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Sylvester's criterion

From Wikipedia, the free encyclopedia
Criterion for positivity of a Hermitian matrix

In mathematics,Sylvester’s criterion is anecessary and sufficient criterion to determine whether aHermitian matrix ispositive-definite, due toJames Joseph Sylvester.

Sylvester's criterion states that an ×n Hermitian matrixM is positive-definiteif and only if all the following matrices have a positivedeterminant:

In other words, all of theleadingprincipal minors must be positive. By using appropriate permutations of rows and columns ofM, it can also be shown that the positivity ofany nested sequence ofn principal minors ofM is equivalent toM being positive-definite.[1]

An analogous theorem holds for characterizingpositive-semidefinite Hermitian matrices, except that it is no longer sufficient to consider only theleading principal minors as illustrated by the Hermitian matrix

A Hermitian matrixM is positive-semidefinite if and only ifallprincipal minors ofM are nonnegative.[2][3]

Proof for the case of positive definite matrices

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SupposeMn{\displaystyle M_{n}} isn×n{\displaystyle n\times n} Hermitian matrixMn=Mn{\displaystyle M_{n}^{\dagger }=M_{n}}. LetMk,k=1,n{\displaystyle M_{k},k=1,\ldots n} be the leading principal minor matrices, i.e. thek×k{\displaystyle k\times k} upper left corner matrices. It will be shown that ifMn{\displaystyle M_{n}} is positive definite, then the principal minors are positive; that is,detMk>0{\displaystyle \det M_{k}>0} for allk{\displaystyle k}.

Mk{\displaystyle M_{k}} is positive definite. Indeed, choosing

we can notice that0<xMnx=xMkx.{\displaystyle 0<x^{\dagger }M_{n}x={\vec {x}}^{\dagger }M_{k}{\vec {x}}.} Equivalently, the eigenvalues ofMk{\displaystyle M_{k}} are positive, and this implies thatdetMk>0{\displaystyle \det M_{k}>0} since the determinant is the product of the eigenvalues.

To prove the reverse implication, we useinduction. The general form of an(n+1)×(n+1){\displaystyle (n+1)\times (n+1)} Hermitian matrix is

whereMn{\displaystyle M_{n}} is ann×n{\displaystyle n\times n} Hermitian matrix,v{\displaystyle {\vec {v}}} is a vector andd{\displaystyle d} is a real constant.

Suppose the criterion holds forMn{\displaystyle M_{n}}. Assuming that all the principal minors ofMn+1{\displaystyle M_{n+1}} are positive implies thatdetMn+1>0{\displaystyle \det M_{n+1}>0},detMn>0{\displaystyle \det M_{n}>0}, and thatMn{\displaystyle M_{n}} is positive definite by the inductive hypothesis. Denote

then

xMn+1x=xMnx+xn+1xv+x¯n+1vx+d|xn+1|2{\displaystyle x^{\dagger }M_{n+1}x={\vec {x}}^{\dagger }M_{n}{\vec {x}}+x_{n+1}{\vec {x}}^{\dagger }{\vec {v}}+{\bar {x}}_{n+1}{\vec {v}}^{\dagger }{\vec {x}}+d|x_{n+1}|^{2}}

By completing the squares, this last expression is equal to

(x+vMn1x¯n+1)Mn(x+xn+1Mn1v)|xn+1|2vMn1v+d|xn+1|2{\displaystyle ({\vec {x}}^{\dagger }+{\vec {v}}^{\dagger }M_{n}^{-1}{\bar {x}}_{n+1})M_{n}({\vec {x}}+x_{n+1}M_{n}^{-1}{\vec {v}})-|x_{n+1}|^{2}{\vec {v}}^{\dagger }M_{n}^{-1}{\vec {v}}+d|x_{n+1}|^{2}}
=(x+c)Mn(x+c)+|xn+1|2(dvMn1v){\displaystyle =({\vec {x}}+{\vec {c}})^{\dagger }M_{n}({\vec {x}}+{\vec {c}})+|x_{n+1}|^{2}(d-{\vec {v}}^{\dagger }M_{n}^{-1}{\vec {v}})}

wherec=xn+1Mn1v{\displaystyle {\vec {c}}=x_{n+1}M_{n}^{-1}{\vec {v}}} (note thatMn1{\displaystyle M_{n}^{-1}} exists because the eigenvalues ofMn{\displaystyle M_{n}} are all positive.) The first term is positive by the inductive hypothesis. We now examine the sign of the second term. By using theblock matrix determinant formula

det(ABCD)=detAdet(DCA1B){\displaystyle \det \left({\begin{array}{cc}A&B\\C&D\end{array}}\right)=\det A\det(D-CA^{-1}B)}

on(){\displaystyle (*)} we obtain

detMn+1=detMn(dvMn1v)>0{\displaystyle \det M_{n+1}=\det M_{n}(d-{\vec {v}}^{\dagger }M_{n}^{-1}{\vec {v}})>0}, which impliesdvMn1v>0{\displaystyle d-{\vec {v}}^{\dagger }M_{n}^{-1}{\vec {v}}>0}.

Consequently,xMn+1x>0.{\displaystyle x^{\dagger }M_{n+1}x>0.}

Proof for the case of positive semidefinite matrices

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LetMn{\displaystyle M_{n}} be ann xn Hermitian matrix. SupposeMn{\displaystyle M_{n}} is semidefinite. Essentially the same proof as for the case thatMn{\displaystyle M_{n}} is strictly positive definite shows that all principal minors (not necessarily the leading principal minors) are non-negative.

For the reverse implication, it suffices to show that ifMn{\displaystyle M_{n}} has all non-negative principal minors, then for allt>0, all leading principal minors of the Hermitian matrixMn+tIn{\displaystyle M_{n}+tI_{n}} are strictly positive, whereIn{\displaystyle I_{n}} is thenxnidentity matrix. Indeed, from the positive definite case, we would know that the matricesMn+tIn{\displaystyle M_{n}+tI_{n}} are strictly positive definite. Since the limit of positive definite matrices is always positive semidefinite, we can taket0{\displaystyle t\to 0} to conclude.

To show this, letMk{\displaystyle M_{k}} be thekth leading principal submatrix ofMn.{\displaystyle M_{n}.} We know thatqk(t)=det(Mk+tIk){\displaystyle q_{k}(t)=\det(M_{k}+tI_{k})} is a polynomial int, related to the characteristic polynomialpMk(t){\displaystyle p_{M_{k}}(t)} viaqk(t)=(1)kpMk(t).{\displaystyle q_{k}(t)=(-1)^{k}p_{M_{k}}(-t).} We use the identity inCharacteristic polynomial#Properties to writeqk(t)=j=0ktkjtr(jMk),{\displaystyle q_{k}(t)=\sum _{j=0}^{k}t^{k-j}\operatorname {tr} \left(\textstyle \bigwedge ^{j}M_{k}\right),}wheretr(jMk){\textstyle \operatorname {tr} \left(\bigwedge ^{j}M_{k}\right)} is the trace of thejth exterior power ofMk.{\displaystyle M_{k}.}

Fromthe definition of minors, we know that the entries in the matrix expansion ofjMk{\displaystyle \bigwedge ^{j}M_{k}} (forj > 0) are just the minors ofMk.{\displaystyle M_{k}.} In particular, the diagonal entries are the principal minors ofMk{\displaystyle M_{k}}, which of course are also principal minors ofMn{\displaystyle M_{n}}, and are thus non-negative. Since the trace of a matrix is the sum of the diagonal entries, it follows thattr(jMk)0.{\displaystyle \operatorname {tr} \left(\textstyle \bigwedge ^{j}M_{k}\right)\geq 0.} Thus the coefficient oftkj{\displaystyle t^{k-j}} inqk(t){\displaystyle q_{k}(t)} is non-negative for allj > 0. Forj = 0, it is clear that the coefficient is 1. In particular,qk(t)>0{\displaystyle q_{k}(t)>0} for allt > 0, which is what was required to show.

Notes

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  1. ^Horn, Roger A.; Johnson, Charles R. (1985),Matrix Analysis,Cambridge University Press,ISBN 978-0-521-38632-6. See Theorem 7.2.5.
  2. ^Carl D. Meyer, Matrix Analysis and Applied Linear Algebra. See section7.6 Positive Definite Matrices, page 566
  3. ^Prussing, John E. (1986),"The Principal Minor Test for Semidefinite Matrices"(PDF),Journal of Guidance, Control, and Dynamics,9 (1):121–122,Bibcode:1986JGCD....9..121P,doi:10.2514/3.20077, archived fromthe original(PDF) on 2017-01-07, retrieved2017-09-28

References

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