Duality of finite-dimensional vector spaces

CHAPTER I Duality of finite-dimensional vector spaces 1 Dual space Let E be a finite-dimensional vector space over a field K The vector space of linear maps E K is denoted by E, so E = L(E, K) This vector space is called the dual space of E Its elements are called linear forms on E For ϕ E and u E, let ϕ, u = ϕ(u) K 11 Proposition For ϕ, ψ E, u, v E and α, β K, 1 αϕ + βψ, u = α ϕ, u + β ψ, u 2 ϕ, uα + vβ = ϕ, u α + ϕ, v β 3 If ϕ, u = 0 for all u E, then ϕ = 0 4 If ϕ, u = 0 for all ϕ E, then u = 0 Proof 1 This follows from the definition of the sum and the scalar multiplication in E 2 This follows from the fact that ϕ is a linear map 3 This is the definition of the zero map 4 If u 0, then u is the first element in some basis of E Let (u, e 2,, e n ) be such a basis From the construction principle for linear maps, it follows that there is a linear form ϕ E which maps u to 1 and the other basis vectors to 0 This proves the contrapositive of implication 4 11 Dual basis Let e = (e 1,, e n ) be a basis of E By the construction principle for linear maps, there exists for all i = 1,, n a linear form e i E which maps e i to 1 and the other basis vectors to 0, ie, { e i, e 1 if i = j, j = δ ij = 0 if i j Thus, for u = n j=1 e ju j E we have n e i, u = e i, e j u j = j=1 n δ ij u j = u i, which shows that the linear form e i maps every vector of E to its i-th coordinate with respect to the basis e Therefore, for all u E, n (1) u = e i e i, u n, 12 Theorem The sequence e = (e 1,, e n) is a basis of E In particular, dime = dime Proof The elements e 1,, e n are linearly independent: Let n α ie i n 0 = α i e i, e j = j=1 n α i e i, e j = 1 n α i δ ij = α j = 0 For j = 1,,

2 ORTHOGONALITY 2 The sequence e spans E : Let us show that for all ϕ E, n (2) ϕ = ϕ, e i e i It suffices to show that both sides take the same value on every vector u E From (1), it follows that n n ϕ, u = ϕ, e j e j, u = ϕ, e j e j, u On the other hand, j=1 n ϕ, e i e i, u = j=1 n ϕ, e i e i, u 12 Bidual space Every vector u E defines a linear form ev u : E K by Thus, ev u (E ) Moreover, the map ev u (ϕ) = ϕ(u) ev: E E ; is linear since for u, v E, α, β K and ϕ E, u ev u ev uα+vβ (ϕ) = ϕ(uα + vβ) = ϕ(u)α + ϕ(v)β = ev u (ϕ)α + ev v (ϕ)β 13 Theorem The map ev: E E is a vector space isomorphism Proof First, we show that ev is injective: if u E is such that ev u = 0, then ϕ, u = 0 for all ϕ E, hence u = 0 by Proposition 11 Therefore, Kerev = 0 and ev is injective On the other hand, applying Theorem 12 twice, we find Since ev is injective, it is therefore also surjective dime = dime = dim E The isomorphism ev is natural, inasmuch as it does not depend on the choice of bases in E and E It is used to identify these two vector spaces Thus, for u E, it is agreed that u = ev u E Therefore, u, ϕ is defined for u E and ϕ E, and we have u, ϕ = ev u, ϕ = ϕ(u) = ϕ, u 2 Orthogonality For every subspace V E, the orthogonal subspace V 0 E is defined by It is easily checked that V 0 is a subspace of E 21 Proposition dimv 0 = dime dimv V 0 = {ϕ E ϕ, u = 0 for all u V } Proof Let (e 1,, e r ) be a basis of V, which we extend into a basis e = (e 1,, e r, e r+1,, e n ) of E Consider the last n r elements of the dual basis e = (e 1,, e r,, e n) We shall show: V 0 = span e r+1,, e n The proposition follows, since the right side is a subspace of dimension n r, as e r+1,, e n are linearly independent Equation (2) shows that every ϕ V 0 is a linear combination of e r+1,, e n Therefore, V 0 span e r+1,, e n

3 TRANSPOSITION 3 On the other hand, every v V is a linear combination of (e 1,, e r ), hence equation (1) shows that e i, v = 0 for i = r + 1,, n Therefore, e r+1,, e n V 0, hence span e r+1,, e n V 0 22 Corollary For every subspace V E, V 00 = V (under the identification of E and E, since V 00 E ) Proof By definition, V 00 = {u E u, ϕ = 0 for all ϕ V 0 } Since ϕ, u = 0 for u V and ϕ V 0, we have V V 00 Now, the preceding proposition and Theorem 12 yield dimv 00 = dime dimv 0 = dime dimv 0 = dimv Therefore, the inclusion V V 00 is an equality 3 Transposition Let A: E F be a linear map between finite-dimensional vector spaces The transpose map is defined by A t : F E A t (ϕ) = ϕ A for ϕ F 31 Proposition A t is the only linear map from F to E such that A t (ϕ), u = ϕ, A(u) for all ϕ F and all u E Proof For ϕ F and u E we have A t (ϕ), u = ϕ A, u = ϕ A(u) = ϕ, A(u), hence A t satisfies the property Suppose B: F E is another linear map with the same property Then for all ϕ F and all u E, B(ϕ), u = ϕ, A(u) = A t (ϕ), u, hence B(ϕ) = A t (ϕ) and therefore B = A t Now, let e and f be arbitrary bases of E and F respectively, and let e, f be the dual bases 32 Proposition The matrices of A t and A are related as follows: Proof Let f (A) e = (a ij ) 1 i m, ie, m (1) A(e j ) = f i a ij and e (A t ) f = (a ij ) 1 i n, ie, 1 j m A t (f j ) = e (At ) f = f (A) t e for j = 1,, n n e i a ij for j = 1,, m

3 TRANSPOSITION 4 We have to prove a ij = a ji for i = 1,, n and j = 1,, m Substituting A t (fj ) for ϕ in equation (2) of section 11, we obtain A t (f j ) = n A t (fj ), e i e i for j = 1,, m, hence a ij = A t (f j ), e i By Proposition 31, it follows that a ij = f j, A(e i ) On the other hand, we may compute the right side of this last equation by using equation (1), which yields f j, A(e i ) = f n j, n f k a ki = fj, f k a ki = a ji k=1 Thus, a ij = a ji for i = 1,, n and j = 1,, m k=1 33 Proposition 1 For A, B: E F and α, β K, 2 For A: E F and B: F G, 3 For A: E F, (αa + βb) t = αa t + βb t (B A) t = A t B t (A t ) t = A (under the identifications E = E and F = F, since (A t ) t : E F ) Proof These properties can be proved either by reduction to the corresponding properties of matrices (by using arbitrary bases of the relevant spaces), or by using the characterization of transpose maps in Proposition 31

CHAPTER II Quotient spaces 1 Definition Let E be an arbitrary vector space over a field K and let V E be an arbitrary subspace For x E let x + V = {x + v v V } E 11 Lemma For x, y E, the equation x + V = y + V is equivalent to x y V Proof Suppose first x+v = y+v Since x = x+0 x+v, we have x y+v, hence x = y+v for some v V and therefore x y = v V Conversely, suppose x y = v V For all w V we then have x + w = y + (v + w) y + V, hence x + V y + V We also have y + w = x + (w v) x + V, hence y + V x + V, and therefore x + V = y + V Let E/V = {x + V x E} An addition and a scalar multiplication are defined on this set by (x + V ) + (y + V ) = (x + y) + V for x, y E, (x + V )α = xα + V for x E and α K It is easily checked that these operations are well-defined and endow the set E/V with a K-vector space structure In view of the definitions above, the map defined by ε V : E E/V ε V (x) = x + V is a surjective linear map It is called the canonical epimorphism of E onto E/V Its kernel is the set of x E such that x + V = 0 + V The lemma shows that this condition holds if and only if x V ; therefore Kerε V = V 12 Proposition Suppose E is finite-dimensional; then Moreover, dime/v = dime dim V (a) if (e 1,, e n ) is a basis of E which extends a basis (e 1,, e r ) of V, then (e r+1 +V,, e n +V ) is a basis of E/V ; (b) if (e 1,, e r ) is a basis of V and u r+1,, u n are vectors of E such that (u r+1 +V,, u n +V ) is a basis of E/V, then (e 1,, e r, u r+1,, u n ) is a basis of E 5

2 INDUCED LINEAR MAPS 6 Proof Since Kerε V = V and Imε V = E/V, the relation between the dimensions of the kernel and the image of a linear map yields dime = dimv + dime/v (a) Since dime/v = dime dimv = n r, it suffices to prove that the sequence (e r+1 +V,, e n +V ) spans E/V Every x E is a linear combination of e 1,, e n If then x = e 1 x 1 + + e n x n, x + V = ε V (x) = ε V (e 1 )x 1 + + ε V (e n )x n Now, ε V (e i ) = 0 for i = 1,, r since e i V, hence x + V = (e r+1 + V )x r+1 + + (e n + V )x n (b) Since dime = dimv + dim E/V, it suffices to prove that e 1,, e r, u r+1,, u n are linearly independent Suppose (1) e 1 α 1 + + e r α r + u r+1 β r+1 + + u n β n = 0 Taking the image of each side under ε V, we obtain (u r+1 + V )β r+1 + + (u n + V )β n = 0, hence β r+1 = = β n = 0 Therefore, by (1) it follows that α 1 = = α r = 0, since (e 1,, e r ) is a basis of V 2 Induced linear maps Let A: E F be a linear map and let V E be a subspace A linear map A: E/V F is said to be induced by A if In other words, A(x + V ) = A(x) for all x E A ε V = A This condition is also expressed as follows: the diagram E ε V E/V is commutative or commutes A A F 21 Proposition A linear map A: E F induces a linear map if and only if V KerA A: E/V F Proof If A ε V = A, then A(v) = 0 for all v V since V = Kerε V For the converse, suppose V KerA and consider x, x E If x + V = x + V, then x x KerA, hence A(x x ) = 0 and therefore We may then define a map A: E/V F by A(x) = A(x ) A(x + V ) = A(x) for all x E It is easily checked that the map A thus defined is linear

2 INDUCED LINEAR MAPS 7 Now, consider a linear map A: E F and subspaces V E and W F A linear map is said to be induced by A if A: E/V F/W A(x + V ) = A(x) + W for all x E or, alternatively, if A ε V = ε W A This condition can also be expressed by saying that the following diagram is commutative: E A F ε V E/V A ε W F/W or that A makes the above diagram commute 22 Corollary The linear map A: E F induces a linear map A: E/V F/W if and only if A(V ) W Proof Consider the linear map A = ε W A: E F/W The preceding proposition shows that there exists a linear map A such that A = A ε V if and only if V KerA In view of the definition of A, this condition is equivalent to A(V ) Kerε W = W If A(V ) W, we may restrict the linear map A to the subspace V and get a linear map A V : V W We aim to compare the matrices of A, of A V and of A: E/V F/W Consider a basis e = (e 1,, e r ) of V, a basis e = (e r+1 + V,, e n + V ) of E/V and the sequence e = (e 1,, e r, e r+1,, e n ) which, according to Proposition 12, is a basis of E Similarly, let f = (f 1,, f s ) be a basis of W, f = (f s+1 + W,, f m + W) a basis of F/W and consider the basis f = (f 1,, f s, f s+1,, f m ) of F 23 Proposition f(a) e = ( f (A V ) e 0 f (A) e ) Proof Let A(e j ) = m f ia ij for j = 1,, n, so that f(a) e = (a ij ) 1 i m For j = 1,, r we have e j V, hence A(e j ) W Therefore, A(e j ) is a linear combination of f 1,, f s, hence a ij = 0 for j = 1,, r and i = s + 1,, m Moreover, A(e j ) = A V (e j ) for j = 1,, n, hence A V (e j ) = s f ia ij and therefore On the other hand, for j = r + 1,, n, f (A V ) e A(e j + V ) = A(e j ) + W = = (a ij ) 1 i s 1 j r m (f i + W)a ij For i = 1,, s, we have f i W, hence f i +W = 0 The preceding equation thus yields A(e j +V ) = m i=s+1 (f i + W)a ij for j = r + 1,, n, hence f (A) e = (a ij ) s+1 i m r+

3 TRIANGULATION 8 The proof is thus complete In particular, if F = E and W = V, e(a) e = ( e (A V ) e 0 e(a) e and since the determinant of a block-triangular matrix is the product of the determinants of the blocks the following corollary follows: 24 Corollary deta = deta V deta and Pc A = Pc A V Pc A 3 Triangulation 31 Proposition Let A: E E be a linear operator on a finite-dimensional vector space E There exists a basis e of E such that the matrix e (A) e is upper triangular if and only if the characteristic polynomial Pc A decomposes into a product of factors of degree 1 Proof If the matrix e (A) e is upper triangular, let e(a) e = λ 1 0 λ n then Pc A (X) = (X λ 1 )(X λ n ) Conversely, suppose Pc A decomposes into a product of factors of degree 1 We argue by induction on the dimension of E If dim E = 1, there is nothing to prove since every matrix of order 1 is upper triangular Let dim E = n By hypothesis, the characteristic polynomial of A has a root λ Let e 1 E be an eigenvector of A with eigenvalue λ and let V = e 1 K be the vector space spanned by e 1 We have hence A induces a linear operator, A(V ) = A(e 1 )K = e 1 λk V, A: E/V E/V By Corollary 24, we have Pc A (X) = (X λ)pc A (X), hence the characteristic polynomial of A decomposes into a product of factors of degree 1 By induction, there exists a basis e = (e 2 + V,, e n + V ) of E/V such that the matrix e (A) e is upper triangular Proposition 12 shows that the sequence e = (e 1, e 2,, e n ) is a basis of E Moreover, by Proposition 23, ( ) λ e(a) e = 0 e(a) e Since the matrix e (A) e is upper triangular, so is e (A) e )

CHAPTER III Tensor product 1 Tensor product of vector spaces Let E 1,, E n, F be vector spaces over a field K Recall that a map is called n-linear or multilinear if f(x 1,, x i 1, x i α + x i α, x i+1,, x n ) = f : E 1 E n F f(x 1,, x i 1, x i, x i+1,, x n )α + f(x 1,, x i 1, x i, x i+1,, x n )α for each i = 1,, n and for all x 1 E 1,, x i 1 E i 1, x i, x i E i, x i+1 E i+1,, x n E n and α, α K In other words, for each i = 1,, n and for all x 1 E 1,, x i 1 E i 1, x i+1 E i+1,, x n E n fixed, the map f(x 1,, x i 1,, x i+1,, x n ): E i F which carries x E i to f(x 1,, x i 1, x, x i+1,, x n ) is linear For example, all the products for which the distributive law holds are bilinear maps Clearly, if f : E 1 E n F is multilinear and A: F G is linear, then the composite map A f : E 1 E n G is multilinear It turns out that, for given E 1,, E n, there exists a universal linear map : E 1 E n E 1 E n, called the tensor product of E 1,, E n, from which all the multilinear maps originating in E 1 E n can be derived The goal of this section is to define this universal map 11 Universal property Let E 1,, E n be vector spaces over a field K A pair (f, F) consisting of a K-vector space F and a multilinear map f : E 1 E n F is called a universal product of E 1,, E n if the following property (called the universal property) holds: for every K-vector space G and every multilinear map g: E 1 E n G there exists a unique linear map A: F G such that g = A f or, in other words, which makes the following diagram commute: E 1 E n f F g A G The existence of a universal product is far from obvious By contrast, the uniqueness (up to isomorphism) easily follows from the definition by abstract nonsense 11 Proposition Suppose (f, F) and (f, F ) are two universal products of E 1,, E n Then there is one and only one vector space isomorphism ϕ: F F such that the following diagram 9

1 TENSOR PRODUCT OF VECTOR SPACES 10 commutes: E 1 E n f F f ϕ F Proof The existence of a unique linear map ϕ: F F such that f = ϕ f follows from the universal property of (f, F) We have to show that ϕ is bijective To achieve this goal, observe that the universal property of (f, F ) yields a linear map ϕ : F F such that f = ϕ f Now, the following diagrams commute: E 1 E n f F E 1 E n f F f Id F F f ϕ ϕ F In view of the uniqueness condition in the universal property of (f, F), it follows that ϕ ϕ = Id F Similarly, the universal property of (f, F ) shows that ϕ ϕ = Id F Therefore, ϕ and ϕ are reciprocal bijections This proposition shows that a universal product of E 1,, E n, if it exists, can be considered as unique We shall refer to this universal product as the tensor product of E 1,, E n and denote it as (, E 1 E n ) The image of an n-tuple (x 1,, x n ) E 1 E n under is denoted x 1 x n Our next goal is to prove the existence of the tensor product We consider first the case where n = 2 and then derive the general case by associativity 12 The tensor product of two vector spaces 12 Proposition Let µ: K m K n K m n be defined by µ ( ) (x i ) 1 i m, (y j ) = (xi y j ) 1 i m The pair (µ, K m n ) is a universal product of K m and K n Proof Denoting the elements of K m and K n as column vectors, we have µ(x, y) = x y t for x K m and y K n, where denotes the matrix multiplication K m 1 K 1 n K m n Therefore, the bilinearity of µ follows from the properties of the matrix multiplication Let (c i ) 1 i m (resp (c j ) ) be the canonical basis of K m (resp K n ) Let also e ij = µ(c i, c j ) for i = 1,, m and j = 1,, n Thus, e ij K m n is the matrix whose only non-zero entry is a 1 at the i-th row and j-th column The family (e ij ) 1 i m is a basis of K m n Therefore, given any bilinear map B: K m K n F, we may use the construction principle to obtain a linear map ϕ: K m n F such that ϕ(e ij ) = B(c i, c j ) For all x = hence x 1 x m K m and y = µ(x, y) = 1 i m y 1 y n K n, we have x = m c ix i, y = n j=1 c j y j, x i y j µ(c i, c j ) = 1 i m x i y j e ij

1 TENSOR PRODUCT OF VECTOR SPACES 11 and B(x, y) = 1 i m x i y j B(c i, c j ) = 1 i m ( x i y j ϕ(e ij ) = ϕ 1 i m x i y j e ij ) Therefore, B = ϕ µ Uniqueness of the linear map ϕ: K m n F such that B = ϕ µ is clear, since this equation yields B(c i, c j) = ϕ ( µ(c i, c j) ) = ϕ(e ij ), and a linear map is uniquely determined by the image of a basis Thus, (µ, K m n ) is a universal product of K m and K n 13 Theorem Any two vector spaces over a field K have a tensor product Proof We give the proof for finite-dimensional vector spaces only If dime = m and dimf = n, we may identify E to K m and F to K n by the choice of arbitrary bases of E and of F (by mapping every vector in E (resp F) to the m-tuple (resp n-tuple) of its coordinates with respect to the chosen basis) The preceding proposition then shows that (µ, K m n ) is a universal tensor product of E and F Note that, even though bases of E and F were chosen in the proof of the theorem, the tensor product itself does not depend on the choice of bases (but its identification with (µ, K m n ) does) It is a canonical construction To simplify the notation, we suppress the bilinear map in the tensor product notation (, E F), and refer to the tensor product of E and F simply as E F 14 Corollary If e = (e i ) 1 i m is a basis of E and f = (f j ) is a basis of F, then (e i f j ) 1 i m is a basis of E F In particular, dim(e F) = dim E dimf Proof It was observed in the proof of Proposition 12 that ( µ(c i, c j )) 1 i m is a basis of K m n If E is identified with K m and F with K n by means of the bases e and f, then e and f correspond to the canonical bases of K m and K n respectively, hence µ(c i, c j ) = e i f j The corollary shows that every element in E F has the form 1 i m α ij e i f j for some α ij K Grouping terms, we obtain n ( m ) α ij e i f j = α ij e i f j 1 i m j=1 hence every element in E F can be written in the form x 1 y 1 + + x r y r for suitable x 1,, x r E and y 1,, y r F (and r N) Note that this expression is not unique: for instance, (e 1 + e 2 ) f 1 + (e 1 e 2 ) f 2 = e 1 (f 1 + f 2 ) + e 2 (f 1 f 2 ) 13 The tensor product of three or more vector spaces Let E 1, E 2, E 3 be vector spaces over K We may consider the tensor products E 1 E 2 and (E 1 E 2 ) E 3 The map : E 1 E 2 E 3 (E 1 E 2 ) E 3 defined by (x 1, x 2, x 3 ) = (x 1 x 2 ) x 3 is 3-linear 15 Proposition The pair (, (E 1 E 2 ) E 3 ) is a universal product of E1, E 2 and E 3

1 TENSOR PRODUCT OF VECTOR SPACES 12 Proof Let T : E 1 E 2 E 3 F be a 3-linear map For all x 3 E 3, the map T(,, x 3 ): E 1 E 2 F is bilinear, hence there is a unique linear map ϕ x3 : E 1 E 2 F such that ϕ x3 (x 1 x 2 ) = T(x 1, x 2, x 3 ) for all x 1 E 1, x 2 E 2 For x 1 E 1, x 2 E 2, x 3, x 3 E 3 and α, α K, we have hence T(x 1, x 2, αx 3 + α x 3 ) = αt(x 1, x 2, x 3 ) + α T(x 1, x 2, x 3 ), ϕ αx3+α x 3 (x 1 x 2 ) = αϕ x3 (x 1 x 2 ) + α ϕ x 3 (x 1 x 2 ) Since E 1 E 2 has a basis consisting of products x 1 x 2, by Corollary 14, and since a linear map is uniquely determined by the image of a basis, it follows that (1) ϕ αx3+α x 3 = αϕ x 3 + α ϕ x 3 Consider the map Φ: (E 1 E 2 ) E 3 F defined by Φ(ξ, x 3 ) = ϕ x3 (ξ) for ξ E 1 E 2 and x 3 E 3 By (1) and since each map ϕ x3 is linear, the map Φ is bilinear Therefore, the universal property of tensor products yields a linear map such that In particular, ϕ: (E 1 E 2 ) E 3 F ϕ(ξ x 3 ) = Φ(ξ, x 3 ) for all ξ E 1 E 2, x 3 E 3 ϕ ( (x 1 x 2 ) x 3 ) = ϕx3 (x 1 x 2 ) = T(x 1, x 2, x 3 ), hence T = ϕ Uniqueness of the map ϕ satisfying this equation is clear, since every element in (E 1 E 2 ) E 3 can be written as i (x i y i ) z i = i (x i, y i, z i ) for some x i E 1, y i E 2 and z i E 3 If ϕ and ϕ satisfy T = ϕ = ϕ, we must have ( ) ϕ (x i y i ) z i = T(x i, y i, z i ) = ϕ ( (x i y i ) z i ), i i hence ϕ = ϕ i The proposition above proves the existence of the tensor product E 1 E 2 E 3, and shows that there is a canonical isomorphism E 1 E 2 E 3 = (E 1 E 2 ) E 3 Obviously, we also have E 1 E 2 E 3 = E 1 (E 2 E 3 ) By induction, it follows that the tensor product of arbitrarily many vector spaces exists, and may be defined by associativity The following corollary also follows by induction on the number of factors: 16 Corollary Let (e iji ) 1 ji n i be a basis of E i, for i = 1,, r Then is a basis of E 1 E r In particular, (e 1j1 e 2j2 e rjr ) 1 j1 n 1 1 j r n r dime 1 E r = dime 1 dime r

1 TENSOR PRODUCT OF VECTOR SPACES 13 14 Example Let E 1,, E n be finite-dimensional K-vector spaces and let F be an arbitrary K-vector space Denote by M(E 1 E n, F) the set of all multilinear maps f : E 1 E n F It is easily checked that M(E 1 E n, F) is a subspace of the space F(E 1 E n, F) of all mappings E 1 E n F Let E i be the dual space of E i for i = 1,, n For y F and ϕ 1 E 1,, ϕ n E n, define by Clearly, the map θ y,ϕ1,,ϕ n is multilinear, hence Moreover, it is also clear that θ y,ϕ1,,ϕ n : E 1 E n F θ y,ϕ1,,ϕ n (x 1,, x n ) = y ϕ 1, x 1 ϕ n, x n θ y,ϕ1,,ϕ n M(E 1 E n, F) (y, ϕ 1,, ϕ n ) θ y,ϕ1,,ϕ n is a multilinear map F E 1 E n M(E 1 E n, F) We denote it by θ By the universal property of tensor products, θ induces a linear map Θ: F E 1 E n M(E 1 E n, F) 17 Theorem The map Θ is an isomorphism of vector spaces Proof We have to show that Θ is one-to-one and onto To achieve this, we construct a reciprocal map For i = 1,, n, let (e iji ) 1 ji m i be a basis of E i, and let (e ij i ) 1 ji m i be the dual basis For every multilinear map f M(E 1 E n, F), define Ψ(f) = f(e 1j1,, e njn ) e 1j 1 e nj n F E1 E n 1 j 1 m 1 1 j n m n The map Ψ: M(E 1 E n, F) F E1 E n is linear To prove that Ψ and Θ are reciprocal bijections, we shall show 1 that Ψ Θ is the identity on F E1 En and Θ Ψ is the identity on M(E 1 E n, F) Every element in F E1 E n can be written in the form ξ = y j1 j n e 1j 1 e nj n For this ξ, we have 1 j 1 m 1 1 j n m n Θ(ξ) = θ yj1 jn,e 1j 1,,e njn hence Θ(ξ)(e 1k1,, e nkn ) = 1 j 1 m 1 1 j n m n y j1 j n e 1j 1, e 1k1 e nj n, e nkn 1 j 1 m 1 1 j n m n = y k1 k n for all k 1 = 1,, m 1,, k n = 1,, m n 1 The notation makes the following proof hardly readable The reader is warmly encouraged to check the statement by him/herself!

2 FORMALISM 14 Now, by definition of Ψ we have Ψ Θ(ξ) = Θ(ξ)(e 1k1,, e nkn ) e 1k 1 e nk n 1 j 1 m 1 1 j n m n = 1 j 1 m 1 1 j n m n = ξ y k1 k n e 1k 1 e nk n Therefore, Ψ Θ is the identity on F E1 En Now, let f M(E 1 E n, F) We have Θ Ψ(f) = θ f(e1j1,,e njn ),e 1j,,e, 1 njn hence Θ Ψ(f)(e 1k1,, e nkn ) = 1 j 1 m 1 1 j n m n f(e 1j1,, e njn ) e 1j 1, e 1k1 e nj n, e nkn 1 j 1 m 1 1 j n m n = f(e 1k1,, e nkn ) for all k 1 = 1,, n 1,, k n = 1,, m n Since multilinear maps are uniquely determined by their values on basis elements, it follows that Θ Ψ(f) = f, hence Θ Ψ is the identity on M(E 1 E n, F) Note that the isomorphism Θ is canonical, even though its reciprocal isomorphism Ψ is defined above through the choice of bases of E 1,, E n We may therefore use Θ to identify F E 1 E n = M(E 1 E n, F) This provides an alternative definition for the tensor product of vector spaces, at least when all but one of the factors are finite-dimensional 21 Covariant and contravariant behaviour 2 Formalism Let E be a finite-dimensional vector space over a field K, and let (e i ) 1 i n be a basis of E It is sometimes convenient to write the coordinates of vectors x E with upper indices instead of lower indices 2 We thus write n x = e i x i This notation is simplified by the Einstein convention: on every index appearing twice in a formula, the summation is understood: x = e i x i (= x i e i ) For notational coherence, the entries in any change-of-basis matrix receive an upper index and a lower index: if (e α) 1 α n is another basis of E we write (1) e α = e i a i α (instead of e α = n e ia iα ) So, (a i α ) 1 i n are the coordinates of e α with respect to (e i) 1 i n The upper index is the row index and the lower index is the column index If (b α i ) 1 i,α n is the inverse matrix, we have (by definition of the inverse matrix) a i α bβ i = δβ α and a i α bα j = δi j 2 Of course, upper indices should not be confused with exponents There should not be any confusion, since in (multi)linear algebra exponents appear very exceptionally

2 FORMALISM 15 where δj i (resp δβ α) is the Kronecker symbol, ie, δj i = 1 if i = j and δi j = 0 if i j We also have (2) e i = e α bα i Now, consider a vector x = e i x i = e αx α Substituting e i a i α for e α, we obtain x = e i a i αx α, hence (3) Similarly, (4) x i = a i α x α x α = b α i x i It is a crucial observation that, although (3) and (4) look like (1) and (2), these equations are essentially different: multiplying by the matrix a i α transforms e i into e α while it has the reverse effect on coordinates, transforming x α into x i Now, consider the dual bases (e i ) 1 i n and (e α) 1 α n (These are bases of the dual space E ) Equation (2) in section 11 of Chapter I yields hence, taking into account (1) and (2), (5) e α = e i e α, e i and e i = e α e i, e α e α = e i bα i and e i = e α ai α In view of these equations, notational coherence dictates to denote e i for e i and e α for e α ; then (5) can be rewritten as (6) e α = e i b α i and e i = e α a i α These equations are exactly like (4) and (3) To complete the picture, consider a linear form ϕ E Equation (2) of section 11 of Chapter I yields ϕ = e i ϕ, e i = e α ϕ, e α Therefore, the coordinates (ϕ i ) 1 i n (resp (ϕ α) 1 α n ) of ϕ with respect to the basis (e i ) 1 i n (resp (e α ) 1 α n ) of E are given by Using (1) and (2), we get (7) ϕ i = ϕ, e i, ϕ α = ϕ, e α ϕ α = ϕ, e i a i α = ϕ i a i α and ϕ i = ϕ, e α b α i = ϕ αb α i, which should be compared to (3) and (4) Thus, coordinates of linear forms behave under a change of basis exactly like basis vectors: this is the covariant behaviour By contrast, coordinates of vectors and dual basis vectors behave in the opposite way to basis vectors: they display the so-called contravariant behaviour 22 Covariant and contravariant tensors For p and q any natural numbers, let Tp(E) q = } E {{ E } E } {{ E } p q if p, q are not both zero, and T0 0(E) = K The elements in T p q (E) are called p-covariant and q-contravariant tensors on E This is a shameless abuse of terminology, since the elements in Tp(E) q do not vary at all It is motivated by the fact that the coordinates of elements in Tp q (E) have p covariant (ie, lower) indices and q contravariant (ie, upper) indices, as we proceed to show Let (e i ) 1 i n be a basis of E, and let (e i ) 1 i n be the dual basis, which is a basis of E Corollary 16 shows that (e i1 e ip e j1 e jq ) 1 i1,,j q n is a basis of Tp(E), q hence every element in Tp(E) q can be written as (8) t = t j1 jq e i1 e ip e j1 e jq

2 FORMALISM 16 Let (e α) 1 α n be another basis of E and (e α ) 1 α n be its dual basis, and suppose the bases are related by (9) hence e α = e ia i α and e i = e α bα i e α = e i b α i and e i = e α a i α, as in the preceding section Substituting in (8), we get t = t j1 jq (e α 1 a i1 α 1 ) (e α p a ip α p ) (e β 1 b β1 j 1 ) (e β q b βq j q ) = ( t j1 jq a i1 α 1 a ip α p b β1 j 1 b βq j q ) e α 1 e α p e β 1 e β q Therefore, the coordinates t β 1 β q α 1 α p of t T q p(e) with respect to the basis (e α 1 e α p e β 1 e β q ) 1 α1,,β q n are given by (10) t β 1 β q α 1 α p = t j1 jq a i1 α 1 a ip α p b β1 j 1 b βq j q This shows that the lower indices are indeed covariant while the upper indices are contravariant (as they should) Similarly, we have t j1 jq = t β 1 β q α 1 α p b α1 i 1 b αp i p a j1 β 1 a jq β q From a formal point of view, the change of basis formula (10) is the only condition for an indexed array to be the coordinates of a p-covariant and q-contravariant tensor, as the following tensoriality criterion shows 21 Theorem Suppose that to each basis e = (e i ) 1 i n of E is attached an array of scalars ( j u(e) 1 j q ) 1 i 1,,j Knp+q q n There exists a tensor in Tp q (E) whose coordinates with respect to each basis e are u(e) j1 jq if and only if the arrays attached to any two bases e, e related by (9) satisfy u(e ) β1 βq α 1 α p = u(e) j1 jq a i1 α 1 a ip α p b β1 j 1 b βq j q Proof The if part readily follows from the computation leading to equation (10) To prove the only if part, pick an arbitrary basis ẽ of E and define a tensor u Tp q (E) by u = u(ẽ) j1 jq ẽ i1 ẽ ip ẽ j1 ẽ jq By definition, ( u(ẽ) j1 jq )1 i 1,,j q n is the array of coordinates of u with respect to the basis (ẽi1 ẽ ip ẽ j1 ẽ jq ) 1 i1,,j q n We have to show that for any basis ê = (ê α ) 1 α n, the coordinates of u with respect to (ê α1 ê αp ê β1 ê βq ) 1 α1,,β q n are ( u(ê) α β1 βq 1 α p )1 α 1,,β Suppose q n ê α = ẽ i a i α and ẽ i = ê α b α i, hence êα = ẽ i b α i and ẽ i = ê α a i α Then by (10) the coordinates of u with respect to (ê α1 ê αp ê β1 ê βq ) 1 α1,,β q n are ( u(ẽ) j1 jq a i1 α 1 a ip α p b β1 j 1 b βq j q )1 α 1,,β q n By hypothesis, this array is ( u(ê) β1 βq α 1 α p )1 α 1,,β q n 23 Tensors on euclidean spaces In this section, we assume that the vector space E is finite-dimensional and endowed with a scalar product ( ): E E K, ie, a symmetric bilinear form which is non-degenerate in the sense that (x y) = 0 for all y E implies x = 0 Using the scalar product, we define a linear map Φ: E E

2 FORMALISM 17 by mapping every vector x E to the linear form y (x y), so Φ(x), y = (x y) (= (y x)) for x, y E The non-degeneracy condition on the scalar product means that Φ is injective, hence it is bijective since dim E = dime, by Theorem 12 of Chapter I We may thus use Φ to identify E and E, hence also T q p (E) with T p+q(e) or T p+q (E) In the rest of this section, we show how this identification works at the level of coordinates Let (e i ) 1 i n be an arbitrary basis of E For i, j = 1,, n, let g ij = (e i e j ) K The matrix (g ij ) 1 i,j n is the Gram matrix of the scalar product with respect to the basis (e i ) 1 i n Its entries are the coordinates of the 2-covariant tensor g = (e i e j )e i e j E E which corresponds to the scalar product (or to the map Φ) under the identification of Theorem 17: M(E E, K) = E E (= M(E, E )) The tensor g is called the metric tensor on the euclidean space E For i = 1,, n, the basis vector e i E is identified with Φ(e i ) E By equation (2) of section 11 of Chapter I, we have hence Φ(e i ) = e j Φ(e i ), e j = e j (e i e j ) = g ij e j, (11) e i = g ij e j under the identification E = E Consequently, any p-covariant and q-contravariant tensor is identified to t = t j1 jq e i1 e ip e j1 e jq T q p(e) t j1 jq e i1 e ip (g j1k 1 e k1 ) (g jqk q e kq ) = ( t j1 jq g j1k 1 g jqk q ) e i 1 e ip e k1 e kq We may therefore define the (p + q)-covariant coordinates of t as follows: t i1 i pj 1 j q = t k1 kq g k1j 1 g kqj q To define the (p + q)-contravariant coordinates of t, we use the inverse of the Gram matrix (g ij ) 1 i,j n (The Gram matrix is invertible because the scalar product is non-degenerate) Let (g ij ) 1 i,j n be the inverse matrix From equation (11), we derive hence e i = g ij e j, t = t j1 jq (g i1k1 e k1 ) (g ipkp e kp ) e j1 e jq Therefore, the (p + q)-contravariant coordinates of t are = ( t j1 jq g i1k1 g ipkp) e k1 e kp e j1 e jq t i1 ipj1 jq = t j1 jq k 1 k p g k1i1 g kpip

3 TENSOR PRODUCT OF LINEAR MAPS 18 3 Tensor product of linear maps 31 Theorem For i = 1,, n, let A i : E i F i be a linear map There is a linear map such that (1) for all x 1 E 1,, x n E n A 1 A n : E 1 E n F 1 F n (A 1 A n )(x 1 x n ) = A 1 (x 1 ) A n (x n ) Proof This follows from the universal property of tensor products, applied to the multilinear map E 1 E n F 1 F n which carries (x 1,, x n ) E 1 E n to A 1 (x 1 ) A n (x n ) Clearly, the linear map A 1 A n is uniquely determined by condition (1), since every element in E 1 E n is a sum of products x 1 x n 32 Proposition 1 If A i : E i F i and B i : F i G i are linear maps, for i = 1,, n, then (B 1 A 1 ) (B n A n ) = (B 1 B n ) (A 1 A n ) 2 Id E1 Id En = Id E1 E n 3 If A i is injective (resp surjective) for all i = 1,, n, then A 1 A n is injective (resp surjective) Proof and 1 For x 1 E 1,, x n E n, we have (B 1 A 1 ) (B n A n )(x 1 x n ) = B 1 A 1 (x 1 ) B n A n (x n ) (B 1 B n ) (A 1 A n )(x 1 x n ) = (B 1 B n ) ( A 1 (x 1 ) A n (x n ) ) 2 For x 1 E 1,, x n E n we have = B 1 A 1 (x 1 ) B n A n (x n ) (Id E1 Id En )(x 1 x n ) = Id E1 (x 1 ) Id En (x n ) = x 1 x n = Id E1 E n (x 1 x n ) 3 If A i is injective (resp surjective), then there is a linear map B i : F i E i such that B i A i = Id Ei (resp A i B i = Id Fi ), hence (B 1 B n ) (A 1 A n ) = Id E1 E n (resp (A 1 A n ) (B 1 B n ) = Id F1 F n ) by the first two parts of the proposition It follows that A 1 A n is injective (resp surjective)