Appel W. Mathematics for Physics and Physicists

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Differential of map w ith values in R

587

B.1.c Tensor n otation

In relativity theory, physicists usually use a single letter to denote a vector, and

use exponents over this letter to indicate the coordinates. Using this tensor

notation (see Chapter 16), we can express the relations above as follows:

f (x + h) = f (x) + h

∂

f + o(h) and d f = ∂

f dx

Writ ing f

,µ

def

= ∂

f , we also have

f (x + h) = f (x) + h

,µ

and d f = f

,µ

B.2

Differential of map with values in RR

It is very easy to generalize the previous results for vector-valued functions

of a vector var iable. Let thus f : R

→ R

be a map with vector values.

All previous notions and results apply to each coordinate of f . We therefore

denote

f ( a) =







( a)







for all a ∈ R

Then, if each coordinate of f has continuous partial derivatives at a ∈ R

we can write down (b.1) for each component:

( a + h) = f

( a) +

j=1

∂ f

∂ x

( a) · h

+ o( h), i = 1, . . . , p.

These relations can be written in matrix notation as follows:







( a + h)













( a)













∂ f

∂ x

( a) ···

∂ f

∂ x

( a)

∂ f

∂ x

( a) ···

∂ f

∂ x

( a)



















+ o( h).

(b.2)

Hence the following deﬁnition:

DEFINITION B.2 (Differential) Let f : R

→ R

be a map s uch that each

coordinate has continuous partial derivatives at a ∈ R

. The differential

588 Elementary reminders of differential calculus

of f at a is the linear map d f

: R

→ R

such that

d f



















∂ f

∂ x

( a) ···

∂ f

∂ x

( a)

∂ f

∂ x

( a) ···

∂ f

∂ x

( a)



















The differential of f , denoted d f , is the map

d f : R

−→ L (R

, R

), a 7−→ d f

Thus, the matrix representing the differential d f

at a is the matrix

partial derivatives of f . The relation (b.2) may also be written in the form

f ( a + h) = f ( a) + d f

.h + o

h→0

(h).

Using tensor notation, this becomes

(x + h) = f

(x) + h

∂

+ o

h→0

(h) and d f

= ∂

B.3

Lagrange multipliers

Using linear forms, it is easy to establish the validity of the method of

Lagrange multipliers.

We start with an elementary lemma of linear algebra.

LEMMA B.3 Let ψ and ϕ

, . . . , ϕ

be linear forms deﬁned on a vector space E of

dimension n. If the forms (ϕ

, . . . , ϕ

) are linearly independent and if

i=1

Ker ϕ

⊂ Ker ψ,

then ψ is a linear combination of the ϕ

 Exercise B.1 Prove this lemma (see page 591 ).

Now consider the following problem. We wish to ﬁnd the extrema of a

real-valued function f (x

, . . . , x

) on E = R

. If is differentiable, we know

that a necessary condition (which is not sufﬁcient) for f to be extremal at a =

, . . . , a

) is that

If p = n, it is calle d the Jacobian matrix of f .

Lagrange multipliers 589

d f

= 0, that is,

∂ f

∂ x

( a) = 0 ∀i ∈ [[1, n]].

Now let us make the problem more complicated: we are looking for ex-

trema of f on a (fairly regular) subs et of R

, such as a curve in R

, or

a surface, or some higher-dimensional analogue (what mathematicians call a

differentiable s ubvariety of E. Assume that this set is of dimension n−k and

is deﬁned by k equations (represented by differentiable functions), namely,

S :







(1)

, . . . , x

) = 0

(k)

, . . . , x

) = 0.

(b.3)

In order that the set S deﬁned by those equations be indeed “an (n − k)-

dimensional subvariety, ” it is necessary that the differentials of the equations

(1)

, . . . , C

(k)

be linearly independent linear forms at every point in S :



(1)

, . . . , dC

(k)



are linearly independent for all x ∈ S .

We are now interested in ﬁnding the points on a subvariety S where f

has an extremum.

THEOREM B.4 (Lagrange multipliers) Let f be a real-valued differentiable func-

tion on E = R

, and let S be a differential subvariety of E deﬁned by equa-

tions (b.3). Then, in order for a ∈ S to be an extremum for f restricted to S , it is

necessary that there exist real constants λ

, . . . , λ

such that

d f

= λ

·dC

(1)

+ ···+ λ

·dC

(k)

Proof. Assume that a ∈ S is an extremum of f restricted to S , and let U be the

tangent space to S at a. This tangent space is an (n −k)-dimensional vector space,

where a ∈ U is the origin.

If a is an e xtremum of f on S , we nece ssarily have d f

. h = 0 for any vector h

tangent to S (if we move inﬁnitesimally on S from a, the values of f cannot increase

or decrease in any direction). Thus we have U ⊂ Ker d f

On the other hand, U is deﬁned by the intersection of the tangent planes to t he

subvarietie s with equations C

(i)

( x) = 0 for 1 ¶ i ¶ k (because S is the intersection

of those). Hence we have U =

i=1

Ker dC

(i)

, and therefore

i=1

Ker dC

(i)

⊂ Ker d f

There only remains to apply Lemma B .3.

In order to ﬁnd the points a ∈ S where f may have an extremum,

one solves the system of linear equations d f

· dC

(i)

, where λ

are

unknowns, together wit h the equations expressing that a ∈ A . In other

590 Elementary reminders of differential calculus

words, the s ystem to solve is











∂ f

∂ x

( a) −

i=1

∂ C

(i)

∂ x

( a) = 0,

∂ f

∂ x

( a) −

i=1

∂ C

(i)

∂ x

( a) = 0

and











(1)

( a) = 0,

(k)

( a) = 0,

(b.4)

where there are n + k equations, with n + k unknowns, the coordinates of

a = (a

, . . . , a

) and the auxiliary unknowns λ

, . . . , λ

, called the Lagrange

multipliers. Those auxiliary unknowns have t he advantage of preserving sym-

metry among all variab les . Moreover, they sometimes have physical signif-

icance (for instance, temperature in thermodynamics is the inverse of a La-

grange multiplier).

Physicists usually proceed as follows: introduce k multipliers λ

, . . . , λ

and construct an auxiliary function

F ( x)

def

= f ( x) −

i=1

(i)

( x).

Look then for points a ∈ R

giving free ex trema of F , that is, where the

n varia bles x

, . . . , x

are independent (do not necessarily lie on S ). The

condition that the differential of F vanishes gives the ﬁrst set of equations

in (b.4), and of course the solutions depend on the additional parameters

. The values for those are found at the end, using the constraints on the

problem, namely, the second set of equations in (b.4).

Example B.5 Let S be a surface in R

with equation C (x, y, z) = 0. Let r

∈ R

be a po int

not in S , and consider the problem of ﬁnding the points of S closest to r

, that is, those

minimizing the function f ( x) = kx − r

, with the constraint C ( x) = 0. The auxiliar y

function is F ( x) = kx − r

−λC ( x), with differential (at a point a ∈ R

) given by

= 2 ( a − r

|·) −λ



grad C ( a)





in other words,

. h = 2( a − r

) · h −λ grad C ( a) · h for all h ∈ R

The free extrema of F are the points a such that ( a − r

) is colinear with the gradient of C

at a. Using the condition C ( a) = 0, the values of λ are found. Since grad C ( a) is a vector

perpendicular to the surface S , the geometric interpretation of the necessary condition is t hat

points of S closest to r

are t h o se for which ( a − r

) is perpe ndicular to the surface at the

point a.

Of course, this is a necessary condition, but not necessari ly a sufﬁcient condition (the

distance may be maximal at the points found in this manner, or there may be a saddle point).

Remark B.6 Introducting the constraint multiplied by an additional parameter in the auxiliary

function in order to satisfy this constraint is a trick used in electromagnetism in order to ﬁx

the gauge. In the Lagrangian formulation of electromagnetism, the goal is indeed to minimize

Solution of exercise 591

the action gi ven by the time integral of t he Lagrangian under the constraint that the p otential

four-vector has a ﬁxed gauge. A gauge-ﬁxing term, for instance λ(∂ · A)

in Lorenz gauge, is

added to the Lagrangian density (see [21, 49]).

SOLUTION

 Solution of exercise B .1 on page 588. By linear algebra, there exist additional linear forms

k+1

, . . . ,ϕ

such that (ϕ

, . . . , ϕ

) is a basis of E

∗

, since the given forms are linearly inde-

pendent. Let (e

, . . . , e

) be the dual basis. By the deﬁnition of a basis, there are constants

, 1 ¶ i ¶ n, such that ψ =

i=1

. Indeed, we have λ

= ψ(e

) using th e for-

mula ϕ

ℓ

) = δ

mℓ

deﬁning the dual basis. Let j = i + 1, . . . , n. The same formula

gives e

∈

1¶i¶k

Ker ϕ

⊂ Ker ψ by assumption, hence λ

= 0 for j > i. This implies

f ∈ Vect(ϕ

, . . . , ϕ

), as desired.

Appendix

Matrices

In this shor t appendix, we recall the formalism of duality (duality bracket or pairing)

and show how it can be used easily to recover the formulas of change of basis, and to

better understand their origin; this can be considered as a warm-up for the ch apter

on the formalism of tensors, which generalizes these results.

C.1

Duality

Let E be a ﬁnite-dimensional vector space over the ﬁeld K = R or C, of

dimension n.

DEFINITION C.1 A linear form on E is a linear map from E to K. The dual

space of E is the space E

∗

= L (E, K) of linear forms on E. For ϕ ∈ E

∗

linear form and x ∈ E a vector, we denote

〈ϕ, x〉

def

= ϕ(x).

The map 〈·, ·〉 : E × E

∗

→ K is called the dual ity bracket or duality pairing.

DEFINITION C.2 (Dual basis) Let E = (e

, . . . , e

) be a basis of E. There is a

unique family E

∗

= (e

∗

, . . . , e

∗

) of linear forms on E such that

∀i, j ∈ [[1, n]]



∗

, e



= δ

i j

The family E

∗

is a basis of E

∗

, and is called the dual basis of E .

For x =

i=1

, we have x



∗

, x



for all i ∈ [[1, n]]. The elements

of the dual ba sis E

∗

are also called coordinate forms.

594 Matrices

C.2

Application to matrix representation

We will now explain h ow the formalism of linear forms ca n be used to clar-

ify the use of matrices to represent vectors, and how to use them to remember

easily the formulas of change of basis.

C.2.a Matrix representing a family of vectors

DEFINITION C.3 Let E = (e

, . . . , e

) be a basis of E and let (x

, . . . , x

) be q

vectors in E. Th e matr ix representing the family ( x

)

1¶ j¶ q

in the basis E

is the (n ×q)-matrix M = (a

i j

)

1¶i¶n

1¶j¶q

such that

for all j ∈ [[1, q]], the column coefﬁcients a

1 j

, . .. , a

n j

are

the coefﬁcients of x

expressed in the basis E.

In other words, the vector x

is expressed in the ba sis E by the linear combi-

nation

i=1

i j

M =







··· a

1 j

··· a

2 j

··· a

n j

··· a







↑

vector x

“seen in E”

Using th e dual basis, we can state the formula

i j

= e

∗

) = 〈e

∗

, x

〉

for all i ∈ [[1, n]] and j ∈ [[1, q]]. We can express the deﬁnition M =



〈e

∗

, x

〉



i j

symbolically in the form

M = mat

, . . . , x

) = 〈e

∗

|x〉 .

C.2.b Matrix of a linear map

As before, let E be a K vector space with a basis B = (b

, . . . , b

); in addition,

let F be another K-vector space with basis C = (c

, . . . , c

). We denote C

∗

, . . . , c

∗

) the basis dual to B.

Application to matrix representation 595

Let f ∈ L (E, F ) be a linear map from E to F , and let A = mat

B,C

( f )

be the matrix representing the map f in the basis B and C. Recall that, by

deﬁnition, th is means t hat the ﬁrst column of A contains the coefﬁcients

expressing the image f (b

) of the ﬁrst basis vector of B in the basis C, and so

on: the j-th column contains the coefﬁcients of f (b

) in the basis C.

It follows that for i ∈ [[1, p]] and j ∈ [[1, n]], the coefﬁcient A

i j

is the i-th

coordinate of f (e

) in the basis C, that is, we have

i j

= c

∗



f ( b

)





∗

, f ( b

)



This is denoted also

A =



∗



f ·b



. (∗)

C.2.c Change of basis

Now let B

′

= (b

′

, . . . , b

′

) be another basis of the space E. The change of

basis matrix from B to B

′

, denoted P = Pass(B, B

′

), is deﬁned as follows:

the columns of P represent the vectors of B

′

expressed in the basis B. In

other words, we have

i j

= i-th coordinate of b

′

= b

∗

′

for all (i, j) ∈ [[1, n]].

We denote P = Pass(B, B

′

) =



∗



′



Notice that the inverse of the matrix P , which is simply the change of basis

matrix from B

′

to B, is



∗



′



−1



′∗





C.2.d Change of basis formula

Let again B = (b

, . . . , b

) and B

′

= (b

′

, . . . , b

′

) be two bases of E, and let

x ∈ E be a vector. T he cha nge of basis formula

mat

(x) = Pass(B, B

′

) ·mat

′

(x)

expresses the relation between the coordinates for x in each basis. In abbrevi-

ated form, we write

X = P X

′

or, with the previous notation,

〈b

∗

|x〉 =



∗



′





′∗





596 Matrices

To remember this formula, it sufﬁces to use the relation

Id =

k=1



′





′∗



or its short form

Id =



′





′∗



(c.1)

to recover t he formula.

Let now f : E → F be a linear map. In addition to the basis B and B

′

E, let C and C

′

be basis of F , and let

M = mat

B,C

( f ) and M

′

= mat

′

( f ).

If we denote by P and Q respectively the change of b asis matrices for E and

F , i.e.,

P = Pass(B, B

′

) =



∗



′



and Q = Pass(C, C

′

) =



∗



′



the change of basis formula relating M and M

′

can be recovered by w riting

′



′∗



f ·b

′





′∗





| {z }

−1



∗



f ·b



| {z }



∗



′



| {z }

= Q

−1

M P (c.2)

One should remember that (c.1) provides a way to recover without mistake the

change of basis formula for matrices, in the form



′∗



f ·b

′





′∗





∗



f ·b



∗



′



 Exercise C.1 The trace of an endomorphism f ∈ L (E) is deﬁned by the formula

tr ( f ) =

i=1



∗

, f (e

)



in a basis E = (e

, . . . , e

). Show that this deﬁnition is independent of the chosen basis.

C.2.e Case of an orthonormal basis

When E has a euclidean or Hilbert space structure and the basis E =

, . . . , e

) is an orthonormal basis, the coordinate forms e

∗

are associated

with the orthogonal projections on the lines spanned by the vectors e

: we

have

∀x ∈ E ∀i ∈ [[1, n]] e

∗

(x) = (e

|x) .

The formula of the previous section, provided all bases involved are orthonor-

mal, becomes



′



f ·b

′





′







f ·b





′



or, with all indices spelled out precisely



′



f (b

′

)



k=1

ℓ=1



′







f ( b

ℓ

)



ℓ



′

