Lecture Notes on Vector and Tensor Algebra and Analysis, Study notes of Algebra

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Lecture Notes on Vector and Tensor

Algebra and Analysis

Ilya L. Shapiro

Departamento de F´ısica – Instituto Ciˆencias Exatas Universidade Federal de Juiz de Fora, Juiz de Fora, CEP 36036-330, MG, Brazil

Preface

These lecture notes are the result of teaching a half-semester course of tensors for undergraduates in the Department of Physics at the Federal University of Juiz de Fora. The same lectures were also given at the summer school in the Institute of Mathematics in the University of Brasilia, where I was kindly invited by Dra. Maria Em´ılia Guimar˜aes and Dr. Guy Grebot. Furthermore, I have used the first version of these notes to teach students of “scientific initiation” in Juiz de Fora. Usually, in the case of independent study, good students learn the material of the lectures in one semester. Since the lectures have some original didactic elements, we decided to publish them. These lectures are designed for the second-third year undergraduate student and are supposed to help in learning such disciplines of the course of Physics as Classical Mechanics, Electrodynamics, Special and General Relativity. One of my purposes was, e.g., to make derivation of grad, div and rot in the curvilinear coordinates understandable for the student, and this seems to be useful for some of the students of Physics, Mathematics or Engineering. Of course, those students which are going to make career in Mathematics or Theoretical Physics, may and should continue their education using serious books on Differential Geometry like [1]. These notes are nothing but a simple introduction for beginners. As examples of similar books we can indicate [2, 3] and [4], but our treatment of many issues is much more simple. A more sophisticated and modern, but still relatively simple introduction to tensors may be found in [5]. Some books on General Relativity have excellent introduction to tensors, let us just mention famous example [6] and [7]. Some problems included into these notes were taken from the textbooks and collection of problems [8, 9, 10] cited in the Bibliography. It might happen that some problems belong to the books which were not cited there, author wants apologize for this occurrence. In the preparation of these notes I have used, as a starting point, the short course of tensors given in 1977 at Tomsk State University (Russia) by Dr. Veniamin Alexeevich Kuchin, who died soon after that. In part, these notes may be viewed as a natural extension of what he taught us at that time. The preparation of the manuscript would be impossible without an important organizational work of Dr. Flavio Takakura and his generous help in preparing the Figures. I am especially grateful to the following students of our Department: to Raphael Furtado Coelho for typing the first draft and to Flavia Sobreira and Leandro de Castro Guarnieri, who saved these notes from many typing mistakes. The present version of the notes is published due to the kind interest of Prof. Jos´e Abdalla Helay¨el-Neto. We hope that this publication will be useful for some students. On the other hand, I would be very grateful for any observations and recommendations. The correspondence may be send to the electronic address shapiro@f isica.uf jf.br or by mail to the following address:

Ilya L. Shapiro Departamento de F´ısica, ICE, Universidade Federal de Juiz de Fora CEP: 36036-330, Juiz de Fora, MG, Brazil

Preliminary observations and notations

i) It is supposed that the student is familiar with the backgrounds of Calculus, Analytic Geom- etry and Linear Algebra. Sometimes the corresponding information will be repeated in the text of the notes. We do not try to substitute the corresponding courses here, but only supplement them.

ii) In this notes we consider, by default, that the space has dimension 3. However, in some cases we shall refer to an arbitrary dimension of space D, and sometimes consider D = 2, because this is the simplest non-trivial case. The indication of dimension is performed in the form like 3D, that means D = 3.

iii) Some objects with indices will be used below. Latin indices run the values

(a, b, c, ..., i, j, k, l, m, n, ...) = (1, 2 , 3)

in 3D and (a, b, c, ..., i, j, k, l, m, n, ...) = (1, 2 , ..., D)

for an arbitrary D.

Usually, the indices (a, b, c, ...) correspond to the orthonormal basis and to the Cartesian coordinates. The indices (i, j, k, ...) correspond to the an arbitrary (generally non-degenerate) basis and to arbitrary, possibly curvilinear coordinates.

iv) Following standard practice, we denote the set of the elements fi as {fi}. The properties of the elements are indicated after the vertical line. For example,

E = { e | e = 2n, n ∈ N }

means the set of even natural numbers. The comment may follow after the comma. For example,

{ e | e = 2n, n ∈ N , n ≤ 3 } = { 2 , 4 , 6 }.

v) The repeated upper and lower indices imply summation (Einstein convention). For example,

aibi =

∑^ D

i=

aibi = a^1 b 1 + a^2 b 2 + ... + aDbD

for the D-dimensional case. It is important that the summation (umbral) index i here can be renamed in an arbitrary way, e.g. Cii = Cjj = Ckk = ....

This is completely similar to the change of the notation for the variable of integration in a definite integral: (^) ∫ b a

f (x)dx =

∫ (^) b

a

f (y)dy.

where, also, the name of the variable does not matter.

vi) We use the notations like a′i^ for the components of the vectors corresponding to the coor- dinates x′i. In general, the same objects may be also denoted as ai′ and xi′. In the text we do not distinguish these two notations, and one has to assume, e.g., that a′i^ = ai′. The same holds for any tensor indices, of course.

vii) We use abbreviation: Def. ≡ Definition

viii) The exercises are dispersed in the text, and the reader is advised to solve them in order of appearance. Some exercises are in fact simple theorems which will be consequently used, many of them are very important statements. Most of the exercises are very simple, and they suppose to consume small time. But, there are a few exercises marked by *. Those are presumably more difficult, therefore they are recommended only for the students which are going to specialize in Mathematics or Theoretical Physics.

Exercise 4. Prove that the expansion (1.1) is unique. That is, for any given basis and for any vector a, its contravariant components ai^ are defined in a unique way.

Observation: It is important that the contravariant components of the vector ai^ do change when we turn from one basis to another. Let us consider, along with the original basis ei , another basis e′ i. Since each vector of the new basis belongs to the same space, it can be expanded using the original basis as

e′ i = ∧ji′ ej. (1.2)

Then, the uniqueness of the expansion of the vector a leads to the following relation:

a = aiei = aj′ ej′ = aj′ ∧ij′ ei ,

and hence

ai^ = aj′ ∧ij′. (1.3)

This is a very important relation, because it shows us how the contravariant components of the vector transform from one basis to another.

Similarly, we can make the transformation inverse to (1.2). It is easy to see that the relation between the contravariant components can be also written as

ak ′ = (∧−^1 )k ′ l al^ ,

where the matrix (∧−^1 ) is inverse to ∧:

(∧−^1 )k l ′· ∧li′ = δk i′′ and ∧li′ ·(∧−^1 )i k′ = δlk

In the last formula we introduced what is called the Kronecker symbol.

δil =

1 if i = j 0 if i = j

Exercise 5. Verify and discuss the relations between the formulas

ei′^ = ∧ji′ ej , ei = (∧−^1 )k ′ i ek′

and ai′ = (∧−^1 )i k′ ak^ , aj^ = ∧ji′ ai′.

Observation: We can interpret the relations (1.3) in a different way. Suppose we have a set of three numbers which characterize some physical, geometrical or other quantity. We can identify whether these numbers are components of a vector, by looking at their transformation rule. They form vector if and only if they transform according to (1.3).

Now we are in a position to introduce the coordinates of a point in the 3D space. Let us remark that the generalization to the general case of an arbitrary D is straightforward.

Def. 5. Three vectors ei, together with some initial point 0 form a system of coordinates. For any point M we can introduce a vector.

r = 0M = xiei , (1.5)

which is called radius-vector of this point. The coefficients xi^ of the expansion (1.5) are called the coordinates of the point M. Indeed, the coordinates are the contravariant components of the radius-vector r.

Observation: Of course, the coordinates xi^ of a point depend on the choice of the basis {ei}, and also on the choice of the initial point O. Similarly to the Eq. (1.5), we can expand the same vector r using another basis

r = xiei = xj′ ej′^ = xj′. ∧ij′ ei

Then, according to (1.3) xi^ = xj′ ∧ij′. The last formula has an important consequence. Taking the partial derivatives, we arrive at the relations

∧ij′ = ∂xi ∂xj′^ and^ (∧

− (^1) )k′ l =^

∂xk′ ∂xl^. Now we can understand better the sense of the matrix ∧. In particular, the relation

(∧−^1 )k l ′· ∧ik′ = ∂x

k′ ∂xl

∂xi ∂xk′^ = ∂x

i ∂xl^ = δli

is nothing but the chain rule for the partial derivatives.

1.2 Scalar, vector and tensor fields

Def. 6. The function ϕ(x) ≡ ϕ(xi) is called scalar field or simply scalar if it does not transform under the change of coordinates.

ϕ(x) = ϕ′(x′). (1.6)

Observation 1. Geometrically, this means that when we change the coordinates, the value of the function ϕ(xi) remains the same in a given geometric point.

Due to the great importance of this definition, let us give clarifying example in 1D. Consider some function, e.g. y = x^2. The plot is parabola, as we all know. Now, let us change the variables x′^ = x + 1. The function y = (x′)^2 , obviously, represents another parabola. In order to preserve the plot intact, we need to modify the form of the function, that is to go from ϕ to ϕ′. Then, the new function y = (x′^ − 1)^2 will represent the original parabola. Two formulas y = x^2 and y = (x′^ − 1)^2

of particles in a stationary flux of a fluid etc. Later on we shall consider particular examples of the transformation of vectors under rotations and inversions.

Observation 3. The scalar and vector fields can be considered as examples of the more general objects called tensors. Tensors are also defined through their transformation rules.

Def. 8. The set of 3n^ functions {ai^1 ...in^ (x)} is called contravariant tensor of rank n, if these functions transform, under xi^ → x′i, as

ai′^1 ...i′^ n^ (x′) = ∂x

i′ 1 ∂xj^1 ...

∂xi′^ n ∂xjn^ a

j 1 ...jn (^) (x). (1.10)

Observation 4. According to the above definitions the scalar and contravariant vector fields are nothing but the particular cases of the contravariant tensor field. Scalar is a tensor of rank 0, and vector is a tensor of rank 1.

Exercise 9. Show that the product ai^ · bj^ is a contravariant second rank tensor, if both ai and bi^ are contravariant vectors.

1.3 Orthonormal basis and Cartesian coordinates

Def. 9. Scalar product of two vectors a and b is defined in a usual way

(a, b) = |a| · |b| · cos(â , b)

where |a|, |b| are absolute values of the vectors a, b and (â , b) is the angle between these two vectors. Of course, the scalar product has the properties familiar from the Analytic Geometry, such as linearity (a, α 1 b 1 + α 2 b 2 ) = α 1 (a, b 1 ) + α 2 (a, b 2 )

and symmetry (a, b) = (b, a).

Observation. Sometimes, we shall use a different notation for the scalar product

(a, b) ≡ a · b.

Until now, we considered an arbitrary basis {ei}. Indeed, we requested the vectors {ei} to be point-independent, but the length of each of these vectors and the angles between them were restricted only by the requirement of their linear independence. It proves useful to define a special basis, which corresponds to the conventional Cartesian coordinates.

Def. 10. Special orthonormal basis {nˆa} is the one with

(ˆna, nˆb) = δab =

1 if a = b 0 if a = b

Exercise 10. Making transformations of the basis vectors, verify that the change of coordi- nates

x′^ = x √^ +^ y 2

• 3 , y′^ = x √^ −^ y 2

does not modify the type of coordinates x′, y′, which remains Cartesian.

Exercise 11. Using the geometric intuition, discuss the form of coordinate transformations which do not violate the orthonormality of the basis.

Notations. We shall denote the coordinates corresponding to the {nˆa} basis as Xa. With a few exceptions, we reserve the indices a, b, c, d, ... for these (Cartesian) coordinates, and use indices i, j, k, l, m, ... for an arbitrary coordinates. For example, the radius-vector r in 3D can be presented using different coordinates:

r = xiei = Xa^ nˆa = xnˆx + y nˆy + z nˆz. (1.12)

Indeed, the last equality is identity, because

X^1 = x , X^2 = y , X^3 = z.

Sometimes, we shall also use notations

ˆnx = ˆn 1 = ˆi , nˆy = ˆn 2 = ˆj , nˆz = ˆn 3 = kˆ.

Observation 1. One can express the vectors of an arbitrary basis {ei} as linear combinations of the components of the orthonormal basis

ei = ∂X

a ∂xi^ nˆa.

An inverse relation has the form

ˆna = ∂xj ∂Xa^ ej^. Exercise 12. Prove the last formulas using the uniqueness of the components of the vector r for a given basis.

Observation 2. It is easy to see that

Xa^ nˆa = xiei = xi^ · ∂Xa ∂xi^ nˆa^ =⇒^ X

a (^) = ∂Xa ∂xi^ ·^ x

i (^). (1.13)

Similarly,

xi^ = ∂x

i ∂Xa^ ·^ X

a (^). (1.14)

Indeed, these relations holds only because the basis vectors ei do not depend on the space-time points and the matrices ∂Xa ∂xi^ and^

∂xi ∂Xa

Thus, ej′ = ∂xj

′ ∂xl^ e l. Using the standard way of dealing with vector transformations, we obtain

aj′^ ej ′ = aj′ ∂xj′ ∂xl^ e

l (^) =⇒ al = ∂x′j ∂xl^ a

′ j.

1.4 Covariant and mixed vectors and tensors

Now we are in a position to define covariant vectors, tensors and mixed tensors.

Def. 13. The set of 3 functions {Ai(x)} form covariant vector field, if they transform from one coordinate system to another as

A′ i(x′) = ∂xj ∂x′i^ Aj^ (x)^.^ (1.17)

The set of 3n^ functions {Ai 1 i 2 ...in (x)} form covariant tensor of rank n if they transform from one coordinate system to another as

A′ i 1 i 2 ...in (x′) = ∂x

j 1 ∂x′i^1

∂xj^2 ∂x′i^2

... ∂x

jn ∂x′in^ Aj 1 j 2 ...jn (x).

Exercise 15. Show that the product of two covariant vectors Ai(x)Bj (x) transforms as a second rank covariant tensor. Discuss whether any second rank covariant tensor can be presented as such a product.

Hint. Try to evaluate the number of independent functions in both cases. Def. 14. The set of 3n+m^ functions {Bi 1 ...in^ j^1 ...jm^ (x)} form tensor of the type (m, n) (an- other possible names are mixed tensor of covariant rank n and contravariant rank m or simply (m, n)-tensor), if these functions transform, under the change of coordinate basis, as

Bi′ 1 ...i′ n^ j^1 ′...j′^ m^ (x′) = ∂x

j′ 1 ∂xl^1

... ∂x

j′m ∂xlm

∂xk^1 ∂x′i^1

... ∂x

kn ∂x′in^ Bk 1 ...kn^ l^1 ...lm^ (x).

Exercise 16. Verify that the scalar, co- and contravariant vectors are, correspondingly, (0, 0)- tensor, (0, 1)-tensor and (1, 0)-tensor.

Exercise 17. Prove that if the Kronecker symbol transforms as a mixed (1, 1) tensor, then in any coordinates xi^ it has the same form

δij =

1 if i = j 0 if i = j

Observation. This property is very important, for it enables us to use the Kronecker symbol in any coordinates.

Exercise 18. Show that the product of co- and contra-variant vectors Ai(x)Bj (x) transforms as a (1, 1)-type mixed tensor.

α 0 x

x

y y M

,

,

Figure 1.1: Illustration of the 2D rotation to the angle α.

Observation. The importance of tensors is due to the fact that they offer the opportunity of the coordinate-independent description of geometrical and physical laws. Any tensor can be viewed as a geometric object, independent on coordinates. Let us consider, as an example, the mixed (1, 1)-type tensor with components Aji. The tensor (as a geometric object) can be presented as a contraction of Aji with the basis vectors

A = Aji ei^ ⊗ ej. (1.18)

The operation ⊗ is called “direct product”, it indicates that the basis for the tensor A is composed by the products of the type ei^ ⊗ ej. In other words, the tensor A is a linear combination of such “direct products”. The most important observation is that the Eq. (1.18) transforms as a scalar. Hence, despite the components of a tensor are dependent on the choice of the basis, the tensor itself is coordinate-independent.

Exercise 19. Discuss the last observation for the case of a vector.

1.5 Orthogonal transformations

Let us consider important particular cases of the global coordinate transformations, which are called orthogonal. The orthogonal transformations may be classified to rotations, inversions (parity transformations) and their combinations. First we consider the rotations in the 2D space with the initial Cartesian coordinates (x, y). Suppose another Cartesian coordinates (x′, y′) have the same origin as (x, y) and the difference is the rotation angle α. Then, the same point M (see the Figure 1.1) has coordinates (x,y) and (x′, y′), and the relation between the coordinates is

x = x′^ cos α − y′^ sin α y = x′^ sin α + y′^ cos α

Exercise 20. Check that the inverse transformation has the form or rotation to the same

(iii) * Write all three matrices ∧ˆz (α), ˆ∧x (γ), and ˆ∧y (β) in 3D for the case of infinitesimal angles α, β, γ. Find all three non-trivial commutators [ ˆ∧z (α) , ∧ˆy (β)

]

= ˆ∧z (α) ˆ∧y (β) − ∧ˆy (β) ˆ∧z (α) , [ ∧ˆy (β) , ∧ˆx (γ)

]

= ˆ∧y (β) ˆ∧x (γ) − ∧ˆx (γ) ˆ∧y (β) , [ ∧ˆz (α) , ˆ∧x (γ)

]

= ˆ∧z (α) ˆ∧x (γ) − ∧ˆx (γ) ˆ∧z (α). (1.25)

of these matrices for α = β = γ. Try to establish the relations between these commutators and the matrices ˆ∧ themselves.

Hint. When making commutations keep only the first order in the angle α. In this way, one may replace sin α  α and cos α  1.

(iv) Verify that the determinant det ˆ∧z^ (α) = 1. Prove that the determinant is equal to one for any rotation matrix.

Observation. Since for the orthogonal matrix ˆ∧ we have the equality

∧ˆ−^1 = ˆ∧T^ ,

we can take determinant and arrive at det ˆ∧ = det ˆ∧−^1 and therefore det ˆ∧ = ±1. As far as any rotation matrix has determinant equal to one, there must be some other orthogonal matrices with the determinant equal to −1. An examples of such matrices are, e.g., the following:

ˆπz =

 , πˆ =

The last matrix corresponds to the transformation which is called inversion. This transformation means that we simultaneously change the direction of all the axis to the opposite ones. One can prove that an arbitrary orthogonal matrix can be presented as a product of inversion and rotation matrices.

Exercise 24. (i) Following the pattern of Eq. (1.26), construct ˆπy and ˆπx. (ii) Find rotation matrices which transform ˆπz into ˆπx and ˆπ into ˆπx. (iii) Suggest geometric interpretation of the matrices ˆπx, ˆπy and ˆπz.

Chapter 2

Operations over tensors, metric tensor

In this Chapter we shall define operations over tensors. These operations may involve one or two tensors. The most important point is that the result is always a tensor.

Operation 1. Summation is defined only for tensors of the same type. Sum of two ten- sors of type (m, n) is also a tensor type (m, n), which has components equal to the sums of the corresponding components of the tensors-summands.

(A + B)i 1 ...in^ j^1 ...jm^ = Ai 1 ...in^ j^1 ...jm^ + Bi 1 ...in^ j^1 ...jm^. (2.1)

Exercise 1. Prove that the components of (A + B) in (2.1) form a tensor, if Ai 1 ...in^ j^1 ...jm and Bi 1 ...in^ j^1 ...jm^ form tensors.

Hint. First try this proof for vectors and scalars. Exercise 2. Prove the commutativity and associativity of the tensor summation. Hint. Use the definition of tensor Def. 1.14.

Operation 2. Multiplication of a tensor by number produces the tensor of the same type. This operation is equivalent to the multiplication of all tensor components to the same number α.

(αA)i 1 ...in^ j^1 ...jm^ = α · Ai 1 ...in^ j^1 ...jm^. (2.2)

Exercise 3. Prove that the components of (αA) in (2.2) form a tensor, if Ai 1 ...in^ j^1 ...jm^ are components of a tensor.

Operation 3. Multiplication of 2 tensors is defined for a couple of tensors of any type. The product of a (m, n)-tensor and a (t, s)-tensor results in the (m + t, n + s)-tensor, e.g.:

Ai 1 ...in^ j^1 ...jm^ · Cl 1 ...ls^ k^1 ...kt^ = Di 1 ...in^ j^1 ...jm^ l 1 ...ls^ k^1 ...kt^. (2.3)

We remark that the order of indices is important here, because, say, aij may be different from aji.

Exercise 4. Prove, by checking the transformation law, that the product of the covariant vector ai^ and the mixed tensor bkj is a mixed (2, 1) - type tensor.

4. The distance between 2 points: M 1 (xi) and M 2 (yi) is defined by

S^212 = gij (xi^ − yi)(xj^ − yj^ ). (2.7)

Proof. Since gij is (2, 0) - tensor and (xi^ − yi) is (0, 1) - tensor (contravariant vector), S 122 is a scalar. Therefore S 122 is the same in any coordinate system. In particular, we can use the nˆa basis, where the (Cartesian) coordinates of the two points are M 1 (Xa) and M 2 (Y a). Making the transformation, we obtain

S 122 = gij (xi^ − yi)(xj^ − yj^ ) = gab(Xa^ − Y a)(Xb^ − Y b) =

= δab(Xa^ − Y a)(Xb^ − Y b) = (X^1 − Y 1 )^2 + (X^2 − Y 2 )^2 + (X^3 − Y 3 )^2 ,

that is the standard expression for the square of the distance between the two points in Cartesian coordinates.

Exercise 7. Check, by direct inspection of the transformationlaw, that the double internal product gij zizj^ is a scalar. In the proof of the Property 4, we used this for zi^ = xi^ − yi.

Exercise 8. Use the definition of the metric and the relations

xi^ = ∂x

i ∂Xa^ Xa^ and ei = ∂X

a ∂xi^ nˆa

to prove the Property 4, starting from S^212 = gab(Xa^ − Y a)(Xb^ − Y b).

Def. 2. The conjugated metric is defined as

gij^ = (ei, ej^ ) where (ei, ek) = δki.

Indeed, ei^ are the vectors of the basis conjugated to ek (see Def. 1.11).

Exercise 9. Prove that gij^ is a contravariant second rank tensor.

Theorem 1.

gikgkj = δij (2.8)

Due to this theorem, the conjugated metric gij^ is conventionally called inverse metric.

Proof. For the special orthonormal basis ˆna^ = ˆna we have gab^ = δab^ and gbc = δbc. Of course, in this case the two metrics are inverse matrices

gab^ · gbc = δac.

Now we can use the result of the Excersize 9. The last equality can be transformed to an arbitrary basis ei, by multiplying both sides by (^) ∂X∂xia and ∂X ∂xjc and inserting the identity matrix (in the parenthesis) as follows

∂xi ∂Xa^ g

ab (^) · gbc^ ∂Xc ∂xj^ =^

∂xi ∂Xa^ g

ab (^ ∂xk ∂Xb

∂Xd ∂xk

gdc ∂Xc ∂xj^ (2.9)

The last expression can be rewritten as

∂xi ∂Xa^ g

ab ∂xk ∂Xb^ ·^

∂Xd ∂xk^ gdc

∂Xc ∂xj^ =^ g

ik (^) gkj.

But, at the same time the same expression (2.9) can be presented in other form

∂xi ∂Xa^ g

ab (^) · gbc^ ∂Xc ∂xj^ =^

∂xi ∂Xa^ δ ca^ ∂Xc ∂xj^ =^

∂xi ∂Xa

∂Xa ∂xj^ =^ δ

ij

that completes the proof.

Operation 5. Raising and lowering indices of a tensor. This operation consists in taking an appropriate internal product of a given tensor and the corresponding metric tensor.

Examples. Lowering the index:

Ai(x) = gij Aj^ (x) , Bik(x) = gij Bj k(x).

Raising the index: Cl(x) = glj^ Cj (x) , Dik(x) = gij^ Dj k(x).

Exercise 10. Prove the relations:

ei gij^ = ej^ , ek^ gkl = el. (2.10)

Exercise 11 *. Try to solve the previous Exercise in several distinct ways.

The following Theorem clarifies the geometric sense of the metric and its relation with the change of basis / coordinate system. Furthermore, it has serious practical importance and will be extensively used below.

Theorem 2. Consider the metric gij corresponding to the basis {ek} and to the coordinates xl. The following relations between the determinants of the metric and of the matrix of the transformation to Cartesian coordinates holds:

g = det (gij ) = det

( (^) ∂Xa ∂xk

, g−^1 = det (gkl) = det

( (^) ∂xl ∂Xb

Proof. The first observation is that, in the Cartesian coordinates the metric is identity matrix gab = δab and therefore det (gab) = 1. Next, since the metric is a tensor,

gij = ∂X

a ∂xi

∂Xb ∂xj^ gab^.

Using the known property of determinants (you are going to prove this relation in the next Chapter as an Exercise) det (A · B) = det (A) · det (B)