218-1923-2|pages=383ff.}}
*{{cite book |title=Inequalities: A Journey into Linear Analysis
|first=D. J. H.|last=Garling
|publisher=Cambridge|year=2007|isbn=0-521-69973-8|page=233}}
*{{cite book |title=Functional Analysis
|first1=Frigyes|last1=Riesz|first2=Béla|last2=Szőkefalvi-Nagy
|publisher=Dover|year=1990|isbn=0-486-66289-6|page=176}}
*{{MathWorld|title=Hadamard's Inequality|urlname=HadamardsInequality}}
 
==Further reading==
*{{cite book |title=Inequalities
|first1=Edwin F|last1=Beckenbach|first2=Richard Ernest|last2=Bellman
|publisher=Springer|year=1965|page=64}}
 
{{DEFAULTSORT:Hadamard's Inequality}}
[[Category:Inequalities]]
[[Category:Determinants]]

Latest revision as of 23:00, 27 October 2014

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