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A metric for covariance matrices

A Metric for Covariance Matrices

Wolfgang F¨o rstner and Boudewijn Moonen

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Carl Friedrich Gauss

Abstract

The paper presents a metric for positive de?nite covariance matrices.It is a natural expression involving traces and joint eigenvalues of the matrices.It is shown to be the distance coming from a canonical invariant Riemannian metric on the space Sym +(n,R )of real symmetric positive de?nite matrices

In contrast to known measures,collected e.g.in Grafarend 1972,the metric is invariant under a?ne transformations and inversion.It can be used for evaluating covariance matrices or for optimization of measurement designs.

Keywords:Covariance matrices,metric,Lie groups,Riemannian manifolds,exponential map-ping,symmetric spaces

1Background

The optimization of geodetic networks is a classical problem that has gained large attention in the 70s.

1972E.W.Grafarend put together the current knowledge of network design,datum transfor-mations and arti?cial covariance matrices using covariance functions in his classical monograph

[?];see also [?].One critical part was the development of a suitable measure for comparing two covariance matrices.Grafarend listed a dozen measures.Assuming a completely isotropic network,represented by a unit matrix as covariance matrix,the measures depended on the eigenvalues of the covariance matrix.

1983,11years later,at the Aalborg workshop on ’Survey Control Networks’Schmidt [?]used these measures for ?nding optimal networks.The visualization of the error ellipses for a single point,leading to the same deviation from an ideal covariance structure revealed de?ciencies of these measures,as e.g.the trace of the eigenvalues of the covariance matrix as quality measure ?]emphasized by the authors

would allow a totally ?at error ellipse to be as good as a circular ellipse,more even,as good as the ?at error ellipse rotated by 900.

Based on an information theoretic point of view,where the information of

a Gaussian variable increases with ln σ2,the ?rst author guessed the squared sum d 2= i ln 2λi of the logarithms

of the eigenvalues to be a better measure,as deviations in both directions would be punished the same amount if measured in percent,i.e.relative to the given variances.He formulated the conjecture that the distance measure d would be a metric.Only in case d would be a metric,comparing two covariance matrices A and B with covariance matrix C would allow to state one of the two to be better than the other.Extensive simulations by K.Ballein [?]substantiated this conjecture as no case was found where the triangle inequality was violated.

1995,12years later,taking up the problem within image processing,the ?rst author proved the validity of the conjecture for 2×2-matrices [?].For this case the measure already had been proposed by V.V.Kavrajski [?]for evaluating the isotropy of map projections.However,the proof could not be generalized to higher https://www.sodocs.net/doc/2611409275.html,ing classical results from linear algebra and di?erential geometry the second author proved the distance d to be a metric for general positive de?nite symmetric matrices.An extended proof can be found in [?].

This paper states the problem and presents the two proofs for 2×2-matrices and for the general case.Giving two proofs for n =2may be justi?ed by the two very di?erent approaches to the problem.

2Motivation

Comparing covariance matrices is a basic task in mensuration design.The idea,going back to Baarda 1973[?]is to compare the variances of arbitrary functions f =e T x on one hand determined with a given covariance matrix C and on the other hand determined with a reference or criterion matrix H .One requirement would be the variance σ2(C )f

of f when calculated with C to be always smaller than the variance σ2(H )f of f when calculated with H .This means:

e T Ce ≤e T He for all e =0

or the Raleigh ratio

0≤λ(e )=e T Ce e T He

≤1for all e =0.The maximum λfrom 1/2?λ(e )/?e =0?λHe ?Ce =(λH ?C )e =0results in the maximum eigenvalue λmax (CH ?1)from the generalized eigenvalue problem

|λH ?C |=0,(1)Observe that λe T He ?e T Ce =e T (λH ?C )e =0for e =0only is ful?lled if (??)holds.The eigenvalues of (??)are non-negative if the two matrices are positive semide?nite.

This suggests the eigenvalues of CH ?1to capture the di?erence in form of C and H completely.The requirement λmax ≤1can be visualized by stating that the (error)ellipsoid x T C ?1x =c lies completely within the (error)ellipsoid x T H ?1x =c .

The statistical interpretation of the ellipses results from the assumption,motivated by the principle of maximum entropy,that the stochastical variables are normally distributed,thus having density:

p (x )=1 (2π)n det Σ

e ?12x T Σ?1x

with covariance matrix Σ.Isolines of constant density are ellipses with semiaxes proportional to the square roots of the eigenvalues of Σ.The ratio

λmax =max e σ2(C )f σ2(H )

f thus gives the worst case for the ratio of the variances when calculated with the covariances C and H respectively.

Instead of requiring the worst precision to be better than a speci?cation one also could require the covariance matrix C to be closest to H in some sense.Let us for a moment assume H =I .Simple examples for measuring the di?erence in form of C compared to I are the trace

tr C =

n i =1λi (C )(2)or the determinant

det C =n

i =1λi (C ).(3)

These classical measures are invariant with respect to rotations (??)or a?ne transformations (??)of the coordinate system.Visualizing covariance matrices of equal trace or determinant can use the eigenvalues.Restricting to n =2in a 2D-coordinate system (λ1,λ2)covariance matrices of equal trace tr(C )=c tr are characterized by the straight line λ1=c tr ?λ2or σ21=c tr ?σ22.Covariance matrices of equal determinant det(C )=c det are determined by the hyperbola λ1=c det /λ2or σ21=c det /σ22.Obviously in both cases covariance matrices with very ?at form of the corresponding error ellipse e T Ce =c are allowed.E.g.,if one requires c tr =2then the pair (0.02,1.98)with a ratio of semiaxes 1.98/0.02=7is evaluated as being similar to the unit circle.The determinant measure is even more unfavourable.When requiring c det =1even a pair (0.02,50.0)with ratio of semiaxes 50is called similar to the unit circle.

However,it would be desirable that the similarity between two covariance matrices re?ects the deviation in variance in both directions according to the ratio of the variances.Thus deviations in variance by a factor f should be evaluated equally as a deviation by a factor 1/f ,of course a factor f =1indicating no di?erence.Thus other measures capturing the anisotropy,such as (1?λ1)2+(1?λ2)2,not being invariant to inversion,cannot be used.

The conditions can be ful?lled by using the sum of the squared logarithms of the eigenvalues.Thus we propose the distance measure

d (A ,B )= n i =1ln 2λi (A ,B )

(4)

between symmetric positive de?nite matrices A and B ,with the eigenvalues λi (A ,B )from |λA ?B |=0.The logarithm guarantees,that deviations are measured as factors,whereas the squaring guarantees factors f and 1/f being evaluated equally.Summing squares is done in close resemblance with the Euclidean metric.

This note wants to discuss the properties of d :

?d is invariant with respect to a?ne transformations of the coordinate system.

?d is invariant with respect to an inversion of the matrices.

?It is claimed that d is a metric .Thus

(i)positivity:d(A,B)≥0,and d(A,B)=0only if A=B.

(ii)symmetry:d(A,B)=d(B,A),

(iii)triangle inequality:d(A,B)+d(A,C)≥d(B,C).

The proof for n=2is given in the next Section.The proof for general n is sketched in the subsequent Sections??–??.

3Invariance Properties

3.1A?ne Transformations

Assume the n×n matrix X to be regular.Then the distance d(A,B)of the transformed matrices

A=XAX T B=XBX T

is invariant w.r.t.X.

Proof:We immediately obtain:

λ(A,B)=λ(XAX T,XBX T)=λ(XAX T(XBX T)?1)

=λ(XAX T(X T)?1B?1X?1)=λ(XAB?1X?1)

=λ(AB?1)=λ(A,B).

Comment:A and B can be interpreted as covariance matrices of y=Xx in case A and B are the covariance matrices of x.Changing coordinate system does not change the evaluation of covariance matrices.Obviously,this invariance only relies on the properties of the eigenvalues, and actually was the basis for Baarda’s evaluation scheme using so-called S-transformations. 3.2Inversion

The distance is invariant under inversion of the matrices.

Proof:We obtain

d2(A?1,B?1)=d2(A?1B)=

i=1

lnλi(A?1B)

i=1

ln[λ?1i(AB?1)]

i=1

?lnλi(AB?1)

i=1

lnλi(AB?1)

=d2(AB?1)

=d2(A,B).

Comment:A?1and B?1can be interpreted as weight matrices of x if one choosesσ2o=1. Here essential use is made of the propertyλ(A)=λ?1(A?1).The proof shows,that also individual inversions of eigenvalues do not change the value of distance measure,as required.

3.3d is a Distance Measure

We show that d is a distance measure,thus the?rst two criteria for a metric hold in general. ad1d≥0is trivial from the de?nition of d,keeping in mind,that the eigenvalues all are positive.Proof of d=0?A=B:

←:If A=B then d=0.

Proof:Fromλ(AB?1)=λ(I)followsλi=1for all i,thus d=0.

→:If d=0then A=B.

Proof:From d=0followsλi(AB?1)=1for all i,thus AB?1=I from which

A=B follows.

ad2As(AB?1)?1=BA?1symmetry follows from the inversion invariance.

3.4Triangle Inequality

For d providing a metric on the symmetric positive de?nite matrices the triangle inequality must hold.

Assume three n×n matrices with the following structure:

?The?rst matrix is the unit matrix:

A=I.

?The second matrix is a diagonal matrix with entries e b i thus

B=Diag(e b i).

?The third matrix is a general matrix with eigenvalues e c i and modal matrix R,thus

C=R Diag(e c i)R T.

This setup can be chosen without loss of generality,as A and B can be orthogonalized simulta-neously[?].

The triangle inequality can be written in the following form and reveals three terms

s .

=d(A,B)+d(A,C)?d(B,C)=d c+d b?d a≥0.(5)

The idea of the proof is the following:

(i)We?rst use the fact that d b and d c are independent on the rotation R.

s(R)=d c+d b?d a(R).

(ii)In case R=I then the correctness of(??)results from the triangle inequality in R n.This even holds for any permutation P(i)of the indices i of the eigenvaluesλi of BC?1.There exists a permutation P max for which d a is maximum,thus s(R)is a minimum.

(iii)We then want to show,and this is the crucial part,that any rotation R=I leads to a decrease of d a(R),thus to an increase of s(R)keeping the triangle inequality to hold. 3.4.1Distances d c and d b

The distances d c and d b are given by

d2c=

i=1

b2i,d2b=

i=1

c2i.

The special de?nition of the matrices B and C now shows to be useful.The last expression results from the fact that the eigenvalues of CA?1=C are independent on rotations R.

3.4.2Triangle Inequality for No Rotation

In case of no rotations the eigenvalues of BC?1are e b i/e c i.Therefore the distance d a yields

d2a=

i=1

e b i

e c i

i=1

(b i?c i)2.

With the vectors b=(b i)and c=(c i)the triangle inequality in R n yields

|c|+|b|?|b?c|≥0

i=1

c2i+

i=1

b2i?

i=1

(b i?c i)2≥0.

holds.

For any permutation P(i)we also get

i=1

P(i)

i=1

b2i?

i=1

(b i?c P(i))2≥0.(6)

which guarantees that there is a permutation P max(i)for which s in(??)is minimum.

3.4.3d is Metric for2×2-Matrices

We now want to show that the triangle inequality holds for2×2matrices.Thus we only need to show that s(R(φ))is monotonous withφin[0..π/2],or equivalently that d a(R)is monotonous. We assume(observe the change of notation in the entries b i and c i of the matrices)

b10

0b2

,b1>0,b2>0.(7)

With

x=sinφ(8) the rotation R(x)=R(φ)is represented as

R(x)= √

1?x2x

√

1?x2

thus only values x∈[0,1]need to be investigated.

With the diagonal matrix Diag(c1,c2),containing the positive eigenvalues

c1>0,c2>0,(9) this leads to the general matrix

C=R

c10

0c2

R T=

c1x2+c2(1?x2)?x

√

1?x2(c2?c1)

√

1?x2(c2?c1)c1(1?x2)+c2x2

The eigenvalues of CB?1are(from Maple)

λ1=u(x)+

v(x)

2b1b2

≥0,λ2=u(x)?

v(x)

2b1b2

≥0

with the discriminant

v(x)=u(x)2?w≥0

and

u(x)=(b1c1+c2b2)(1?x2)+(b1c2+b2c1)x2≥0,(10)

w=4b1b2c1c2≥0,(11) the last inequality holding due to(??)(??).The distance

d a(x)=

ln2λ1(x)+ln2λ2(x),

which is dependent on x,has?rst derivative

?d a(x)?x =2

x(b2?b1)(c2?c1)

v(x)d a(x)

u(x)?

v(x)

2b1b2

?ln u(x)+

v(x)

2b1b2

x(b2?b1)(c2?c1)ln

u(x)?

v(x)

u(x)+

v(x)

v(x)d a(x)

.(12)

For?xed b1,b2,c1and c2this expression does not change sign in x∈[0,1].This is because the discriminant v(x)=u2(x)?w(x)(cf.(??))is always positive,due to

v(0)=(b1c1?b2c2)2≥0

v(1)=(b2c1?b1c2)2≥0

?v(x)

=?4x(b2?b1)(c2?c1)u(x)

with u(x)≥0(cf.(??))thus v(x)being monotonous.Furthermore,v(x)is always smaller than u2(cf.(??),(??)),thus the logarithmic expression always negative.As the triangle equation is ful?lled at the extremes of the interval[0,1]it is ful?lled for all x,thus for all rotations.

Comment:When substituting x=sinφ(cf.(??))the?rst derivative(??)is of the form ?d a(φ)/?φ=sinφf(φ)with a symmetric function f(φ)=f(?φ).Thus the derivative is skew symmetric w.r.t.(0,0),indication d a to be symmetric d a(φ)=d a(?φ),which is to be expected.

3.4.4d is Claimed to be a Metric for n×n-Matrices

The proof of the metric properties of d for2×2matrices suggests that in the general case of n×n matrices any rotation away from the worst permutation of the indices(cf.(??))results in an increase of the value s.The proof for the case n=2can be used to show,that,starting with the worst permutation of the indices,any single rotation around one of the axes leads to a monotonous change of s.Therefore,for proving the case of general n,there would have to be shown that any combination of two rotations away from the worst permutation leads to a monotonous change of s allowing to reach any permutation by a rotation while increasing s(R). Completing this line of proof has not been achieved so far.

4Restating the Problem

In der K¨u rze liegt die W¨u rze.

Deutscher Volksmund

Let

M (n,R ):={A =(a ij )|1≤i,j ≤n ,a ij ∈R }

be the space of real n ×n -matrices,and let

S +:=Sym +(n,R ):= A ∈M (n,R ) A =A T ,A >0

be the subspace of real,symmetric,positive de?nite matrices.Recall that any symmetric matrix A can be substituted into a function f :R ?→R which gives a symmetric matrix f (A )commuting with all matrices commuting with A .In particular,a symmetric matrix A has an exponential exp (A ),and a symmetric positive de?nite matrix A has a logarithm ln (A ),and these assignments are inverse to each other.A symmetric positive de?nite matrix

A also has a unique square root √A which is of the same type.De?ne,for A ,

B ∈Sym +(n,R ),their

distance d (A ,B )≥0by

d 2(A ,B ):=tr ln 2 √A ?1B √A ?1 ,

(13)where tr denotes the trace.In particular,this shows that

d (A ,B )≥0

,d (A ,B )=0??A =B .In more down-to-earth terms:

d (A ,B )=

n i =1ln 2λi (A ,B ),(14)

where λ1(A ,B ),...,λn (A ,B )are the joint eigenvalues of A and B ,i.e.the roots of the equation

det(λA ?B )=0.

This is the proposal of [?],i.e.of equation (??)above.(To see why these two de?nitions coincide,note that λA ?B =√A (λE ?√A ?1B √A ?1)√A ,

so that the joint eigenvalues

λi (A ,B )are just the eigenvalues of the real symmetric positive de?nite matrix √A ?1B √A ?1;in particular,they are positive real numbers and so the de?nition (??)makes sense.)The equation (??)shows that d is invariant under congruence transformations with X ∈GL (n,R ),where GL (n,R )is the group of regular linear transformations of R n :

?X ∈GL (n,R ):d (A ,B )=d (XAX T ,XBX T )

(15)(since det (λA ?B )and det X (λA ?B )X T have the same roots);this is not easily seen from

de?nition (??).It also shows that

d (A ,B )=d (B ,A ),d (A ,B )=d (A ?1,B ?1).

5The results

One then has

Theorem1.The map d de?nes a distance on the space Sym+(n,R),i.e there holds

(i)Positivity:d(A,B)≥0,and d(A,B)=0??A=B

(ii)Symmetry:d(A,B)=d(B,A)

(iii)Triangle inequality:d(A,C)≤d(A,B)+d(B,C)

for all A,B,C∈Sym+(n,R).Moreover,d has the following invariances:

(iv)It is invariant under congruence transformations,i.e.

d(A,B)=d(XAX T,XBX T)

for all A,B∈Sym+(n,R),X∈GL(n,R)

(v)It is invariant under inversion,i.e.

d(A,B)=d(A?1,B?1)

for all A,B∈Sym+(n,R)

The same conclusions hold for the space SSym(n,R)of real symmetric positive de?nite matrices of determinant one,when one replaces the general linear group GL(n,R)with the special linear group SL(n,R),the n×n–matrices of determinant one,and the space of real symmetric matrices Sym(n,R)with the space Sym0(n,R)of real symmetric traceless matrices.

Remark1.We use here the terminology“distance”in contrast to the standard terminology “metric”in order to avoid confusion with the notion of“Riemannian metric”,which is going to play a r?o le soon.

The case n=2is already interesting;see Remark??below.

All the properties except property(iv),the triangle inequality,are more or less obvious from the de?nition(see above),but the triangle inequality is not.In fact,the theorem will be the consequence of a more general theorem as follows.

The most important geometric way distances arise is as associated distances to Riemannian metrics on manifolds;the Riemannian metric,as an in?nitesimal description of length is used to de?ne the length of paths by integration,and the distance between two points then arises as the greatest lower bound on the length of paths joining the two points.More precisely,if M is a di?erentiable manifold(in what follows,“di?erentiable”will always mean“in?nitely many times di?erentiable”,i.e.of class C∞),a Riemannian metric is the assignment to any p∈M of a Euclidean scalar product ?|? p in the tangent space T p M depending di?erentiably on p. Technically,it is a di?erentiable positive de?nite section of the second symmetric power S2T?M of the cotangent bundle,or a positive de?nite symmetric2-tensor.In classical terms,it is given in local coordinates(U,x)as the“square of the line element”or“?rst fundamental form”

d s2=g ij(x)d x i d x j(16) (Einstein summation convention:repeated lower and upper indices ar

e summed over).Here the g ij are di?erentiable functions(the metric coe?cients)subjected to the transformation rule

g ij(x)=g kl(y(x))?y k

?x i

?y l

?x j

A di?erentiable manifold together with a Riemannian metric is called a Riemannian manifold .Given a piecewise di?erentiable path c :[a,b ]?→M in M ,its length L [c ]is de?ned to be

L [c ]:=

b a ˙

c (t ) c (t )

d t ,

where for X ∈T p M we have X p := X |X p ,the Euclidean norm associated to the scalar product in T p M given by the Riemannian metric.In local coordinates

L [c ]=

b a g ij (

c (t ))˙c i (t )˙c j (t )

d t.

Given p ,q ∈M ,the distance d (p,q )associated to a given Riemannian metric then is de?ned to be

d (p,q ):=inf c

L [c ],(17)the in?mum running over all piecewise di?erentiable paths c joining p to q .This de?nes indeed a distance:

Proposition.The distance de?ned by (??)on a connected Riemannian manifold is a metric in the sense of metric spaces,i.e.de?nes a map d :M ×M ?→R satisfying

(i)d (p,q )≥0,d (p,q )=0??p =q

(ii)d (p,q )=d (q,p )

(iii)d (p,r )≤d (p,q )+d (q,r ).

An indication of proof will be given in the next section.

So the central issue here is the fact that a Riemannian metric is the di?erential substrate of a distance and,in turn,de?nes a distance by integration.This is the most important way of constructing distances,which is the fundamental discovery of Gau?and Riemann .In our case,this paradigm is realized in the following way.

The space Sym +(n,R )is a di?erentiable manifold of dimension n (n +1)/2,more speci?cally,it is an open cone in the vector space

Sym (n,R ):= A ∈M (n,R ) A =A T

of all real symmetric n ×n -matrices.Thus the tangent space T A Sym +(n,R )to Sym +(n,R )at a point A ∈Sym +(n,R )is just given as

T A Sym +(n,R )=Sym (n,R ).

The tangent space T A SSym +(n,R )to SSym (n,R )at a point A ∈SSym (n,R )is just given as

T A SSym (n,R )=Sym 0(n,R ):={X ∈Sym (n,R )|tr (X )=0},

the space of traceless symmetric matrices.

Now note that there is a natural action of GL (n,R )on S +,namely,as already referred to above,by congruence transformations:X ∈GL (n,R )acts via A →XAX T .If one regards A as the matrix corresponding to a bilinear form with respect to a given basis,this action represents a change of basis.This action is transitive,and the isotropy subgroup at E ∈S +is just the orthogonal group O (n ):

X ∈GL (n,R ) XEX T =E = X ∈GL (n,R ) XX T =E =O (n ),

and so S +can be identi?ed with the homogeneous space GL (n,R )/O (n )upon which GL (n,R )acts by left translations (its geometric signi?cance being that its points parametrize the possible scalar products in R n )?].

In general,a homogeneous space is a di?erentiable manifold with a transitive action of a Lie group G ,whence it has a representation as a quotient M =G/H with H a closed subgroup of G .The most natural Riemannian metrics in this case are then those for which the group G acts by isometries,or,in other words,which are invariant under the action of G ;e.g.the classical geometries –the Euclidean,the elliptic,and the hyperbolic geometry –arise in this manner.Looking out for these metrics,Theorem ??will be a consequence of the following theorem :Theorem 2.(i)The Riemannian metrics g on Sym +(n,R )invariant under congruence

transformations with matrices X ∈GL (n,R )are in one-to-one-correspondence with posi-tive de?nite quadratic forms Q on T E Sym +(n,R )=Sym (n,R )invariant under conjuga-tion with orthogonal matrices,the correspondence being given by

g A (X ,Y )=B (√A ?1X √A ?1,√A ?1Y √A ?1),

where A ∈Sym +(n,R ),X ,Y ∈Sym (n,R )=T A Sym +(n,R ),and B is the symmetric positive bilinear form corresponding to Q

(ii)The corresponding distance d Q is invariant under congruence transformations and inver-

sion,i.e.satis?es

d Q (A ,B )=d Q (XAX T ,XBX T )

and

d Q (A ,B )=d Q (A ?1,B ?1)

for all A ,B ∈Sym +(n,R ),X ∈GL (n,R ),and is given by d 2Q (A ,B )=14Q ln √A ?1B √A ?1 .(18)

(iii)In particular,the distance in Theorem ??is given by the invariant Riemannian metric

corresponding to the canonical non–degenerate bilinear form ?]

?X ,Y ∈M (n,R ):B (X ,Y ):=4tr (XY )

on M (n,R ),restricted to Sym (n,R )=T E Sym (n,R )+,i.e.to the quadratic form

?X ∈Sym (n,R ):Q (X ):=4tr X 2

on Sym (n,R ).As a Riemannian metric it is,in classical notation,

d s 2=4tr √X ?1d X √X ?1 2 =4tr X ?1d X 2 (19)

where X =(X ij )is the matrix of the natural coordinates on Sym +(n,R )and d X =(dX ij )is a matrix of di?erentials.

The same conclusions hold for the space SSym (n,R )of real symmetric positive de?nite matrices of determinant one,when one replaces the general linear group GL (n,R )with the special linear group SL (n,R ),the n ×n –matrices of determinant one,and the space of real symmetric matrices Sym (n,R )with the space Sym 0(n,R )of real symmetric traceless matrices.

A concrete map p :G ?→S +achieving this identi?cation is given by p (A ):=AA T ;it is surjective and satis?es p (XA )=X p (A )X T .The fact that p is an identi?cation map is then equivalent to the polar decomposi-tion :any regular matrix can be uniquely written as the product of a positive de?nite symmetric matrix and an orthogonal matrix.This generalizes the representation z =re i?of a nonzero complex number z .

?]the famous Cartan-Killing-form of Lie group theory

Remark2.Although the expression(??)appears to be explicit in the coordinates,it seems to be of no use for analyzing the properties of the corresponding Riemannian metric,since the operations of inverting,squaring,and taking the trace gives,in the general case,untractable expressions.In particular,it apparently is of no help in deriving the expression(??)for the associated distance by direct elementary means.

There is,however,one interesting case where it can be checked to give a very classical expression; this is the case n=2.In this case,one has

SSym+(2,R)=

x y

y z

x,y,z∈R,xz?y2=1,x>0

a hyperboloid in3-space.This is classically known as a candidate for a model of the hyperbolic plane.In fact,in this case,one may show by explicit computation that the metric(??)restricts to the classical hyperbolic metric,and that the corresponding distance just gives one of the classical formulas for the hyperbolic distance.For details,see[?].

Of course,the next question is which invariant metrics there are.Also this question can be answered:

Addendum.(i)The positive de?nite quadratic forms Q on Sym(n,R)invariant under con-jugation with orthogonal matrices are of the form

Q(X)=αtr

+β

tr(X)

,α>0,β>?

(ii)The positive de?nite quadratic forms Q on Sym0(n,R)invariant under conjugation with orthogonal matrices are unique up to a positive scalar and hence of the form

Q(X)=αtr

,α>0.

In particular,the Riemannian metric(??)corresponds to the caseα=1,β=0.Since from the point of this classi?cation all these metrics stand on an equal footing,it would be interesting to know by which naturality requirements this choice can be singled out.

6The proofs

To put this result into proper perspective and to cut a long story short,let us very brie?y summarize why Theorem??,and consequently Theorem??,are true.First,however,we indicate a proof of the Proposition above,since it is on this Proposition that our approach to the triangle equality for the distance de?ned by(??)is based.

The fact that d(p,q)≥0and the symmetry of d are immediate from the de?nitions.There remains to show d(p,q)=0=?p=q and the triangle inequality.

For given p∈M,choose a coordinate neighbourhood U～=R n around p such that p corresponds to0∈R n.We then have the expression(??)for the given metric in U.Moreover,we have in U the standard Euclidean metric

d s2E:=δij d x i d x j=

i=1

(dx i)2.

Let ? denote the norm belonging to the given Riemannian metric in U and|?|the norm given by the standard Euclidean metric.For r>0let

B(p;r):={x∈R n||x|≤r}

be the standard closed ball with radius r around p=0,and

S(p;r):={x∈R n||x|=r}

its boundary,the sphere of radius r around p∈R n.

As a continuous function U×R n?→R the norm ? takes its minimum m>0and its maximum M>0on the compact set B(p;1)×S(p;1).It follows that we have

?q∈B(p;1),X∈R n:m|X|≤ X q≤M|X|

by homogeneity of the norm,and so by integrating and taking the in?mum

?q∈B(p;1):md E(p,q)≤d(p,q)≤Md E(p,q)(20) where d E(p,q)=|q?p|is the Euclidean distance.If q∈B(p;1),then any path c joining p to q meets the boundary S(p;1)in some point r,from which follows L[c]≥L[c ]≥d(p,r)≥m –where c denotes the part of c joining p to r for the?rst time,say–whence d(p,q)≥m.In other words,if d(p,q)

For the triangle inequality,let c be a path joining p to q and d a path joining q to r.Let c?d be the composite path joining p to r.Then L[c?d]=L[c]+L[d].Taking the in?mum on the left hand side over all paths joining p to r gives d(p,r)≤L[c]+L[d].Taking on the right hand side?rst the in?mum over all paths joining p to q and subsequently over all the paths joining q to r then gives d(p,r)≤d(p,q)+d(q,r),which is the triangle inequality.

Remark3.In particular,(??)shows that the metric topology induced by the distance d on a connected Riemannian manifold coincides with the given manifold topology.

Now to the proof of Theorem??.Recall the terminology of[?],Chapter X:Let G be a Lie group with Lie algebra g,H?G a closed Lie subgroup corresponding to the Lie subalgebra h?g.Let M be the homogeneous space M=G/H.Then G operates as a symmetry group on M by left translations.M has the distinguished point o=eH=H corresponding to the coset of the unit element e∈G with tangent space T o M=g/h.This homogeneous space is called reductive if g splits as a direct vector space sum g=h⊕m for a linear subspace m?g such that m is invariant under the adjoint action Ad:H?→GL(g).Then canonically T o M=m. In our situation,G=GL(n,R),H=O(n).Then g=M(n,R),the full n×n-matrices,and h=Asym(n,R),the antisymmetric matrices.As is well known,

M(n,R)=Asym(n,R)⊕Sym(n,R),

since any matrix X splits into the sum of its antisymmetric and symmetric part via

X=X?X T

X+X T

The adjoint action of O∈O(n)on M(n,R)is given by X→OXO?1=OXO T and clearly preserves Sym(n,R).So Sym+(n,R)is a reductive homogeneous space.

We now have the following facts from the general theory:

a)On a reductive homogeneous space there is a distinguished connection invariant under the action of G,called the natural torsion free connection in[?].It is uniquely characterized by the following properties([?],Chapter X,Theorem2.10)

–It is G –invariant

–Its geodesics through o ∈M are the orbits of o under the one-parameter subgroups of G ,i.e.of the form t →exp(tX )·o for some X ∈g ,where exp :g ?→G is the exponential mapping of Lie group theory

–It is torsion free

In particular,with this connection M becomes an a?ne locally symmetric space ,i.e.the geodesic symmetries at a point of M given by in?ection in the geodesics locally preserve the connection (loc.cit Chapter XI,Theorem 1.1).If M is simply connected,M is even an a?ne symmetric space ,i.e.the geodesic symmetries extend to globally de?ned transformations of M preserving the connection (loc.cit.,Chapter XI,Theorem 1.2).By homogeneity,these are determined by the geodesic symmetry s at o .In our case M =Sym +(n,R ),M is even contractible,hence simply connected,and so with the natural torsion free connection an a?ne symmetric space.We have o =E ,the n ×n unit matrix.For G =GL (n,R ),the exponential mapping of Lie group theory is given by the “naive matrix exponential”e X = ∞k =0t k X k /k !.So the geodesics are t →exp (t X )E exp (t X )T =e 2t X ,where X ∈Sym (n,R ),and s is given by s (X )=X ?1.b)The Riemannian metrics g on M invariant under the action of G are in one-to-one-corres-pondence with positive de?nite quadratic forms Q on m invariant under the adjoint action of H (loc.cit,Chapter X ,Corollary 3.2),the correspondence being given by

?X ∈T o M =m :g o (X,X )=Q (X ).

This is intuitively obvious,since we can translate o to any point of M by operating on it with an element g ∈G .

c)All G-invariant Riemannian metrics on M (there may be none)have the natural torsion free connection as their Levi-Civit`a connection (loc.cit ,Chapter XI,Theorem 3.3).In particular,such a metric makes M into a Riemannian (locally)symmetric space,i.e.the geodesic sym-metries are isometries,and the exponential map of Riemannian Geometry at o ,Exp o :T o M =m ?→M is given by the exponential map of Lie group theory for G :

?X ∈m :Exp o (tX )=exp (tX )·o.

Collecting these results,we now can come to terms with formula (??).First we see that Part (i)of Theorem ??is a standard result in the theory of homogeneous spaces.Furthermore,S +,being a Riemannian symmetric space with the metric (??),is complete (loc.cit ,Chapter XI,Theorem 6.4),the exponential mapping Exp E of Riemannian geometry is related to the exponential mapping exp :S ?→S +,S =T E S +from Lie theory and the matrix exponential e X via Exp E (X )=exp(2X )=e 2X and is a di?eomorphism §].

Having reached this point,here is the showdown.Since,by general theory,the Riemannian exponential mapping is a radial isometry,we get for the square of the distance d Q :

d 2Q (A ,B )=d 2Q (E ,√A ?1B √A ?1)

since d Q is invariant under congruences by (??),

=Q (12

exp ?1(√A ?1B √A ?1))§]

The fact that the naive matrix exponential is a di?eomorphism,whence S +is complete,can be seen by elementary means in the case under consideration.The main point is that it coincides with the exponential mapping coming from Riemannian Geometry (up to scaling with a factor of 2).

since Exp E is a radial isometry,

Q(ln(

√

A?1B

√

A?1)),

and this is just equation(??).In particular,from this one directly reads o?that the distance is invariant under inversion,as claimed.Of course,the invariances in question are for the particular case corresponding to(??)read o?easily from the classical form(??)of the Riemannian metric. On the other hand,we see that the invariance under inversion comes from the structural facts that S+is a symmetric space,and that the geodesic symmetry at E,which on general grounds must be an isometry,is just given by matrix inversion(see a)above).

One should add that these arguments are general and pertain to the situation of a symmetric space of the non–compact type;for this,see[?].

The representation of the orthogonal group O(n)on the symmetric matrices by conjugation is not irreducible,but decomposes as

Sym(n,R)=Sym0(n,R)⊕d(n,R),(21)

where d(n,R)are the scalar diagonal matrices.It is easy to see that both summands are invariant under conjugation with orthogonal matrices,and it can be shown that both parts are irreducible representations of O(n).From this it is standard to derive the Addendum.In the geometric framework of symmetric spaces,this describes the decomposition of the holonomy representation and correspondingly the canonical de Rham–decomposition

Sym+(n,R)～=SSym(n,R)×R+

of the symmetric space Sym(n,R)into irreducible factors.This is a direct product of Riemannian manifolds,i.e.the metric on the product is just the product of the metrics on the individual factors,that is given by the Pythagorean description.Thus it su?ces to classify the invariant metrics on the individual factors,which accounts for the Addendum.

Thus,it transpires that the theorems above follow from the basics of Lie group theory and Di?erential Geometry and so should be clear to the experts.The main results upon which it is based appeared originally in the literature in[?].All in all,it follows in a quite straightforward manner from the albeit rather elaborated machinery of modern Di?erential Geometry and the theory of symmetric spaces.In conclusion,it might therefore be still interesting to give a more elementary derivation of the result,as was done above in the case n=2.

As a general reference for Di?erential Geometry and the theory of symmetric spaces I recommend [?],[?](which,however,make quite a terse reading).A detailed exposition[?]covering all the necessary prerequisites is under construction;the purpose of this paper is to introduce the non–experts to all the basic notions of Di?erential Geometry and to expand the brief arguments just sketched.

References

[1]Baarda,W.(1973):S-Transformations and Criterion Matrices,Band5der Reihe1.

Netherlands Geodetic Commission,1973.

[2]Ballein,K.,1985,Untersuchung der Dreiecksungleichung beim Vergleich von Kovarianz-

matrizen,pers¨o nliche Mitteilung,1985

[3]F¨o rstner,W.,A Metric for Comparing Symmetric Positive De?nite Matrices,Note,

August1995

[4]Fukunaga,K.,Introduction to Statistical Pattern Recognition,Academic Press,1972

[5]Grafarend,E.W.(1972):Genauigkeitsmasse geod¨a tischer Netze.DGK A73,Bayerische

Akademie der Wissenschaften,M¨u nchen,1972.

[6]Grafarend,E.W.(1974):Optimization of geodetic networks.Bolletino di geodesia e

Scienze A?ni,33:351–406,1974.

[7]Grafarend,E.W.and Niermann,A.(1994):Beste echte Zylinderabbildungen,Kar-

tographische Nachrichten,3:103–107,1984.

[8]Kavrajski,V.V.,Ausgew¨a hlte Werke,Bd.I:Allgemeine Theorie der kartographischen

Abbildungen,Bd.2.Kegel-und Zylinderabbildungen(russ.),GVSMP,Moskau,1958,zitiert in[?]

[9]Kobayashi,S.&Nomizu,K.,Foundations of Di?erential Geometry,Vol.I,Interscience

Publishers1963

[10]Kobayashi,S.&Nomizu,K.,Foundations of Di?erential Geometry,Vol.II,Interscience

Publishers1969

[11]Moonen,B,Notes on Di?erential Geometry,

ftp://ftp.uni-bonn.cd/pub/staff/moonen/symmc.ps.gz

[12]Nomizu,K.,Invariant a?ne connections on homogeneous spaces,Amer.J.Math,76

(1954),33-65

[13]Schmitt,G.(1983):Optimization of Control Networks–State of the Art.In:Borre,K.;

Welsch,W.M.(Eds.),Proc.Survey Control Networks,pages373–380,Aalborg University Centre,1983.

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路由协议的优先级,以及管理距离AD和metric的区别

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英语选修六课文翻译第五单元word版本

英语选修六课文翻译第五单元

英语选修六课文翻译第五单元 reading An exciting job I have the greatest job in the world. travel to unusual places and work alongside people from all over the world sometimes working outdoors sometimes in an office sometimes using scientific equipment and sometimes meeting local people and tourists I am never bored although my job is occasionally dangerous I don't mind because danger excites me and makes me feel alive However the most important thing about my job is that I heIp protect ordinary people from one of the most powerful forces on earth-the volcano. I was appointed as a volcanologist working for the Hawaiian Volcano Observatory (HVO) twenty years ago My job is collecting information for a database about Mount KiLauea which is one of the most active volcanoes in Hawaii Having collected and evaluated the information I help oyher scientists to predict where lava from the path of the lava can be warned to leave their houses Unfortunately we cannot move their homes out of the way and many houses have been covered with lava or burned to the ground. When boiling rock erupts from a volcano and crashes back to earth, it causes less damage than you might imagine. This is because no one lives near the top of Mount Kilauea, where the rocks fall. The lava that flows slowly like a wave down the mountain causes far more damage because it buries everything in its path under the molten rock. However, the eruption itself is really exciting to watch and I shall never forget my first sight of one. It was in the second week after I arrived in Hawaii. Having worked hard all day, I went to bed early. I was fast asleep when suddenly my bed began shaking and I heard a strange sound, like a railway train passing my window. Having experienced quite a few earthquakes in Hawaii already, I didn't take much notice. I was about to go back to sleep when suddenly my bedroom became as bright as day. I ran out of the house into the back garden where I could see Mount Kilauea in the distance. There had been an eruption from the side of the mountain and red hot lava was fountaining hundreds of metres into the air. It was an absolutely fantastic sight. The day after this eruption I was lucky enough to have a much closer look at it. Two other scientists and I were driven up the mountain and dropped as close as possible to the crater that had been formed duing the eruption. Having earlier collected special clothes from the observatory, we put them on before we went any closer. All three of us looked like spacemen. We had white protective suits that covered our whole body, helmets,big boots and special gloves. It was not easy to walk in these suits, but we slowly made our way to the edge of the crater and looked down into the red, boiling centre. The other two climbed down into the crater to collect some lava for later study, but this being my first experience, I stayed at the top and watched them.

实验7 修改RIP的Metric值

实验七修改RIP的Metric值试验目的： 1、掌握修改RIP的Metric值的方法试验设备： RG-RSR20 二台、网线若干试验拓扑图：实验步骤及要求： 1、配置各台路由器用户名和IP地址，并且使用ping命令确认各路由器的直连口的互通性。 2、按照拓扑配置RIPv2路由协议，关闭自动汇总。 3、查看R2的路由表： R2#show ip route R 1.1.1.0/24 [120/1] via 12.1.1.1, 00:00:28, FastEthernet 0/1 C 2.2.2.0/24 is directly connected, Loopback 0 C 2.2.2.2/32 is local host. C 12.1.1.0/24 is directly connected, FastEthernet 0/1 C 12.1.1.2/32 is local host. 4、修改 RIP的 Metric 值 R1(config)#access-list 1 permit 1.1.1.0 0.0.0.255

R1(config)#router rip R1(config-router)#offset-list 1 out 10 fastEthernet 0/0 R1(config-router)# 5、查看R2的路由表 R2#show ip route R 1.1.1.0/24 [120/11] via 12.1.1.1, 00:00:16, FastEthernet 0/1 C 2.2.2.0/24 is directly connected, Loopback 0 C 2.2.2.2/32 is local host. C 12.1.1.0/24 is directly connected, FastEthernet 0/1 C 12.1.1.2/32 is local host. 6、试验完成

管理距离(AD)度量值(metric)

管理距离（AD）和度量值（Metric） 2008-10-11 21:00:39| 分类：Ccna|举报|字号订阅 R1#show ip route ...省略 R 10.2.0.0[120/1] via 10.1.1.2,00:00:21,Serial0/0 C 10.3.0.0 is directly connected,Serial0/1 ############################################################## ####### 在输出中，首先显示路由条目各种类型的简写，如“C”为直连网络，“S”为静态路由。以上面粗体的路由为例： “R”-------------------------表示这条路由是“RIP”协议学习得到的； “10.2.0.0”-----------------是目的网络； “[120/1]”-------------------是管理距离（Administrative Distance,AD）/ 度量值（Metric）； “via 10.1.1.2”-------------是指到达目的网络的下一跳路由器IP地址； “00:00:21”-----------------是指路由器最近一次得知路由到现在的时间； “Serial 0/0”----------------是指到达下一跳应从哪个端口出去。技术要点：管理距离（AD）：用来表示路由器可能从多种途径获得同一路由，例如，一个路由器要获得“10.2.0.0/24”网络的路由，可以来自RIP，也可以是静态路由。不同途径获得的路由可能采取不同的路径到达目的网络，为了区分不同路由协议的可信度，用管理距离加以表示。

人教版英语选修六Unit5 the power of nature(An exciting Job)

高二英语教学设计 Book6 Unit 5 Reading An Exciting Job 1.教学目标（Teaching Goals）: a. To know how to read some words and phrases. b. To grasp and remember the detailed information of the reading material . c. To understand the general idea of the passage. d. To develop some basic reading skills. 2.教学重难点: a.. To understand the general idea of the passage. b. To develop some basic reading skills. Step I Lead-in and Pre-reading Let’s share a movie T: What’s happened in the movie? S: A volcano was erupting. All of them felt frightened/surprised/astonished/scared…… T: What do you think of volcano eruption and what can we do about it? S: A volcano eruption can do great damage to human beings. It seems that we human beings are powerless in front of these natural forces. But it can be predicted and damage can be reduced. T: Who will do this kind of job and what do you think of the job? S: volcanologist. It’s dangerous. T: I think it’s exciting. Ok, this class, let’s learn An Exciting Job. At first, I want to show you the goals of this class Step ⅡPre-reading Let the students take out their papers and check them in groups, and then write their answers on the blackboard (Self-learning) some words and phrases：volcano, erupt, alongside, appoint, equipment, volcanologist, database, evaluate, excite, fantastic, fountain, absolutely, unfortunately, potential, be compared with..., protect...from..., be appointed as, burn to the ground, be about to do sth., make one’s way. Check their answers and then let them lead the reading. Step III Fast-reading 这是一篇记叙文，一位火山学家的自述。作者首先介绍了他的工作性质，说明他热爱该项工作的主要原因是能帮助人们免遭火山袭击。然后，作者介绍了和另外二位科学家一道来到火山口的经历。最后，作者表达了他对自己工作的热情。许多年后，火山对他的吸引力依然不减。 Skimming Ⅰ.Read the passage and answer: (Group4) 1. Does the writer like his job?( Yes.) 2. Where is Mount Kilauea? (It is in Hawaii) 3. What is the volcanologist wearing when getting close to the crater? (He is wearing white protective suits that covered his whole body, helmets, big boots and

Unit5 Reading An Exciting Job(说课稿)

Unit5 Reading An Exciting Job 说课稿 Liu Baowei Part 1 My understanding of this lesson The analysis of the teaching material:This lesson is a reading passage. It plays a very important part in the English teaching of this unit. It tells us the writer’s exciting job as a volcanologist. From studying the passage, students can know the basic knowledge of volcano, and enjoy the occupation as a volcanologist. So here are my teaching goals: volcanologist 1. Ability goal: Enable the students to learn about the powerful natural force-volcano and the work as a volcanologist. 2. Learning ability goal: Help the students learn how to analyze the way the writer describes his exciting job. 3. Emotional goal: Make the Students love the nature and love their jobs. Learn how to express fear and anxiety Teaching important points: sentence structures 1. I was about to go back to sleep when suddenly my bedroom became as bright as day. 2. Having studied volcanoes now for more than twenty years, I am still amazed at their beauty as well as their potential to cause great damage. Teaching difficult points: 1. Use your own words to retell the text. 2. Discuss the natural disasters and their love to future jobs. Something about the S tudents: 1. The Students have known something about volcano but they don’t know the detailed information. 2. They are lack of vocabulary. 3. They don’t often use English to express themselves and communicate with others.

an exciting job 翻译

我的工作是世界上最伟大的工作。我跑的地方是稀罕奇特的地方，我见到的是世界各地有趣味的人们，有时在室外工作，有时在办公室里，有时工作中要用科学仪器，有时要会见当地百姓和旅游人士。但是我从不感到厌烦。虽然我的工作偶尔也有危险，但是我并不在乎，因为危险能激励我，使我感到有活力。然而，最重要的是，通过我的工作能保护人们免遭世界最大的自然威力之一，也就是火山的威胁。我是一名火山学家，在夏威夷火山观测站（HVO）工作。我的主要任务是收集有关基拉韦厄火山的信息，这是夏威夷最活跃的火山之一。收集和评估了这些信息之后，我就帮助其他科学家一起预测下次火山熔岩将往何处流，流速是多少。我们的工作拯救了许多人的生命，因为熔岩要流经之地，老百姓都可以得到离开家园的通知。遗憾的是，我们不可能把他们的家搬离岩浆流过的地方，因此，许多房屋被熔岩淹没，或者焚烧殆尽。当滚烫沸腾的岩石从火山喷发出来并撞回地面时，它所造成的损失比想象的要小些，这是因为在岩石下落的基拉韦厄火山顶附近无人居住。而顺着山坡下流的火山熔岩造成的损失却大得多，这是因为火山岩浆所流经的地方，一切东西都被掩埋在熔岩下面了。然而火山喷发本身的确是很壮观的，我永远也忘不了我第一次看见火山喷发时的情景。那是在我到达夏威夷后的第二个星期。那天辛辛苦苦地干了一整天，我很早就上床睡觉。我在熟睡中突然感到床铺在摇晃，接着我听到一阵奇怪的声音，就好像一列火车从我的窗外行驶一样。因为我在夏威夷曾经经历过多次地震，所以对这种声音我并不在意。我刚要再睡，突然我的卧室亮如白昼。我赶紧跑出房间，来到后花园，在那儿我能远远地看见基拉韦厄火山。在山坡上，火山爆发了，红色发烫的岩浆像喷泉一样，朝天上喷射达几百米高。真是绝妙的奇景！就在这次火山喷发的第二天，我有幸做了一次近距离的观察。我和另外两位科学被送到山顶，在离火山爆发期间形成的火山口最靠近的地方才下车。早先从观测站出发时，就带了一些特制的安全服，于是我们穿上安全服再走近火山口。我们三个人看上去就像宇航员一样，我们都穿着白色的防护服遮住全身，戴上了头盔和特别的手套，还穿了一双大靴子。穿着这些衣服走起路来实在不容易，但我们还是缓缓往火山口的边缘走去，并且向下看到了红红的沸腾的中心。另外，两人攀下火山口，去收集供日后研究用的岩浆，我是第一次经历这样的事，所以留在山顶上观察他们

A metric for covariance matrices

英语选修六课文翻译Unit5 The power of nature An exciting job的课文原文和翻译

rip协议metric是什么意思

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英语选修六课文翻译第五单元word版本

实验7 修改RIP的Metric值

管理距离(AD)度量值(metric)

人教版英语选修六Unit5 the power of nature(An exciting Job)

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an exciting job 翻译

英语选修六课文翻译第五单元

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