DUAL VECTOR SPACES and SCALAR PRODUCTS 1. Dual Vector

DUAL VECTOR SPACES AND SCALAR PRODUCTS

JURGEN¨ TOLKSDORF

NOVEMBER 28, 2012

1. Dual vector spaces Let, respectively, V and W be a real vector spaces of dimension 1 ≤ n < ∞ and 1 ≤ m < ∞. The real vector space of all linear maps from V to W is denoted by m×n HomR(V,W ). It is isomorphic to R when a basis is chosen in V and W . Deﬁnition 1.1. The real vector space ∗ V := HomR(V, R) ≡ {α : V −→ R | linear } (1) of all linear maps from V to R (the latter considered as a real one-dimensional vector space) is called the dual vector space of V . Lemma 1.1. The dimension of V ∗ equals the dimension of V . ∗ Proof. Let ~v1, . . . ,~vn ∈ V be a basis. One may deﬁne n linear maps νi ∈ V as

νi(~vj) := δij (1 ≤ i, j ≤ n) . (2) Remember that any linear map is fully determined by its action on an (arbitrary) P basis. In fact, for ~v = 1≤k≤n λk~vk one gets νi(~v) = λi ∈ R (i = 1, . . . , n). ∗ We prove that ν1, . . . , νn ∈ V are linearly independent. Assume that the vector P ∗ α := 1≤k≤n µkνk ∈ V is the zero map. I.e. ν(~v) = 0 ∈ R holds for all ~v ∈ V . Since this holds for all ~v ∈ V , it follows that µ1 = µ2 = ··· µn = 0 simply by letting ∗ ~v = ~vk, k = 1, . . . , n. Hence, n ≤ dim(V ). ∗ To demonstrate that span(ν1, . . . , νn) = V , we remember that the vector space HomR(V,W ) of all linear maps between real vector spaces V and W (each of finite dimension) equals dim(V )dim(W ). In fact, choosing a basis in V and a basis in W dim(V ) dim(W ) allows to identify V with R and W with R . Hence, HomR(V,W ) can be identified with the real vector space Rdim(V )×dim(W ) of dimension dim(V )dim(W ). In ∗ the special case where W = R one therefore obtains that dim(V ) = dim(V ) as was to be proven. Notation: The elements of V ∗ are referred to as linear functionals, or one-forms. They are also called “dual vectors” or, especially in physics, as “co-vectors”. Accord- ∗ ingly, the basis ν1, . . . , νn ∈ V that is defined by (2) is called the dual basis (co-basis) of the basis ~v1, . . . ,~vn ∈ V .

Notice that the dual basis of some chosen basis ~v1, . . . ,~vn in V is uniquely deﬁned by ~v1, . . . ,~vn ∈ V . Furthermore, in general it only makes sense to speak about the dual of a basis. In contrast, with a single vector ~v ∈ V one cannot, in general, associate a (uniquely determined) dual vector α ∈ V ∗.

1 2 J. TOLKSDORF

n n Example: We may consider V = R with its standard basis ~e1, . . . ,~en ∈ R . Its dual basis is simply deﬁned by the mappings:

∗ n ek : R −→ R (3) (x1, x2, . . . , xn) 7→ xk (k = 1, . . . , n) .

n When the real vector space R of n−tuples is identiﬁed with the real vector space n×1 n ∗ R of matrices of size n × 1, the dual space (R ) is identiﬁed with the real vector 1×n space R of matrices of size 1 × n. That is, one has the isomorphism

t n ∗ 1×n σn :(R ) −→ R X ∗ (4) α = µkek 7→ µ ≡ (µ1, . . . , µk) , 1≤k≤n

P n such that for all ~y ≡ (y1, . . . , yn) = 1≤k≤n yk~ek ∈ R and   y1  .  n×1 y ≡ σn(~y) :=  .  ∈ R (5) yn

n ∗ n one obtains for the action of α ∈ (R ) on R that X α(~y) = µy = µkyk ∈ R . (6) 1≤k≤n

∗ ∗ In particular, ei (ej) = ei ej = δij for all 1 ≤ i, j ≤ n. 1×n n As a warning to not confound R with R we discuss the case n = 2 in some ~ 2 ~ 2 detail. For this consider b1 := ~e1 + ~e2 = (1, 1) ∈ R and b1 := ~e1 − ~e2 = (1, −1) ∈ R 2 ~ ~ as a diﬀerent basis in R . It follows that ~x ≡ (x1, x2) = x1~e1 + x2~e2 = y1b1 + y2b2 = 2 2×1 (y1 + y2, y1 − y2) ∈ R . The coordinate vectors x, y ∈ R read: x y x = 1 , y = 1 . (7) x2 y2

2×1 n Here, the coordinate vector x ∈ R represents ~x ∈ R with respect to the standard n 2×1 basis ~e1,~e2 ∈ R . In contrast, the coordinate vector y ∈ R represents the same 2 ~ ~ 2 ~x ∈ R with respect to the non-standard basis b1, b2 ∈ R . ∗ ∗ 2 ∗ ~ ~ 2 How does the dual basis b1, b2 ∈ (R ) of the basis b1, b2 ∈ R look like? For this one has to solve the linear system of equations that corresponds to the deﬁnition of a ∗ ~ dual basis bi (bj) := δij , 1 ≤ i, j ≤ 2: ∗ ∗ b1(1, 1) = 1 , b1(1, −1) = 0 , ∗ ∗ (8) b2(1, 1) = 0 , b2(1, −1) = 1 .

∗ P ∗ 2 ∗ Clearly, we may write bi = 1≤j≤n βijej ∈ (R ) for i = 1, 2.

Notice the order of the indices and compare this with the corresponding expansion 2 of the non-standard basis with respect to the standard basis of R . EUCLIDEAN SPACES 3

One obtains the system of linear equations

β11 + β12 = 1 ,

β11 − β12 = 0 , (9) β21 + β22 = 0 ,

β21 − β22 = 1 , whose unique solution is given by β11 = β21 = β21 = −β22 = 1/2. Hence,

∗ 1 ∗ ∗ b1 = e1 + e2 , 2 (10) 1 b∗ = e∗ − e∗ . 2 2 1 2 ∗ ∗ ∗ ∗ 2 ∗ The corresponding matrix representations of α = µ1b1 + µ2b2 = λ1e1 + λ2e2 ∈ (R ) 1×2 1×2 read: µ = (µ1, µ2) ∈ R and λ = (λ1, λ2) ∈ R . That is to say, both dual basis are represented by the same matrices: ∗ t ∗ t ∗ ∗ 1×2 e1 ≡ σ2 (e1) = σ2 (b1) ≡ b1 = (1, 0) ∈ R , (11) ∗ t ∗ t ∗ ∗ 1×2 e2 ≡ σ2 (e2) = σ2 (b2) ≡ b2 = (0, 1) ∈ R . It follows that α(~x) = λx = µy ∈ R . (12) Indeed, one has

µy = µ1y1 + µ2y2 1 1 = (λ + λ )(x + x ) + (λ − λ )(x − x ) 2 1 2 1 2 2 1 2 1 2 (13) = λ1x1 + λ2x2 = λx . n ∗ n ∗ Notice that ~ek ∈ R , as opposed to ek ∈ (R ) . It is the matrix representation ∗ 1×n ∗ n ∗ n ek ∈ R of ek ∈ (R ) that looks formally identical to ~ek ∈ R . This is because real n−tuples and real matrices of size 1×n look alike. As the example demonstrates, how- n ∗ ever, the matrix representation of any basis in (R ) formally looks like the standard n n basis of R . Therefore, one has to clearly distinguish between ~x = (x1, . . . , xn) ∈ R ∗ 1×n ∗ and x = (x1, . . . , xn) ∈ R . In the latter case, x is supposed to represent some n ∗ n element in (R ) with respect to some chosen basis. In contrast, ~x = (x1, . . . , xn) ∈ R does not refer to any such a choice of basis. n ∗ Furthermore, a linear form (co-vector) α ∈ (R ) can never be expressed in terms n of an n−tuple, though its action on R can be represented by matrix multiplication: 1×n n ∗ α(~v) = αv. The matrix α ∈ R represents the linear form α ∈ (R ) and the matrix n×1 n ~ ~ n v ∈ R the vector ~v ∈ R with respect to a chosen basis b1,..., bn ∈ R and its ∗ ∗ n ∗ corresponding dual basis b1, . . . , bn ∈ (R ) .

The value λx ∈ R is independent of the choice of the basis used to express α(~x) in terms of matrix multiplication. This is most clearly expressed, for instance, by the equality (12). To see why this this holds true, actually, let ~v1, . . . ,~vn ∈ V be any ∗ ∗ ∗ basis with its dual basis denoted by v1, . . . , vn ∈ V . Also, let ~u1, . . . , ~un ∈ V be ∗ ∗ ∗ another basis with corresponding dual basis u1, . . . , un ∈ V . Let f ∈ AutR(V ) be an isomorphism that is deﬁned by f(~vk) := ~uk for all k = 1, . . . , n ≡ dim(V ) ≥ 1. 4 J. TOLKSDORF

∗ ∗ ∗ Correspondingly, let g ∈ AutR(V ) be an isomorphism on the real vector space V ∗ ∗ ∗ ∗ that is deﬁned by g (vk) := uk for all k = 1, . . . , n = dim(V ). We claim that g∗ is fully determined by f −1. Indeed, from the deﬁnition of a dual basis one infers that ∗ ∗ ∗ ∗ ui (~uj) = g (vi ) f(~vj) = δij = vi (~vj) , for all i, j = 1, . . . , n . (14) P ∗ P ∗ We may rewrite this as follows. Let ~uj = 1≤i≤n aij~vi ∈ V and ui = 1≤j≤n bijvj ∈ ∗ V , with uniquely determined aij, bij ∈ R for all 1 ≤ i, j ≤ n. It follows that for all 1 ≤ i, j ≤ n: ∗ X ∗ ui (~uj) = bikaljvk(~vl) 1≤k,l≤n X = bikaljδkl 1≤k,l≤n (15) X = bikakj 1≤k≤n = δij . n×n ∗ Therefore, the matrix B := (bij)1≤i,j≤n ∈ R , representing the linear map g , must n×n be the inverse of the matrix A := (aij)1≤i,j≤n ∈ R that represents the linear map f, i.e. B = A−1. Hence, the isomorphism g∗ is fully determined by f −1. To also demonstrate this more explicitly, we mention that every linear map g : V −→ W between real vector spaces V and W uniquely determines a correspondingly linear map g∗ : W ∗ −→ V ∗, via g∗(α)(~v) := α(g(~v)), for all α ∈ W and ~v ∈ V . Using this one therefore obtains that g∗ = (f −1)∗. For instance, with respect to our example above this simply means that µy = λA−1Ax = λA−1Ax (16) = λx . Hence, the real number λx ∈ R is indeed independent of the choice of basis.

n×n 1×n Notice that B ∈ R acts from the right on λ ∈ R , according to the rules of matrix multiplication: 1×n µ = λB ∈ R . (17) This explains the already mentioned converse order of summation indices when com- n×1 pared to the matrix multiplication y = Ax ∈ R . One says that the change of a dual basis is contra-gradient to the change of a basis.

2. Inner product spaces Let again V be a real vector space of (ﬁnite) dimension n ≥ 1. Deﬁnition 2.1. A map β : V × V −→ R (18) (~v, ~u) 7→ β(~v, ~u) is called a bilinear form, provided it is linear in both arguments:

(1) β(~v1 + λ~v2, ~u) = β(~v1, ~u) + λβ(~v2, ~u) , EUCLIDEAN SPACES 5

(2) β(~v, ~u1 + λ~u2) = β(~v, ~u1) + λβ(~v, ~u2) , for all ~v1, . . . , ~u2 ∈ V and all λ ∈ R. A bilinear form is said to be symmetric, provided that for all ~v, ~u ∈ V : β(~v, ~u) = β(~u,~v) . (19) A bilinear form is called positive semi-definite if for all ~v ∈ V : β(~v,~v) ≥ 0 . (20) It is called positive definite if furthermore β(~v,~v) = 0 ⇒ ~v = ~0. Finally, a bilinear form is called non-degenerate, if for all ~u ∈ V : β(~v, ~u) = 0 ⇒ ~v = ~0 . (21) A non-degenerated bilinear form is also called an inner product on V . A symmetric and positive definite inner product is called a scalar product on V . Definition 2.2. Let V be a finite dimensional real vector space endowed with a scalar product β. The function k · k : V −→ R (22) ~v 7→ pβ(~v,~v) is called the norm (or length) of ~v ∈ V with respect to β.

Since β is a scalar product it follows for all ~v ∈ V and λ ∈ R that k~vk ≥ 0, whereby k~vk = 0 ⇔ ~v = ~0, and kλ~vk = |λ|k~vk. Most important is the following statement called Schwarz inequality, which we state without proof: Proposition 2.1. Let V be a finite dimensional real vector space endowed with a scalar product β. The induced norm fulfils for all ~v, ~u ∈ V : |β(~v, ~u)| ≤ k~vk k~uk , (23) where the equality holds if and only if ~v = λ~u for some λ ∈ R. Definition 2.3. Let again V be a finite dimensional real vector space endowed with a scalar product β. A linear map f : V −→ V is called symmetric, if for all ~v, ~u ∈ V : β(f(~u),~v) = β(f(~v), ~u) . (24) A symmetric linear map is called positive semi-definite if it fulfills for all ~u ∈ V : β(f(~u), ~u) ≥ 0 . (25) If the inequality is strict for all ~u ∈ V \{~0}, then f is called positive definite. Notice that a positive linear map is always invertible and thus an automorphism on V . Lemma 2.1. An inner product β on V is in one-to-one correspondence with an isomorphism V ' V ∗. Proof. Let β be an inner product on V . We may define f : V −→ V ∗ V −→ (26) ~v 7→ R ~u 7→ β(~v, ~u) . 6 J. TOLKSDORF

This map is linear and injective since β is bilinear and non-degenerated. It therefore is an isomorphism since dim(V ∗) = dim(V ). Conversely, let f : V −→' V ∗ be an isomorphism. We then consider β(~v, ~u) := f(~v)(~u) , (27) for all ~v, ~u ∈ V . The thus deﬁned map β : V ×V → R is bilinear and non-degenerated, ∗ for f is an isomorphism and f(~v) ∈ V is a linear form on V for all ~v ∈ V . Notation: It is common to denote f(~v) = β(~v, ·) ≡ v[ ∈ V ∗ and f −1(α) ≡ α] ∈ V , where f and its inverse f −1 are called musical isomorphisms.

An inner product thus allows to identify a vector space with its dual vector space. This should not be confounded with the fact that every choice of a basis always uniquely determines a corresponding dual basis. For the latter to be (uniquely) determined, however, one needs the whole set of basis vectors, in general. In contrast, an inner product allows to associated with each individual vector a uniquely defined linear form (and conversely). A vector space V together with an inner product β is called an inner product space (V, β). P P Let ~v1, . . . ,~vn ∈ V be a basis. Also, let ~x = 1≤i≤n xi~vi and ~y = 1≤j≤n yj~vj. Since β is bilinear one has n X β(~v, ~u) = xiyjβ(~vi,~vj) i,j=1 n (28) X ≡ gijxiyj . i,j=1 Here, the coefficients gij := β(~vi,~vj) ∈ R (29) n×n give rise to the matrix G ≡ (gij)1≤i,j≤n ∈ R , called the coefficient matrix of the bilinear form β with respect to the chosen basis. Similar to linear maps, a bilinear form β is also uniquely determined with respect to a basis, i.e. by the corresponding coefficient matrix. Let again ~u1, . . . , ~un ∈ V be another basis and f ∈ AutR(V ) be an isomorphism that 0 is defined by f(~vk) := ~uk for all k = 1, . . . , n. We may set gij := β(~ui, ~uj). Accordingly, 0 0 n×n we may denote by G := (gij)1≤i,j≤n ∈ R the coefficient matrix of β with respect to the basis ~u1, . . . , ~un ∈ V . It follows that 0 gij = β(f(~vi), f(~vj)) Pn Pn = β k=1aki~vk, l=1alj~vl n X = akialjβ(~vk,~vl) (30) k,l=1 X = akigklalj . 1≤k,l≤n That is, G0 = AtGA , (31) EUCLIDEAN SPACES 7

t n×n with A ≡ (aik)1≤i,k≤n ∈ R being the transposed matrix of the matrix A ≡ n×n (aki)1≤k,i≤n ∈ R , which represents the automorphism f with respect to the basis ~v1, . . . ,~vn ∈ V . As a consequence, one may express also the action of a bilinear form β in terms of matrix multiplication as β(~x,~y) = xtGy = x0tG0y0 . (32)

t Notice that the matrix product x Gy ∈ R does not depend on the basis ~v1, . . . ,~vn ∈ n×1 V used to represent ~x,~y ∈ V and β by the coordinate vectors x, y ∈ R and the co- n×n 0 −1 0 eﬃcient matrix G ∈ R . This is due to x = A x and similar for y . Furthermore, n×m m×k t t t k×n for all matrices A ∈ R , B ∈ R one has (AB) = B A ∈ R . Also, for all n×n −1t t−1 invertible matrices A ∈ R it follows that A = A .

A bilinear form β is non-degenerated if and only if there exists a basis such that the corresponding coeﬃcient matrix is invertible. Finally, a bilinear form is symmetric if and only if there is a basis such that the corresponding coeﬃcient matrix is symmetric: Gt = G.

Deﬁnition 2.4. Let (V, β) be an inner product space where β is a scalar product. A basis ~v1, . . . ,~vn ∈ V is called orthonormal with respect to β, provided that

β(~vi,~vj) = δij (1 ≤ i, j ≤ n) . (33) Notice that the coeﬃcient matrix of β respect to an orthonormal basis is given by the unit matrix. In this case, one thus has β(~x,~y) = xty.

Without proof we give the following statement: Every finite dimensional vector space V that is endowed with a scalar product β possesses an orthonormal basis. This is proved by what is called the Gram-Schmidt procedure. The latter goes p as follows: First, choose any ~u1 ∈ V \{~0} and put ~v1 := ~u1/ β(~u1, ~u1). Clearly, β(~v1,~v1) = 1. Notice that the definition of ~v1 ∈ V makes sense, for β is supposed to be positive definite. Now take any ~w1 ∈ V \{~0,~v1} and consider ~u2 := ~w1 − β(~w1,~v1)~v1. When the Schwarz inequality is taken into account, one may prove that ~u2 ∈ V \{~0}. p Moreover, β(~v1, ~u2) = 0. Then set ~v2 := ~u1/ β(~u2, ~u2) to obtain β(~vi,~vj) = δij for i, j = 1, 2. To proceed consider any ~w2 ∈ V \{~0,~v1,~v2} and set

~u3 := ~w1 − β(~w1,~v1)~v1 − β(~w1,~v2)~v2 (34)

Again, when the Schwarz inequality is taken into account, one may show that ~u3 6= ~0. p Also, β(~vk, ~u3) = 0 for all k = 1, 2. Then set ~v3 := ~u3/ β(~u3, ~u3) and proceed until one has constructed ~v1, . . . ,~vn ∈ V , such that β(~vi,~vj) = δij for i, j = 1, . . . , n. It is straightforward to demonstrate that these vectors are linearly independent and thus build an orthonormal basis of (V, β). One may therefore conclude that any scalar product is fully characterized by an orthonormal basis. However, the latter is not unique. Deﬁnition 2.5. Let V and W be ﬁnite dimensional real vector spaces and, respectively, βV and βW be a scalar products on V and W . An isomorphism f ∈ HomR(V,W ) is 8 J. TOLKSDORF called an isometry, provided that for all ~x,~y ∈ V :

βW(f(~x), f(~y)) = βV(~x,~y) . (35) An isometry keeps the length ﬁxed. It respects the full structure and therefore may also be viewed as an isomorphism ' (V, βV) −→ (W, βW) . (36)

For W = V , it is straightforwardly checked that an automorphism f ∈ AutR(V ) is n×n an isometry if and only if the matrix A ∈ R , which represents f with respect to an orthonormal basis, fulfills At = A−1 . (37) Matrices satisfying this relation are called orthogonal matrices. They build a group, n×n called orthogonal group, that is denoted by SO(n) ⊂ R . Similarly, the set of all isometries build a group SO(V, β) ⊂ AutR(V ). It follows that any scalar product β is defined by selecting a basis in V , which is considered as being orthonormal. This defines β uniquely up to isometries.

n n Example: We again consider the special case V = R . Since R possesses a n n distinguished basis ~e1, . . . ,~en ∈ R , one also has a distinguished scalar product on R , called the dot product: ~ei · ~ej := δij (1 ≤ i, j ≤ n) . (38) n The dot product is thus the scalar product on R that is defined by considering the standard basis as being orthonormal. Any other scalar product is given by a positive linear map f : V → V , i.e. β(~u,~v) := f(~u) · ~v for all ~u,~v ∈ V. (39) Indeed, β is symmetric since f is supposed to be symmetric with respect to the dot product. Also, β is positive definite since f is positive definite with respect to the Pn dot product. We may set f(~ej) = i=1 gij~ei for all j = 1, . . . , n. It follows that n×n t t G ≡ (gij)1≤i,j≤n ∈ R is symmetric and positive definite: G = G and x Gx > 0 n×1 for all 0 6= x ∈ R . Hence, n n X X β(~ei,~ej) = gik~ek · ~ej = gikδkj = gij . (40) k=1 k=1 ~ −1 ~ ~ n We may set bk := f (~ek) for k = 1, . . . , n. Then, b1,..., bn ∈ R is a basis since n f ∈ AutR(R ). In fact, it constitutes an orthonormal basis with respect to β:

β(~bi,~bj) = δij (1 ≤ i, j ≤ n) . (41) However, n ~ ~ X bi · bj = gikgkj , (42) k=1 ~ ~ n t 2 i.e. the positive matrix deﬁned by (bi · bj)1≤i,j≤n ∈ R coincides with G G = G . Consequently, 2 gij = λi δij (1 ≤ i, j ≤ n) , (43) th where λk ∈ R is strictly positive and denotes the k −eigenvalue of f: f(~ek) = λk~ek ~ n for all k = 1, . . . , n. Hence, bk = ~ek/λk ∈ R . EUCLIDEAN SPACES 9

n We have thus proved that all scalar products on R are determined by (43), i.e. by coefficient matrices of the form  2  λ1 ··· 0  . .  n×n G =  . ··· .  ∈ R . (44) 2 0 ··· λn Notice that the musical isomorphism V −→' V ∗ (45) ~x 7→ x[ with respect to any scalar product β corresponds to the isomorphism n×1 ' 1×n R −→ R (46) x 7→ xt . This correspondency refers to any β−orthonormal basis. Especially this holds true for n V = R and β being given by the dot product. In this case, one simply gets ~x · ~y = x[(~y) = xty , (47) n [ n ∗ for all ~x,~y ∈ R . That is, one my tend to simply identify x ∈ (R ) with the matrix t 1×n n n×1 x ∈ R , whenever the vector ~x ∈ R is identified with the matrix x ∈ R . However, this is not appropriate since neither ~x·~y, nor x[(~y) refers to any basis. Thus, n n×1 one may identify, for instance, the vector ~x ∈ R with the matrix x ∈ R with n t help of any basis in R . However, this yields ~x · ~y = x Gy, where again G is the coefficient matrix of the dot product with respect to the chosen basis. That is, one has [ n ∗ t t 1×n to identify x ∈ (R ) with x G = Gx ∈ R . Notice that in the case considered xtGy is still the matrix representation of the dot product, but with respect to an arbitrary basis. This should not be confounded with xtGy representing an arbitrary scalar product β(~x,~y) with respect to the standard basis.

Max Planck Institute for Mathematics in the Sciences, Leipzig, Germany E-mail address: [email protected]