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单词 ProofOfHadamardsInequality
释义

proof of Hadamard’s inequality


Let’s first prove the second inequality. If A is singular, the thesis istrivially verified, since for a Hermitian positive semidefinite matrix theright-hand side is always nonnegative, all the diagonalMathworldPlanetmath entries being nonnegative. Let’s thus assume det(A)0, which means, A beingHermitian positive semidefinitePlanetmathPlanetmath, det(A)>0. Then no diagonal entry of Acan be 0 (otherwise, since for a Hermitian positive semidefinite matrix, 0λminaiiλmax, λmin and λmax being respectively the minimal and the maximal eigenvalueMathworldPlanetmathPlanetmathPlanetmathPlanetmath, this would imply λmin=0, that is A is singular); for this reason we can define D=diag(d11,d22,,dnn), with dii=aii-12, since all aii+. Let’s furthermore define B=DAD. It’s easy to check that B too is Hermitian positive semidefinite, so its eigenvalues λB are all non-negative (actually, since A=AH and since D is real and diagonal, BH=(DAD)H=DH(DA)H=DHAHDH=DAD=B; on the other hand, for any 𝐱𝟎, 𝐱HB𝐱=𝐱HDAD𝐱=(𝐱HD)A(D𝐱)=(DH𝐱)HA(D𝐱)=(D𝐱)HA(D𝐱)=𝐲HA𝐲0).Moreover, we have obviously bii=diiaiidii=1 so that tr(B)=nand, recalling the geometric-arithmetic mean inequality (http://planetmath.org/ArithmeticGeometricMeansInequality), which holds in thiscase because the eigenvalues of B are all non-negative,

det(B)=i=1nλB(i)(1ni=1nλB(i))n=(1ntr(B))n=1,

and since det(B)=det(D)2det(A)=(i=1naii)-1det(A), we have the thesis.Since det(A)=i=1naii if and only if det(B)=1 and since in the geometric-arithmetic inequality equality holds ifand only if all terms are equal, we must have λB(i)=λB, so that i=1nλB(i)=λBn=1,whence λB=1 (λB having to be non-negative), and sinceB is Hermitian and hence is diagonalizable, we obtain B=I, and so A=D-1BD-1=D-2=diag(a11,a22,,ann). So we canconclude that equality holds if and only if A is diagonal.

Let’s now derive the more general first inequality.Let A be a complex-valued n×n matrix. If A is singular,the thesis is trivially verified. Let’s thus assume det(A)0; then B=AAH is a Hermitian positive semidefinite matrix (actually, BH=(AAH)H=(AH)HAH=AAH=B and, for any 𝐱𝟎,𝐱HB𝐱=𝐱HAAH𝐱=(𝐱HA)(AH𝐱)=(AH𝐱)H(AH𝐱)=𝐲H𝐲=𝐲220). Therefore, the second inequality can be applied to B, yielding:

|det(A)|2=det(A)det(A)=det(A)det(AH)=det(AAH)=
=det(B)i=1nbii=i=1nj=1naij(aH)ji=i=1nj=1naijaij=i=1nj=1n|aij|2.

As we proved above, for det(B) to be equal to i=1nbii, B must be diagonal, which means that k=1naikajk=|aii|2δij. So we can conclude that equality holds if and only if the rows of A are orthogonalMathworldPlanetmathPlanetmath.

References

  • 1 R. A. Horn, C. R. Johnson,Matrix Analysis, Cambridge University Press, 1985
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