Vector spaces - University of Pittsburghsparling/14/20141540/20141540... · Matrix representations for linear transformations If S is a nite set put s(t) = 0, if S 3t6=sand s(s) =

Vector spaces

We fix a field F.A vector space, V, over the field F, is a set V, equipped with:

• An additive operation + : V × V → V, denoted (v, w) → v + w, forany v and w ∈ V, which makes V into an abelian group with additiveidentity denoted 0 ∈ V and additive inverse of v ∈ V denoted −v, foreach v ∈ V.

• An operation of scalar multiplication F × V → V, denoted (λ, v) →λv = vλ, for each λ ∈ F and each v ∈ V, such that, for any v and w inV and any λ and µ in F, we have the relations:

1v = v, 0v = 0, (−1)v = −v,

λ(v + w) = λv + λw,

(λ+ µ)v = λv + µv,

λ(µv) = (λµ)v = µ(λ)v.

Examples of vector spaces

• The trivial vector space over a field F is a set with one element, denoted0, with the operations 0 + 0 = 0 and λ0 = 0, for each λ ∈ F.

• The field F is a vector space over itself, with its usual operations.

• Let S be a set.Then the space FS of all maps from S to F has the natural structure ofa vector space, via the formulas, valid for each f : S → F, g : S → Fand for each λ and µ in F:

(λf + µg)(s) = λf(s) + µg(s), for each s ∈ S.

FS is called the free vector space over S.

1

Constructions of vector spaces

Subspaces

Given a vector space V, and a subset X of V, if X is invariant under theoperations of V, we may restrict these operations to X and then X becomesa vector space in its own right, called a subspace of V.

To verify that X ⊂ V qualifies as a subspace, one need only show thatλx+ µy ∈ X, for any x and y in X and any λ and µ in F.In general a linear combination of vectors of X is a vector x of the formx = Σk

i=1λixi, for some xi ∈ X, i = 1, 2, . . . , k, where λi ∈ F, i = 1, 2, . . . , k,for some positive integer k. Then X is a subspace if and only it contains alllinear combinations of vectors of X.

The zero element of any vector space constitutes a subspace, called the trivialsubspace.

The intersection of any family of subspaces of a vector space V is a sub-space. If a set X is a subset of a vector space V, then the intersection of allsubspaces of V containing X is called [X], the subspace spanned by X. Then[[X]] = [X] and X = [X] if and only if X is a subspace. The space [X] consistsof all possible linear combinations of the vectors of X. Every vector space isspanned by itself. The dimension of a non-trivial vector space V is the small-est cardinality of a set that spans it. The trivial vector space is assigneddimension zero. A space V is called finite dimensional if it is spanned by afinite set: it then has dimension zero if and only if it is trivial and otherwiseits dimension is a positive integer.

If X = x1, x2, . . . xn ⊂ V, then [X] is written [x1, x2, . . . , xn].Then [X] has dimension at most n and has dimension n if and only if thevectors xj, j = 1, 2, . . . , n are linearly independent:

Σnj=1λjxj = 0 with λj ∈ F, for each j = 1, 2, . . . , n, if and only if λj = 0, for j = 1, 2, . . . , n.

The vectors are linearly dependent if and only if they are not linearly inde-pendent, if and only if there is a relation 0 = Σn

j=1λjxj = 0, with all λi ∈ Fand at least one λk non-zero, if and only if the dimension of [X] is less thann.

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If a collection of vectors x1, x2, . . . , xn ⊂ V spans a subspace X, then eitherX = V, or one can enlarge this collection to a list x1, x2, . . . , xn, xn+1 ⊂ V,such that X ⊂ [x1, x2, . . . , xn+1] 6= X: the new vector xn+1 ∈ V just has tobe chosen to lie in the complement of X in V.Then if the x1, x2, . . . , xn are linearly independent, so span a space of di-mension n, so are x1, x2, . . . , xn+1 and they span a space of dimension n+1.

The collection of all n-dimensional subspaces of V is called the Grassmannianof all n-planes in V, Gr(n,V).If V has dimension m, then Gr(n,V) is a space of dimension n(m− n).

Each one-dimensional subspace of V is a set [x] of the form [x] = λx : λ ∈ Ffor some 0 6= x ∈ V.We have [x] = [y], for 0 6= y ∈ V if and only if x and y are linearly dependent.The space of all one-dimensional subspaces of V is called P(V) = Gr(1,V),the projective space of V.

Each two-dimensional subspace of V is a set [x, y] of the form [x, y] =λx + µy : λ ∈ F, µ ∈ F for some linearly independent vectors x and yin V.

We have [x, y] = [p, q] if and only if there is an invertible matrix,a bc d

,

with coefficients in F, such that p = ax+ by, q = cx+ dy, if and only if eachx and y are linearly independent, as are p and q, but any three of the fourvectors x, y, p and q are linearly dependent.

A subset X of V is a subspace if and only if [x, y] ⊂ X, for any x and yin X.

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The direct sum of vector spaces

Let S and X be sets equipped with a surjective map π : X → S, such thatXs = π−1(s) ⊂ S has the structure of a vector space over F, for each s ∈ S.

A section of π is a map f : S → X, such that π f = idS. Then thespace Γ of all sections of π becomes a vector space by fibrewise addition andscalar multiplication: for any f and g in Γ, any λ ∈ F and each s ∈ S:

(f + g)(s) = f(s) + g(s), (λf)(s) = λf(s).

We write Γ = ⊕s∈SXs, called the direct sum of the family of vector spacesXs. An element f of Γ is written f = ⊕s∈Sf(s).The support of a section f ∈ Γ is the set of all s ∈ S such that f(s) = 0.The collection of all elements of Γ with finite support is a subspace of Γ,called the restricted direct sum of the family of spaces Xs : s ∈ S.These two notions of direct sum coincide when S is finite.

In particular if Xs = F for each s ∈ S, then Γ = FS.

Homomorphisms

Let V and W be vector spaces.Then a homomorphism or linear map, T from V to W is a set map T fromV to W, respecting the vector space structure:

T (λx+ µy) = λT (x) + µT (y), for any x and y in V and any λ and µ in F.

Then the image T (X) of a subspace X of V under T is a subspace of W, sinceif p and q lie in T (X) and if λ and µ are in F, then x and y in X exist, withT (x) = p and T (y) = q, so then z = λx+µy lies in X, since X is a subspace,so T (z) = λT (x) + µT (y) = λp + µq ∈ T (X), so T (X) satisfies the subspacecriterion. In particular the image of V itself is a subspace.

• T is a monomorphism if and only if T is injective.

• T is a epimorphism if and only if T is surjective.

• T is an isomorphism if and only if T is invertible, if and only if T isboth an epimorphism and a monomorphism.In this case the inverse map T−1 lies in Hom(W,V).

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• T is called an endomorphism in the case that V = W and an automor-phism if T is an invertible endomorphism.The automorphisms of V form a group, under composition, calledGL(V).

The collection of all homomorphisms, T : V → W itself naturally forms avector space denoted Hom(V,W).

If V,W and X are vector spaces, then the composition map Hom(V,W) ×Hom(W,X), (f, g)→ gf , for any f ∈ Hom(V,W) and any g ∈ Hom(W,X),is bilinear (i.e. linear in each argument).

In the particular case that W = F, we put V∗ = Hom(V,F).Then V∗ is called the dual space of V.There is a natural homomorphism J from V → (V∗)∗ given by the formula,valid for any v ∈ V and any α ∈ V∗:

(Jv)(α) = α(v).

If V is finite dimensional, then J is an isomorphism.

A vector space V over F is finite dimensional if and only if any one of thefollowing three conditions is met:

• There is an epimorphism from FP to V, for some finite set P

• There is a monomorphism from V to FQ, for some finite set Q.

• There is an isomorphism from FS to V for some finite set S.

5

Matrix representations for linear transformations

If S is a finite set put δs(t) = 0, if S 3 t 6= s and δs(s) = 1.Then δs ∈ FS and for any f ∈ FS, we have the unique expression f =∑

s∈S f(s)δs, so δS = δs : s ∈ S forms a basis for FS called the canonicalbasis.If T is also a finite set and the canonical basis for FT is εT = εt : t ∈ T,then a homomorphism M from FS to FT is given by the formulas:

M(δs) =∑t∈T

εtMts, for each s ∈ S,

M(f) =∑t∈T

∑s∈S

εtMtsf(s), for any f ∈ FS,

M(f)(t) =∑s∈S

Mtsf(s), for any f ∈ FS and any t ∈ T.

Here the matrix entry Mts is in F, for each s ∈ S and t ∈ T.Next, if U is another finite set and N a homomorphism from FT to FU, wherethe canonical basis of FU is ζU = ζu : u ∈ U, then we have:

N(εt) =∑u∈U

ζuNut,

(N M)(δs) =∑u∈U

ζu(NM)us,

(NM)us =∑t∈T

NutMts.

This formula gives by definition the matrix product of the matrices repre-senting M and N .

If e is an isomorphism of FS and a vector space V and f is an isomorphismof FT and a vector space W, then the set of vectors E = es = e(δs) : s ∈ Sforms a basis for V, whereas the set of vectors F = ft = f(εt) : t ∈ T formsa basis of W.Then the map M : Hom(V,W) → Hom(FS,FT) given by the composition:M(T ) = f−1 T e, for any T ∈ Hom(V,W) is an isomorphism of vectorspaces.

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For T ∈ Hom(V,W) its image M(T ) is called the matrix of T with respectto the the bases e and f .We have the formula:

T (es) =∑t∈T

ft(M(T ))ts, for any s ∈ S and any T ∈ Hom(V,W).

If e′ and f ′ are also isomorphisms of FS and FT with V and W, respectively,then we have:

M ′(T ) = (f ′)−1 T e′ = ((f ′)−1 f) f−1 T e(e−1 e′) = FM(T )E−1.

Here F = (f ′)−1 f is an automorphism of FT and E = (e′)−1 e, whosematrix representatives are invertible.

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Tensor products

Let X be a vector space over a field F. The tensor algebra of X, denotedT (X), is the associative algebra over F, generated by X and F, such that themultiplication operations of the algebra are bilinear.

More explicitly, we may define a word w in X of length k, a nonnegativeinteger, to be an ordered (k + 1)-tuple w = λ,w1, w2 . . . , wk, where λ ∈ Fand each wi, i = 1, 2, . . . , k lies in X. If also x = µ, x1, x2 . . . , xm is a wordof length m, the product z = wx is the word of length k + m, given byz = ν, z1, z2, . . . , zm+k, where ν = λµ ∈ F, zi = wi ∈ X, for 1 ≤ i ≤ k andzi = xi−k ∈ X, for k + 1 ≤ i ≤ k + m. Then we have (xy)z = x(yz), for allwords x, y and z in X. The tensor algebra T (X) is then the span of all thewords, subject to the conditions than the multiplication of words is bilinearin its arguments and to the requirement that if w = λ,w1, w2 . . . , wk is aword and w′ = λ′, w′1, w′2 . . . , w′k is another word of the same length, thenw = w′ if w′i = λiwi, for all i and some λi ∈ F, where λ′Πk

i=1λ′i = λ.

Note that the word w = λ,w1, w2 . . . , wk is then the (ordered) productof the word λ of length zero and the words 1, wi of length one. Thewords of length zero may be identified with the field F and the words oflength one with the vector space X. Also note that any word with one ormore entries zero gives the zero element of the tensor algebra. We abbrevi-ate the words w, x and z = wx as w = λw1w2 . . . wk, x = µx1x2 . . . xm andz = wx = λµw1w2 . . . wkx1x2 . . . xm.

By definition, for each non-negative integer k, T k(X) is the subspace ofT (X) spanned by the words of length k. Then the algebra product mapsT k(X)×T m(X) into T k+m(X), for any non-negative integers k and m (whereT 0(X) = F and T 1(X) = X.

The full tensor algebra of X, is the quotient of T (X ⊕ X∗) by the relationsαv = vα, for any v ∈ T (X) and any α ∈ T (X∗). In this algebra, the subspacespanned by the words αw, with α ∈ T q(X∗) and with w ∈ T p(X) is denotedT pq (X) and we have that the product maps T pq (X)×T rs (X) into T p+rq+s (X), forany non-negative integers p, q, r and s. When X is finite dimensional, T pq (X)and T qp (X∗) are naturally isomorphic, for any nonnegative integers p and q.The words spanning T pq (X) may be written λw1w2 . . . wpα1α2 . . . αq, whereλ ∈ F, w1, w2, . . . , wp lie in X and α1, α2 . . . , αq lie in X∗.

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If the vector space X has dimension n and if ei, i = 1, 2, . . . , n is a set ofvectors of X that spans X, so forms a basis, then any tensor τ of type (p, 0)can be written uniquely:

τ =n∑

i1=1

n∑i2=1

· · ·n∑

ip=1

τ i1i2...ipei1ei2 . . . eip .

If also ej, i = 1, 2, . . . , n is the dual basis of X∗, so we have ej(ei) = δji , forany 1 ≤ i ≤ n and 1 ≤ j ≤ n, then any tensor υ of type (0, q) can be writtenuniquely:

υ =n∑

j1=1

n∑j2=1

· · ·n∑

jq=1

υj1j2...jqej1ej2 . . . ejq .

Also any tensor φ of type (p, q) can be written uniquely:

φ =n∑

i1=1

n∑i2=1

· · ·n∑

ip=1

n∑j1=1

n∑j2=1

· · ·n∑

jq=1

φi1i2...ipj1j2...jq

ei1ei2 . . . eipej1ej2 . . . ejq .

Here each of the coefficients τ i1i2...ip , υj1j2...jq and φi1i2...ipj1j2...jq

lies in F and eachcoefficient may be chosen arbitrarily.In particular, it follows that the vector space of tensors of type (p, q) is iso-morphic to Fr, where r = np+q.

Finally, if As : s ∈ S is a set of subspaces of X labelled by a totallyordered finite set S, then the tensor product ⊗s∈SA is the subspace of thetensor algebra of X spanned by the words formed from the ordered productof the elements of the image of maps x : S→ X, such that x(s) ∈ As, for alls ∈ S.

9

The Kronecker delta and the trace

Let V and W be vector spaces over F. The vector space V∗⊗W acts naturallyon V, such that for any α ∈ V∗, any v ∈ V and w ∈W, we have:

(α⊗ w)(v) = α(v)w.

This gives an element of Hom(V,W).This induces a natural map, easily seen to be an endomorphism, from V∗⊗Wto Hom(V,W). This map is an isomorphism if V is finite dimensional.

When V and W are both finite dimensional, the following spaces are allnaturally isomorphic:

• Hom(V,W)

• V∗ ⊗W

• W⊗ V∗

• Hom(W∗,V∗)

. Also the dual space of this quartet of spaces has four corresponding versions:

• Hom(W,V)

• W∗ ⊗ V

• V⊗W∗

• Hom(V∗,W∗)

In particular, if V = W, we see that Hom(V,V) is its own dual. This entailsthat there is a natural map E : Hom(V,V) × Hom(V,V) → F, which isbilinear in its arguments. Specifically we have the formulas, valid for any αand β in V∗, v and w in V and any T in Hom(V,V):

E(α⊗ v, β ⊗ w) = β(v)α(w),

E(α⊗ v, T ) = α(T (v)).

Then E is a non-degenerate symmetric bilinear form on Hom(V,V): for anyS and T in Hom(V,V), we have E(S, T ) = E(T, S) and if S ∈ Hom(V,V),then E(S,W ) = 0, for all W ∈ Hom(V,V), if and only if S = 0.

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There is a distinguished element of Hom(V,V), the identity operator δV,defined by the formula, valid for any v ∈ V.

δV(v) = v.

Then δV is called the Kronecker delta tensor for V.Then δV gives a natural element of Hom(V,V), called trV, the trace, givenby the formula, valid for any T ∈ Hom(V,V):

trV(T ) = E(δV, T ).

In particular we have, for any α ∈ V∗ and any v ∈ V the formula:

trV(α⊗ v) = α(v).

Also we have, for any S and T in Hom(V,V):

trV(S T ) = trV(S T ).

Finally we have a key formula:

trV(δV) = n.

Here n is the dimension of V, but here regarded as an element of F, not as apositive integer.

If X is a vector space with dual X∗, the full tensor algebra of X is the ten-sor algebra of X ⊕ X∗. It is usual to quotient this algebra by the relationsαw = wα, for any word w in X and any word α in X∗. This gives the algebraof mixed tensors of X. If a word of this algebra is the concatenation of pletters form X and q from X∗, the word is said to be of type (p, q). Thewords of type (p, 0) are said to be contravariant, the words of type (0, q) co-variant. Then T pq (X) is the subspace of the tensor algebra spanned by words

of type (p, q). Under the tensor product, we have T pq (X)⊗T rs (X) ⊂ T p+rq+s (X),for any nonnegative integers p, q, r and s (Here T 0

0 (X) = F). Note thatT pq (X) = T p0 (X) ⊗ T 0

q (X) = T p(X) ⊗ T q(X∗) and T p0 (X) = T p(X), whereasT 0q (X) = T q(X∗), for all non-negative integers p and q.

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For each positive integer s, denote by Ns the (totally ordered) set Ns =t ∈ N : 1 ≤ t ≤ s. Now let τ be a word in T pq (X), so τ has the formτ = x⊗ α, where x = ⊗k∈Npxk and α = ⊗j∈Nqαj and each xk, k = 1, 2, . . . , pis in X, whereas each αj, j = 1, 2, . . . , q is in X∗. For some integer r with1 ≤ r ≤ min(p, q), let there be given a bijection µ from a subset Ar of Np

with r elements to a subset Br of Nq, also with r elements. Then definetrµ(τ) ∈ T p−rq−r (X) by the formula:

trµ(τ) = λx′ ⊗ α′,

x′ = ⊗k∈Np−Arxk ∈ Tp−r0 (X),

α′ = ⊗j∈Nq−Brαj ∈ T 0q−r(X),

λ = Πri=1αµ(r)(xr) ∈ F.

Note that the maps µ and µ−1 are not, in general, order preserving. Then trµextends naturally to a linear map, still called trµ, from T pq (X) to T p−rq−r (X): itis called the µ trace applied to T pq (X). In the special case that p = q = r = 1,the map µ is the identity and trµ equals the ordinary trace trX, defined above.

Tensors in the one-dimensional case

Let L be a one-dimensional space over a field F. Then the tensor algebraof L is abelian. For any positive integer r, denote by Lr the r-fold tensorproduct of L with itself, with, in particular, L1 = L. Then the dual spaceof L may be denoted L−1 and for any positive integer s, its s-fold tensorproduct with itself may be denoted L−s. Finally we put L0 = F. Then thereis a natural isomorphism of Lr ⊗ Ls with Lr+s, for any integers r and s. Ife is a basis for L, the induced basis for Lr is denoted er, for any integerr. In particular e0 = 1 and e−1 is dual to e, e−1e = 1 and e−1 is a basis forL−1. Then a general tensor T can be written uniquely in the form:

T = Σr∈Ztrer.

Here the map Z → F, r → tr is required to be non-zero for only a finitenumber of integers r.The (associative, commutative) tensor multiplication of T with the tensorU = Σr∈Zuse

s is the tensor:

TU = UT = Σn∈Nvnen, vn = Σr+s=ntrus.

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The rank theorem for homomorphisms

Let V and W be finite dimensional vector spaces over F, with dimensions mand n, respectively and let T : V→W be a linear transformation.Associated to T is its image, T (V), which we have proved above to be asubspace of W.Also associated to T is its kernel ker(T ) = v ∈ V : T (v) = 0.If v and w are in ker(T ), so T (v) = T (w) = 0, then for any λ and µ in F, wehave:

T (λv + µw) = λT (v) + µT (w) = 0.

So λv + µw ∈ ker(T ).So ker(T ) is a subspace of V. T is a monomorphism if and only if ker(T ) = 0.Let E2 = eα, α ∈ S form a basis of ker(T ) (where E2 is empty if ker(T ) =0.This has at most m elements, since any m + 1 vectors of V are linearly de-pendent.Since any n+ 1 vectors of W are linearly dependent, the same is true of anysubspace of W, so T (V) has finite dimension k, for some non-negative integerk ≤ n.

Let f1, f2, . . . , fk ∈ T (V) be a minimal spanning set for T (V) (regarded asthe empty set if T = 0).Then by repeatedly adding new vectors fk+1, fk+2, . . . , whilst maintaininglinear independence, we can extend to a list F = f1, f2, . . . , fn, yielding abasis for W.Let ei ∈ V obey the relations T (ei) = fi, for each i = 1, 2, . . . k.Then if λi ∈ F obey the relation

∑ki=1 λiei = 0, applying T to this equation,

we get:

0 = T (0) = T

(k∑i=1

λiei

)=

k∑i=1

λiT (ei) =k∑i=1

λifi.

But the fi; i ∈ Nk are linearly independent in W, so all the λi must vanish,for i ∈ Nk.So the set E1 = ei; i ∈ Nk is a linearly independent set of vectors in V.

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Now consider the union E of the sets E1 and E2.If∑k

i=1 λiei +∑

α∈S µαeα = 0, for some λi : i ∈ Nk and µα : α ∈ S, thenapplying T , we get the formula:

0 = T (0) = T

(k∑i=1

λiei +∑α∈S

µαeα

)

=k∑i=1

λiT (ei) +∑α∈S

µαT (eα)

=k∑i=1

λifi +∑α∈S

µα(0) =k∑i=1

λifi.

But the fi : i ∈ Nk is linearly independent.So the λi, i = 1, 2, . . . k are all zero.Back substituting, we get: 0 =

∑α∈S µαeα.

But the eα : α ∈ S are linearly independent.So all the µα vanish, for each α ∈ S.So the set E is a linearly independent set.Next let v ∈ V.Then T (v) ∈ T (V), so there is an expression:

T (v) =k∑i=1

vifi, for some vi ∈ F, i = 1, 2, . . . , k.

Put w = v −∑k

i=1 viei ∈ V.Then we have:

Tw = T

(v −

k∑i=1

viei

)= Tv −

k∑i=1

viT (ei)

= Tv −k∑i=1

vifi = 0.

So w ∈ ker(T ), so we have an expression w =∑

α∈Swαeα, for some wα ∈F : α ∈ S. Then we have the formula:

v = w +k∑i=1

viei =∑α∈S

wαeα +k∑i=1

viei.

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So the set E is linearly independent and spans V, so is a basis for V.In particular the cardinality of S is m− k and we have proved the formula:

ν(T ) + ρ(T ) = dim(V).

Here ρ(T ), called the rank of T is the dimension of the image of T and ν(T ),called the nullity of T is the dimension of the kernel of T .

Now by relabeling the elements of E2, we may write E = ei : i ∈ Nm.Now we have the Rank Theorem in two versions:

• With respect to the bases E of V and F of W, the matrix M of T is:

T (ei) =n∑j=1

fjMji, for any i ∈ Nm.

Here we have T (ei) = 0, for i > k and T (ei) = fi, for 1 ≤ i ≤ k, so thematrix Mji has the block form:

Ik 00 0

.

Here Ik is the k × k identity matrix.

• More geometrically, we can decompose V = V1⊕V2 and W = W1⊕W2,where T gives an isomorphism of V1 with W1 = T (V), both spaces ofdimension k and T is zero on V2 = ker(T ).

A corollary is that given any n×m-matrix M , there is an invertible n× n-

matrix E and an invertible m×m matrix F , such that EMF−1 =Ik 00 0

,

for some non-negative integer k ≤ min(m,n) and the number k, called therank of the matrix M , is independent of the choice of the matrices E and F .

15

Monomorphisms, epimorphisms and morphisms

We first describe monomorphisms and epimorphisms, in term of each other.

• If V and W are vector spaces, a map L : V→W is a monomorphism ifand only if there exists a vector space X and a linear map M : W→ X,such that ML is an isomorphism.

• If W and X are vector spaces, a linear map M : W → X is an epi-morphism if and only if there exists a vector space V and a linear mapL : V→W, such that ML is an isomorphism.

Given vector spaces V,W and X, L ∈ Hom(V,W) and M ∈ Hom(W,X),we say that the pair (L,M) forms a dual pair if and only if ML is an iso-morphism. If (L,M) is a dual pair, then L is necessarily a monomorphismand M is necessarily an epimorphism. So given L ∈ Hom(V,W), L is amonomorphism if and only if X and M ∈ Hom(W,X) exist such that (L,M)is a dual pair. Conversely, given M ∈ Hom(W,X), M is an epimorphism ifand only if V and L ∈ Hom(V,W) exist such that (L,M) is a dual pair.

Note that if L is a monomorphism from V to W, then there is a dual pair(L,M) with M : W→ X and ML = idV. Also if M is an epimorphism fromW to V, then there is a dual pair (L,M) with L : V→W and ML = idV.

Now let T : X → Z be any homomorphism from the vector space X tothe vector space Z. Then there exists a factorization T = LM , with M anepimorphism from X to a vector space Y and L a monomorphism from Y toZ. Also there exist a monomorphism S from Y to X and an epimorphism Ufrom Z to Y, such that UTS is an isomorphism.

If T = LM = L′M ′ with M : X → Y, M ′ : X → Y′, L : Y → Z andL′ : Y′ → Z and with L and L′ monomorphisms, whereas M and M ′ areepimorphisms, then there is an isomorphism N : Y→ Y′, such that L′N = Land M ′ = NM . In particular the dimension of Y is independent of allchoices. It is called the rank of T and is bounded by the min(dim(X), dim(Z)),since if dim(Y) > dim(Z), there is no monomorphism from Y to Z and ifdim(Y) > dim(X), there is no epimorphism from X to Y. The rank of T isequal to dim(X) if and only if T is a monomorphism and the rank of T isequal to dim(Z) if and only if T is an epimorphism.

16

Pictures

We represent vectors and tensors by pictures:

This represents a vector of a vector space V.The line represents that the vector space in question is V.For another vector in the same space, we can use a different shape:

Addition and scalar multiplication can be represented naturally:

+ 32

Depending on the context it might be more appropriate to bring the linestogether, to emphasize that they represent only a single vector:

+ 32

Also, depending on the context it might be more appropriate to tie the scalarsto the vector that they act on, in which case, if the lines are brought together,the plus sign becomes superfluous:

32

17

Alternatively, we can tie the scalar to the vector:

2

3

If we have a vector of another vector space W, we use a different colour:

If we have an element of the dual vector space V∗, it is represented by adownward oriented line:

Objects without lines emerging from them represent scalars, or elements ofthe field. So for example the dual pairing of a co-vector and a vector isrepresented by joining the two:

Tensors based on V and V∗ are represented by multiple lines:

• A tensor of type (3, 0):

• A tensor of type (0, 3):

18

• A tensor T of mixed type (3, 3):

T

Tensors act on bunches of vectors and dual vectors, as appropriate to getscalars. So, for example, when the (3, 3) tensor T acts on three vectors v, wand x in V and three dual vectors α, β and γ in V∗, we get the scalar:

βα γ

wv x

T

This scalar, usually written T (v, w, x, α, β, γ) is linear in all seven arguments.

Sometimes we need to permute the arguments of tensors, so for exampleif we permute the first and third vector arguments and cyclically permutethe dual vector arguments of T , we can create the following tensor U :

U = T

In general the tensor U is not the same tensor as T , since its action on thesame vector and dual vector arguments is different.

The (associative) tensor product is just denoted by juxtaposition:

• A tensor product T ⊗ U :

T U

19

Here the tensor product of the (3, 0) tensor T and the (2, 0) tensor Uis the (5, 0) tensor T ⊗ U .

• A mixed tensor product T ⊗W :

T W

This tensor product produces a tensor T ⊗W of mixed type (3, 2).

• It is usual when considering mixed tensors, to assume that tensorsbased on V commute with those based on V∗, so we can also writeT ⊗W , without ambiguity as, for example:

T

W

• The contraction of tensors is represented by joining up the lines.So, for example, if we contract the third argument of T with the firstargument ofW and the second argument of T with the second argumentof W , we get a dual vector:

W

T

Notice in particular that the action of the (3, 3)-tensor T on the vectorsv, w and x and the dual vectors α, β and γ given above is a completecontraction (meaning that the result is a scalar) of the tensor productT ⊗ α⊗ β ⊗ γ ⊗ v ⊗ w ⊗ x.

20

An endomorphism L of a vector space is represented as a (1, 1) tensor:

L

This acts in many different ways, every one linear in each argument, forexample:

• Mapping a vector v to a vector L(v):

v→ L

v

• Mapping a dual vector α to a dual vector α(L):

α→ L

α

• Mapping a vector v and a dual vector α to a scalar α(L(v)):

α ,( )

v → L

α

v

21

• Mapping an endomorphism M to to a scalar tr(LM):

M →L

M

The Kronecker delta tensor, the identity endomorphism, is just representedby a line, so we have the trace of L as:

→L

The trace of the Kronecker delta is n the dimension of the vector space,regarded as an element of the field of scalars F, not as an integer:

n =

22

The elements of the direct sum of two (or more) vector spaces are representedby the merging of their colors to form a new color:

Here the blue/green and the yellow/green endomorphism are injections ofthe blue and yellow vector spaces into the green one, respectively.

An (L,M) pair, with L : V → W an injection and M : W → V a sur-jection, such that ML = idV is written as follows:

, ,L = M =

23

= .

A V basis for a vector space W is represented by a pair of isomorphisms, thebasis and the dual basis:

, .

24

These are required to compose both ways to give the Kronecker delta:

= , = .

25

If the vector space V is Fn, the vectors of V and V∗ are ordered lists of nelements of F:

8−5

4−3

6

1, −4, 2, 7,−2 =

−5,

8,

4,

−3,

6,

20,

−32,

−16,

12,

−24,

−10,

16,

8,

−6,

12,

−35,

56,

28,

−21,

42,

10

−16

−8

6

−12

.

Here the 5 by 5 matrix is called the Kronecker product of the constituentvectors. The dual pairing (also the trace of the matrix) is then:

8−5

4−3

6

1, −4, 2, 7,−2

= 1(−5) + (−4)(8) + 2(4) + 7(−3) + (−2)(6) = −62.

26

The sign of a permutation

For n ∈ N, the permutation group Sn is the group of all bijections from Nn

to itself, with the group multiplication given by composition.Then Sn has n! elements.Note that S1 has just one element, the identity element.

For n ≥ 2, put D(x) = Πi<j(xi − xj), for (commutative) variables x =(x1, x2, . . . xn).For s ∈ Sn, put s(D)(x) = Πi<j(xs(i) − xs(j)).Since s is a bijection, the factors xs(i) − xs(j) are all distinct and not identi-cally zero and each one is of the form εk,m(xk − xm) for some k < m, with kand m in Nn and εk,m = ±1.

There aren(n− 1)

2such factors, the same as in the polynomial D(x), so by

the pigeonhole principle, every factor of D(x) occurs exactly once in s(D)(x),so we arrive at the formula:

s(D)(x)) = ε(s)D(x), ε(s) = ±1.

If also t ∈ Sn and we put yi = xt(i), for 1 ≤ i ≤ n, then we get:

s(D)(y) = ε(s)D(y) = ε(s)t(D)(x) = ε(t)ε(s)D(x)

= Πi<j(ys(i) − ys(j)) = Πi<j(xt(s(i)) − xt(sj)))

= Πi<j(x(ts)(i)) − xts)(j))) = (t s)(D)(x) = ε(t s)D(x).

So the polynomial: (ε(t)ε(s) − ε(t s))D(x) is identically zero, so we inferthe relation, valid for any permutations s and t in Sn:

ε(t)ε(s) = ε(t s).

This shows that the map ε : Sn → Z2, s → ε(s) is a group homomorphism,where Z2 = 1,−1 is the multiplicative group with two elements, 1, theidentity and −1, with (−1)2 = 1.Then ε(s) is called the sign of the permutation s.The permutation s is called even if ε(s) = 1 and odd if ε(s) = −1.

27

Next define the auxiliary polynomials, for any i and j with 1 ≤ i 6= j ≤ n:

Eij(x) = (−1)i+j+1Π1≤k≤n,k 6=i,k 6=j((xi − xk)(xj − xk)),

Fij(x) = Π1≤k<l≤n,k 6=i,l 6=i,k 6=j,l 6=j(xk − xl).

Note that, by inspection, Eij(x) and Fij(x) are symmetric under the inter-change of i and j. Also Fij(x) is independent of the variables xi and xj.Then we have the formula, highlighting the dependence of the polynomialD(x) on the variables xi and xj, with i < j:

D(x) = (xi − xj)Eij(x)Fij(x).

If now s ∈ Sn satisfies s(i) = j, s(j) = i and s(k) = k, for all k with1 ≤ k ≤ n, k 6= i and k 6= j, then we have:

s(D)(x) = (xs(i) − xs(j))Hij(x)Mij(x),

Hij(x) = (−1)i+j+1Π1≤k≤n,k 6=i,k 6=j((xs(i) − xs(k))(xs(j) − xs(k)))

= (−1)i+j+1Π1≤k≤n,k 6=i,k 6=j((xj − xk)(xi − xk)) = Eij(x),

Mij(x) = Π1≤k<l≤n,k 6=i,l 6=i,k 6=j,l 6=j(xs(k)−xs(l)) = Π1≤k<l≤n,k 6=i,l 6=i,k 6=j,l 6=j(xk−xl) = Fij(x).

So we get:

s(D)(x) = (xs(i) − xs(j))Hij(x)Mij(x) = (xj − xi)Eij(x)Fij(x) = −D(x).

This proves that ε(s) = −1.The mapping s is called the (i, j) transposition and is denoted (ij). So themap ε : Sn → Z2 is surjective, when n ≥ 2 (and not when n = 1, since then S1

consists only of the identity element, which is defined to be even ε(1) = 1).For n ≥ 2, the kernel of ε is called the alternating group An, the normalsubgroup of Sn of all even permutations, a subgroup of index two in Sn, sowith 2−1n! elements. Every element of the symmetric group is a productof transpositions, so the sign of the element is therefore positive or negativeaccording to the evenness or oddness of the number of transpositions in theproduct, so the parity of the number of such transpositions is an invariant.

28

Acting on tensors, we can use permutations to rearrange the arguments ofthe tensor. This is achieved by acting on the tensor with suitable productsof the Kronecker delta tensor. For example in the case of S3, its six elementsare represented as:

(1) = , (231) = , (312) = ,

(23) = , (13) = , (12) = .

Here (231) denotes the cyclic permutation 1→ 2→ 3→ 1 and (312) denotesthe cyclic permutation 1→ 3→ 2→ 1.

In general, given a word w = fw1w2 . . . wp in T p(X), for some fixed non-negative integer p and given a permutation σ ∈ Sp, we define:

σ(w) = fwσ(1)wσ(2) . . . wσ(p).

If now ρ is another permutation in Sp, we have:

ρ(σ(w)) = fwρ(σ(1))wρ(σ(2)) . . . wρ(σ(p))

= fw(ρσ)(1))w(ρσ)(2)) . . . wρσ)(p))

= (ρ σ)(w).

Then the assignment w → σ(w) for each word extends naturally to givean endomorphism of T p(X), still denoted σ, such that we have ρ(σ(t)) =(ρ σ)(t), for any permutations ρ and σ in Sp and any t ∈ T p(X). So foreach p, we get a linear representation of the group Sp on the np dimensionalvector space T p(X).

29

The symmetric tensor algebra

The symmetric tensor algebra S(X) of a vector space X is the algebra ob-tained from the tensor algebra T (X), by requiring that all the letters of allwords commute, so the multiplication in this algebra is commutative andindeed is isomorphic to a standard commutative algebra of polynomials inn-variables over the field F, when X has dimension n. So there is a surjec-tive algebra homomorphism from T (X) to S(X), which maps each word toitself. For clarity the symmetric tensor product is often represented by thenotation . The images of the np basis words ei1ei2 . . . eip for T p(X), whereei, i = 1, 2, . . . , n is a basis for X, gives the following basis words for the imageSp(X) of T p(X) in S(X): ei1 ei2 · · · eip , for 1 ≤ i1 ≤ i2 · · · ≤ ip ≤ n.

This basis has

(n+ p− 1

p

)elements.

There is a natural subspace of T (X) that maps isomorphically to the sym-metric algebra: it is the subspace spanned by all words of the form fwp, forw ∈ X and f ∈ F. Explicitly there is an F-linear idempotent symmetrizationmap S : T (X) → T (X), which preserves tensor type and which for a givenpositive integer p, we describe in terms of the action of the symmetric groupSp on T p(X) given above: For each tensor t ∈ T p(X), we define:

S(t) =1

p!

∑σ∈Sp

σ(t).

This gives an endomorphism of T p(X). We prove that it is idempotent:

S2(t) = S

1

p!

∑σ∈Sp

σ(t)

=1

p!

∑ρ∈Sp

∑σ∈Sp

ρ(σ(t)) =1

p!

∑ρ∈Sp

∑σ∈Sp

(ρ σ)(t)

=1

p!

∑ρ∈Sp

∑σ∈Sp

(ρ (ρ−1 σ))(t) =1

p!

∑ρ∈Sp

∑σ∈Sp

σ(t) =∑σ∈Sp

σ(t) = S(t).

Here we used the fact that as σ runs once through all permutations in Sp,so does ρ−1 σ, for each fixed permutation ρ in Sp. The image of S is thencalled the subspace of symmetric tensors. On this space we can define the(associative) symmetric tensor product by the formula A B = S(AB), forany symmetric tensors A and B. This product makes the space of symmetrictensors into an algebra naturally isomorphic to the algebra S(X).

30

For example, given a tensor T in T 3(X), we can represent its symmetrizationS(T ) diagrammatically as follows:

6S(T )=

T+

T+

T

+

T+

T+

T.

Just like any algebra of polynomials, the algebra S(X) carries a naturalcollection of derivations (or first order derivative operators). A derivation Dof S(X) is an endomorphism of S(X), obeying the rules:

• Dc = 0, for any c ∈ F = S0(X),

• D(f + g) = Df +Dg, for any f and g in S(X),

• D(fg) = (Df)g + (Dg)f , for any f and g in S(X).

Then D is determined by its action on S1(X) = X, which is an arbitrarylinear map from X to S(X), so D is determined by an element of S(X)⊗X∗.If D and E are derivations, so is their bracket [D,E] = DE − ED and thisbracket obeys all the usual rules of a Lie bracket (see the section on the Liederivative along a vector field). Indeed we have a natural linear derivativeoperator denoted ∂, which takes S(X) to X⊗S(X) and is determined uniquelyby the derivative formula, valid for any t ∈ X and any positive integer p:

∂ ⊗ tp

p!= t⊗ tp−1

(p− 1)!.

Alternatively the operator ∂ is determined by the formal power series formula(obtained by summing the previous formula over p):

∂ ⊗ et = t⊗ et.

31

Equivalently, we have, acting on the word w = fw1w2 . . . wp, with f ∈ F andeach wi ∈ X, the derivative formula:

∂⊗(fw1w2w3 . . . wp) = fw1⊗w2w3 . . . wp+fw2⊗w1w3 . . . wp+fw3⊗w1w2 . . . wp+· · ·+fwp⊗w1w2w3 . . . wp−1.

Then the general derivation is given by D.∂, where D lies in S(X) ⊗ X∗.Finally, we can compute multiple derivatives, which (since they commutewith each other) give natural maps from S(X) to S(X)⊗S(X); so for example,we have (when p ≥ k) and for any t ∈ X:

∂k ⊗ tp

p!= tk ⊗ tp−k

(p− k)!,

∂k ⊗ et = tk ⊗ et,e∂ ⊗ et = et ⊗ et.

The skew-symmetric tensor algebra: the Grassmann orexterior algebra

The skew-symmetric tensor algebra Ω(X) of a vector space X, also called itsGrassmann algebra, or its exterior algebra, is the algebra obtained from thetensor algebra, T (X), by requiring that all the individual letters of all wordsanti-commute, so the multiplication in this algebra is graded commutative:an element is even if it is a sum of words each with an even number of let-ters and is odd if it is a sum of words each with an odd number of letters.Then even elements commute with everything, whereas odd elements com-mute with even and anti-commute with each other. In particular, the squareof any odd element is zero. There is then a surjective algebra homomor-phism from T (X) to Ω(X), which maps each word to itself. For clarity, theskew-symmetric product is often represented by the notation ∧. The imageof the np basis words ei1ei2 . . . eip for T p(X), where ei, i = 1, 2, . . . , n is abasis for X, gives the following basis words for the image Ωp(X) of T p(X) in

Ω(X): ei1 ∧ ei2 ∧ · · · ∧ eip , for 1 ≤ i1 < i2 · · · < ip ≤ n. This basis has

(n

p

)elements. In particular the Grassmann algebra of X is finite dimensional, of

total dimensionn∑p=0

(n

p

)= 2n.

32

There is a natural subspace of T (X) that maps isomorphically to the Grass-mann algebra. Explicitly there is an F-linear idempotent skew-symmetrizationmap Ω : T (X) → T (X), which preserves tensor type and which for a givenpositive integer p, we describe in terms of the action of the symmetric groupSp on T p(X) given above. For each tensor t ∈ T p(X), we define:

Ω(t) =1

p!

∑σ∈Sp

ε(σ)σ(t).

Here ε(σ) is the sign of the permutation σ. This gives an endomorphismof T p(X). We prove that the endomorphism Ω idempotent: for any fixedpositive integer p and any t ∈ T p(X), we have:

Ω2(t) = Ω

1

p!

∑σ∈Sp

ε(σ)σ(t)

=1

p!

∑ρ∈Sp

ε(ρ)∑σ∈Sp

ε(σ)ρ(σ(t))

=1

p!

∑ρ∈Sp

∑σ∈Sp

ε(ρ σ)(ρ σ)(t) =1

p!

∑ρ∈Sp

∑σ∈Sp

ε(ρ (ρ−1 σ))(ρ (ρ−1 σ))(t)

=1

p!

∑ρ∈Sp

∑σ∈Sp

ε(σ)σ(t) =∑σ∈Sp

σ(t) = Ω(t).

Here we used the fact ε is a group homomorphism and that as σ runs oncethrough all permutations in Sp, so does ρ−1 σ, for each fixed permutationρ in Sp. The image of Ω is then called the subspace of skew-symmetric ten-sors. On this space we can define the skew-symmetric tensor product by theformula A ∧ B = Ω(AB), for any skew-symmetric tensors A and B. Thisproduct makes the space of skew-symmetric tensors into an algebra naturallyisomorphic to the algebra Ω(X).

33

For example, given a tensor T in T 3(X), we can represent its skew-symmetrizationΩ(T ) diagrammatically as follows:

6Ω(T )=

T+

T+

T

−T

−T

−T

.

Just like the symmetric algebra, the Grassmann algebra Ω(X) carries a natu-ral derivative operator. A derivation D of Ω(X) is an endomorphism of Ω(X),obeying the rules:

• Dc = 0, for any c ∈ F = Ω0(X),

• D(f + g) = Df +Dg, for any f and g in Ω(X),

• D(fg) = (Df)g + (−1)pq(Dg)f , for any f ∈ Ωp(X) and g ∈ Ωq(X) andany non-negative integers p and q.

Then D is determined by its action on Ω1(X) = X, which is an arbitrarylinear map from X to Ω(X), so D is determined by an element of Ω⊗X∗. Aderivation D is said to have integral degree k if it maps Ωp(X) to Ωp+k(X),for each non-negative integer p. If D is a derivation of degree p and E aderivation of degree q, so is their bracket [D,E] = DE − (−1)pqED andthis bracket obeys all the usual rules of a graded Lie bracket (see the sectionon differential forms). Indeed we have a natural linear derivation of degreeminus one, denoted δ, which takes Ω(X) to X ⊗ Ω(X) and is determineduniquely by the derivative formula acting on any word w = fw1w2 . . . wk,with f ∈ F and each wi ∈ X, the derivative formula:

δ⊗(fw1w2w3 . . . wk) = fw1⊗w2w3 . . . wk−fw2⊗w1w3 . . . wk+fw3⊗w1w2 . . . wk+· · ·+(−1)k−1fwk⊗w1w2w3 . . . wk−1.

Then the general derivation is given by D.δ, where D lies in Ω(X) ⊗ X∗.Finally, we can compute multiple derivatives, which (since they anti-commutewith each other) give natural maps from Ω(X) to Ω(X)⊗Ω(X). In particularif X has dimension n all derivatives of order n+ 1 or more vanish identically.

34

Determinants

Consider a vector space X with dual space X∗, defined over a field F, withdimension n, a positive integer. Then X acts on Ω(X∗) the exterior algebraof X∗ by the derivation δv, defined for each v ∈ X. The operator δv isthe derivation of degree minus one, which kills Ω0(X∗) = F and acting onΩ1(X∗) = X∗ obeys the relation, valid for any v ∈ X and α ∈ X∗:

δv(α) = α(v).

Consider now the vector space L = Ωn(X∗), a vector space over F of dimen-sion 1, so isomorphic to F. Given n-vectors, v1, v2, . . . , vn in X and ε ∈ L,the following quantity is a well-defined element of F:

D(v1, v2, . . . , vn)(ε) = δvnδvn−1 . . . δv2δv1ε.

This quantity is linear in all (n+1)-variables (v1, v2, . . . , vn, ε). In particular,the quantity D(v1, v2, . . . , vn) gives an element of L−1 the dual space of L(so L−1 is also a vector space of dimension one). Then, from its definition,D(v1, v2, . . . , vn) is linear in each argument and totally skew, since the deriva-tions δv, for v ∈ X pairwise anti-commute and each is linear in v.

Let e = e1 be a basis of L (so e is any non-zero element of L), with dualbasis e−1 of L−1, so we have e−1(e) = 1. Then there is a basis e1, e2, . . . , enfor X, with dual basis e1, e2, . . . , en, a basis of X∗, such that e = e1e2 . . . en.Then we have:

D(e1, e2, . . . , en)(e) = δenδen−1 . . . δe2δe1(e1e2 . . . en) = 1 = e−1(e).

So we have the formula D(e1, e2, . . . en) = e−1.This, in turn, yields the general formula:

D(v1, v2, . . . , vn) = dete(v1, v2, . . . , vn)e−1 = dete(v1, v2, . . . , vn)D(e1, e2, . . . en).

Here the quantity dete(v1, v2, . . . , vn) is an element of F, and is linear ineach vector argument, v1, v2, . . . , vn and totally skew. Then dete is called thedeterminant function, relative to the basis e of L. Under the replacemente→ f , where f is also a basis for L we have detf = r dete, where f = re.The argument above shows that when the vectors vi, i = 1, 2, . . . , n arelinearly independent, so constitute a basis, then D(v1, v2, . . . , vn) is non-zero.Conversely, it is easy to see that if the vectors vi, i = 1, 2, . . . , n are linearlydependent, then D(v1, v2, . . . , vn) is zero. So we have that dete(v1, v2, . . . , vn)vanishes if and only if the vectors vi, i = 1, 2, . . . , n are linearly dependent.

35

Consider now n-elements α1, α2, . . . , αn of the space X∗. Then we have:

α1α2 . . . αn = E(α1, α2, . . . , αn) ∈ L.

Here the quantity E(α1, α2, . . . , αn) is linear in each argument and totallyskew. The arguments given above, show that E(α1, α2, . . . , αn) vanishes ifand only if the αi, i = 1, 2, . . . , n are linearly dependent. Further we maywrite E(α1, α2, . . . , αn) = dete−1(α1, α2, . . . , αn), where dete−1(α1, α2, . . . , αn) ∈F is linear in each argument and totally skew.

Now consider the element of F:

D(v1, v2, . . . , vn)E(α1, α2, . . . , αn) = dete(v1, v2, . . . , vn)dete−1(α1, α2, . . . , αn)

= δvnδvn−1 . . . δv2δv1(α1α2 . . . αn) = Σσ∈Snε(σ)(Πni=1(ασ(i))(vi)

).

Note that this relation is symmetrical between the vector space X and itsdual X∗:

D(v1, v2, . . . , vn)E(α1, α2, . . . , αn) = δvnδvn−1 . . . δv2δv1(α1α2 . . . αn) = δαnδαn−1 . . . δα2δα1(v1v2 . . . vn)

= Σσ∈Snε(σ)(Πni=1vi(ασ(i))

)= Στ∈Snε(τ)

(Πni=1αj(vτ(j))

)= det(αi(vj)).

Also this relation embodies the fact that the one-dimensional vector spacesΩn(X) and Ωn(X∗) are naturally dual.

36

Volume

In a Euclidean space, of finite dimension, denote the inner product of vectorsv and w by v.w.If we represent the elements of the space by column vectors, we can alsorepresent this inner product by w.v = wTv = vTw = v.w.Then the length Lv of a vector v is Lv = |v| =

√v.v =

√vTv.

Here T applied to a matrix denotes its transpose.Then Lv is well-defined and is zero if and only if v = 0.The Cauchy-Schwarz inequality is the inequality |v.w| ≤ |v||w|, with equalityif and only if the vectors v and w are linearly dependent.The real angle θ, with 0 ≤ θ ≤ π, between non-zero vectors v and w is givenby the formula: |v||w| cos(θ) = v.w. The Cauchy-Schwarz inequality impliesthat θ is well-defined. We say that (possibly zero) vectors v and w are or-thogonal or perpendicular, if v.w = 0.The orthogonal group is the group of linear transformations v → L(v), pre-serving length, so for any vector v we have:

|L(v)| = |v|, vTv = vTLTLv.

This gives the defining relation for elements of the orthogonal group:

LLT = LTL = I.

Here I denotes the identity transformation.A reflection is a linear transformation of the form:

v → Tn(v) = v − 2(v.n)n, |n| = 1,

Tn = I − 2n⊗ nT , nTn = 1.

If v is any vector we can write v uniquely in the form v = a + b, where a isperpendicular to the unit vector n and b is parallel to n.Explicitly we have b = (v.n)n and a = v − (v.n)n.Then the reflection Tn acts so that Tn(v) = a− b.There are then two basic facts about the orthogonal group:

• The group is generated by reflections.

• The parity (even/odd) of the number of reflections whose product is agiven element of the orthogonal group is independent of the choice ofthe reflections producing that element.

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More precisely, from the relation AAT = I, we infer that det(A)2 = 1, sodet(A) = ±1. Then every reflection Tn has det(Tn) = −1 and an orthogonaltransformation A is the product of an even number of reflections if and onlyif det(A) = 1 and is otherwise the product of an odd number of reflections.The orthogonal transformations of unit determinant form a subgroup of theorthogonal group, called the rotation group.

If now v and w are vectors, they define a parallelogram in the Euclideanspace. It has area Av,w given by the formula:

Av,w =√

(v.v)(w.w)− (v.w)2.

The Cauchy-Schwarz inequality gives that Av,w is well-defined and Av,w = 0if and only if v and w are linearly dependent, precisely the case where theparallelogram collapses onto a line. More generally, if P = v1, v2, . . . vk isa collection of vectors of the space, then the volume VP of the parallelepipeddetermined by them is:

VP =√

det(vi.vj).

One can prove that VP is well-defined and zero if and only if the vectors ofP are linearly dependent.

We can remove the dependence on the standard basis of Euclidean space, bypassing to a real vector space V with a symmetric bilinear form g : V×V→ R,that is positive definite: for any 0 6= v ∈ V, we have g(v, v) > 0. If nowP = v1, v2, . . . vk is a collection of vectors of the space V, then the volumeVP of the parallelepiped determined by them is:

VP =√

det(g(vi, vj)).

Again VP is well-defined and vanishes if and only if the elements of P arelinearly dependent.

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The metric of V induces a metric still called g on the tensor algebra of V,such that if v1v2 . . . vk and w1w2 . . . wm are words of the algebra, then theirinner product is zero, unless k = m and when k = m we have:

g(v1v2 . . . vk, w1w2 . . . wk) = Πkj=1g(vi, wi).

This in turn induces inner products on the symmetric tensor algebra, suchthat in the case of the symmetric algebra we have:

g(ev, ew) = eg(v,w).

Equivalently, we have, for each nonnegative integer k:

g(vk, wk) = k!g(v, w)k.

Then for example, we have:

g(v1v2, w1w2) =1

16g((v1 + v2)

2 − (v1 − v2)2, (w1 + w2)2 − (w1 − w2)

2)

=1

8

(g(v1 + v2, w1 + w2)

2 − g(v1 + v2, w1 − w2)2 − g(v1 − v2, w1 + w2)

2 + g(v1 − v2, w1 − w2)2)

=1

8(g(v1+v2, w1+w2)−g(v1+v2, w1−w2))(g(v1+v2, w1+w2)+g(v1+v2, w1−w2))

−1

8((g(v1−v2, w1+w2)−g(v1−v2, w1−w2))(g(v1−v2, w1+w2)+g(v1−v2, w1−w2))

=1

2(g(v1 + v2, w1)g(v1 + v2, w2)− g(v1 − v2, w1)g(v1 − v2, w2))

= g(v1, w1)g(v2, w2) + g(v2, w1)g(v1, w2).

The right-hand-side is called a permanent, a variation of the formula for thedeterminant, where all signs in the formula are taken to be positive. In gen-eral, we could use the Laplace or Fourier transform to compute the innerproducts and hence permanents.

In the case of the exterior algebra, we have:

g(v1v2 . . . vk, w1w2 . . . wk) = det(g(vi, wj)).

Note in particular that g(v1v2 . . . vk, v1v2 . . . vk) = det(g(vi, vj)) = V2P . Here

VP is the volume of the parallelepiped determined by the vectors v1, v2, . . . , vk.

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Determinants and linear transformations

Let X and Y be vector spaces (over the same field F) of the same dimensionn. Put ωX = Ωn(X) and ωY = Ωn(Y). These are each one-dimensionalvector spaces. Let L : X → Y be a linear transformation. Then L inducesa natural linear mapping called T (L) mapping the tensor algebra of X tothat of Y, such that if x = fx1x2 . . . xk is a word of T(X), then we haveT (L)(x) = fL(x1)L(x2) . . . L(xk). This mapping is compatible with thepassage to the exterior algebra, so we induce a natural map called Ω(L)mapping Ω(X) to Ω(Y). This mapping preserves degree, so, in particular, inthe highest dimension, it induces a natural map, denoted det(L) from theline ωX to the line ωY. Clearly this map is functorial: if M : Y → Z isalso a linear map with Z a vector space of dimension n over F, then we getdet(M L) = det(M) det(L). Also we have that det(cL) = cn det(L), forany c ∈ F and det(L) 6= 0, if and only if L is an isomorphism. If K and Lare linear maps from X to Y, their relative characteristic polynomial is thepolynomial det(sK+ tL) and is a homogeneous polynomial of total degree n,of the form sn det(K) + · · · + tn det(L). Of interest also is the restriction ofΩ(L) to the n-dimensional space Ωn−1(X). This is called the adjoint mappingof L, and is denoted adj(L) : Ωn−1(X) → Ωn−1(Y). Note that we have theformulas:

X ∧ Ωn−1(X) = Ωn(X),

L(w) ∧ adj(L)(α) = det(L)(w ∧ α), for any w ∈ X and any α ∈ Ωn−1(X).

s

In the special case that Y = X, so L is an endomorphism of X, then det(L)is a multiple of the identity isomorphism of ωX, by an element of F, so maybe regarded canonically as an element of F. Then det gives a homomor-phism from the space of endomorphisms of X under composition to the fieldF under multiplication, whose restriction to the group of isomorphisms of Xis a group homomorphism to the group F∗ of non-zero elements of F undermultiplication. In particular we have det(idX) = 1. Also the characteristicpolynomial det(s idX + tL) is a homogeneous polynomial in the variables sand t of total degree n, of the form sn + tr(L)sn−1t+ · · ·+ tn det(L).

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