Multilinear Algebra [Lecture notes]

18 Pages·2006·0.179 MB·English

by  Jean-Pierre Tignol

CHAPTER I
Duality of finite-dimensional vector spaces
1. Dual space
LetE beafinite-dimensionalvectorspaceoverafieldK. ThevectorspaceoflinearmapsE →K
is denoted by E∗, so
E∗ =L(E,K).
This vectorspace is calledthe dual space ofE. Its elements arecalled linear forms onE. For ϕ∈E∗
and u∈E∗, let
hϕ,ui=ϕ(u)∈K.
1.1. Proposition. For ϕ, ψ ∈E∗, u, v ∈E and α, β ∈K,
1. hαϕ+βψ,ui=αhϕ,ui+βhψ,ui.
2. hϕ,uα+vβi=hϕ,uiα+hϕ,viβ.
3. If hϕ,ui=0 for all u∈E, then ϕ=0.
4. If hϕ,ui=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. Ifu6=0,thenuisthefirstelementinsomebasisofE. Let(u,e ,... ,e )besuchabasis. From
2 n
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.
1.1. Dual basis
Let e=(e ,... ,e ) be a basis ofE. By the constructionprinciple for linearmaps, thereexists for all
1 n
i=1, ... , n a linear form e∗ ∈E∗ which maps e to 1 and the other basis vectors to 0, i.e.,
i i
1 if i=j,
he∗,e i=δ =
i j ij
(0 if i6=j.
n
Thus, for u= e u ∈E we have
j=1 j j
n n
P
he∗,ui= he∗,e iu = δ u =u ,
i i j j ij j i
j=1 j=1
X X
which shows that the linear form e∗ maps every vector of E to its i-th coordinate with respect to the
i
basis e. Therefore, for all u∈E,
n
(1) u= e he∗,ui.
i i
i=1
X
1.2. Theorem. The sequence e∗ =(e∗,... ,e∗) is a basis of E∗. In particular, dimE∗ =dimE.
1 n
Proof. The elements e∗, ... , e∗ are linearly independent: Let n α e∗ = 0. For j = 1, ... ,
1 n i=1 i i
n,
P
n n n
0= α e∗,e = α he∗,e i= α δ =α .
i i j i i j i ij j
DXi=1 E Xi=1 Xi=1
1
2. ORTHOGONALITY 2
The sequence e∗ spans E∗: Let us show that for all ϕ∈E∗,
n
(2) ϕ= hϕ,e ie∗.
i i
i=1
X
It suffices to show that both sides take the same value on every vector u ∈ E. From (1), it follows
that
n n
hϕ,ui= ϕ, e he∗,ui = hϕ,e ihe∗,ui.
j j j j
D Xj=1 E Xj=1
On the other hand,
n n
hϕ,e ie∗,u = hϕ,e ihe∗,ui.
i i i i
DXi=1 E Xi=1
1.2. Bidual space
Every vector u∈E defines a linear form ev : E∗ →K by
u
ev (ϕ)=ϕ(u).
u
Thus, ev ∈(E∗)∗. Moreover,the map
u
ev: E →E∗∗; u7→ev
u
is linear since for u, v ∈E, α, β ∈K and ϕ∈E∗,
ev (ϕ)=ϕ(uα+vβ)=ϕ(u)α+ϕ(v)β =ev (ϕ)α+ev (ϕ)β.
uα+vβ u v
1.3. 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 =0, then hϕ,ui=0 for all
u
ϕ∈E∗, hence u=0 by Proposition 1.1. Therefore, Kerev=0 and ev is injective.
On the other hand, applying Theorem 1.2 twice, we find
dimE∗∗ =dimE∗ =dimE.
Since ev is injective, it is therefore also surjective.
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 ∈E∗∗.
u
Therefore, hu,ϕi is defined for u∈E and ϕ∈E∗, and we have
hu,ϕi=hev ,ϕi=ϕ(u)=hϕ,ui.
u
2. Orthogonality
For every subspace V ⊂E, the orthogonal subspace V0 ⊂E∗ is defined by
V0 ={ϕ∈E∗ |hϕ,ui=0 for all u∈V}.
It is easily checked that V0 is a subspace of E∗.
2.1. Proposition. dimV0 =dimE−dimV.
Proof. Let(e ,... ,e )beabasisofV,whichweextendintoabasise=(e ,... ,e ,e ,... ,e )
1 r 1 r r+1 n
of E. Consider the last n−r elements of the dual basis e∗ =(e∗,... ,e∗,... ,e∗). We shall show:
1 r n
V0 =spanhe∗ ,... ,e∗i.
r+1 n
The proposition follows, since the right side is a subspace of dimension n−r, as e∗ , ... , e∗ are
r+1 n
linearly independent.
Equation (2) shows that every ϕ∈V0 is a linear combination of e∗ , ... , e∗. Therefore,
r+1 n
V0 ⊂spanhe∗ ,... ,e∗i.
r+1 n
3. TRANSPOSITION 3
Onthe other hand,everyv ∈V is alinear combinationof (e ,... ,e ), hence equation(1)showsthat
1 r
he∗,vi=0 for i=r+1,... ,n.
i
Therefore, e∗ , ... , e∗ ∈V0, hence
r+1 n
spanhe∗ ,... ,e∗i⊂V0.
r+1 n
2.2. Corollary. For every subspace V ⊂E,
V00 =V
(under the identification of E and E∗∗, since V00 ⊂E∗∗).
Proof. By definition,
V00 ={u∈E |hu,ϕi=0 for all ϕ∈V0}.
Since hϕ,ui = 0 for u ∈ V and ϕ ∈ V0, we have V ⊂ V00. Now, the preceding proposition and
Theorem 1.2 yield
dimV00 =dimE∗−dimV0 =dimE−dimV0 =dimV.
Therefore, the inclusion V ⊂V00 is an equality.
3. Transposition
Let A: E →F be a linear map between finite-dimensional vector spaces. The transpose map
At: F∗ →E∗
is defined by
At(ϕ)=ϕ◦A for ϕ∈F∗.
3.1. Proposition. At is the only linear map from F∗ to E∗ such that
At(ϕ),u = ϕ,A(u) for all ϕ∈F∗ and all u∈E.
Proof. For ϕ∈F(cid:10)∗ and u(cid:11)∈E(cid:10)we have(cid:11)
At(ϕ),u =hϕ◦A,ui=ϕ◦A(u)= ϕ,A(u) ,
hence At satisfies the prope(cid:10)rty. (cid:11) (cid:10) (cid:11)
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) = At(ϕ),u ,
hence B(ϕ)=At(ϕ) and therefo(cid:10)re B =A(cid:11)t. (cid:10) (cid:11) (cid:10) (cid:11)
Now, let e and f be arbitrary bases of E and F respectively, and let e∗, f∗ be the dual bases.
3.2. Proposition. The matrices of At and A are related as follows:
e∗(At)f∗ =f(A)te.
Proof. Let (A) =(a ) , i.e.,
f e ij 16i6m
16j6n
m
(1) A(e )= f a for j =1,... ,n
j i ij
i=1
X
and e∗(At)f∗ =(a′ij)16i6n, i.e.,
16j6m
n
At(f∗)= e∗a′ for j =1,... ,m.
j i ij
i=1
X
3. TRANSPOSITION 4
We have to prove a′ = a for i = 1, ... , n and j = 1, ... , m. Substituting At(f∗) for ϕ in
ij ji j
equation (2) of section 1.1, we obtain
n
At(f∗)= At(f∗),e e∗ for j =1,... ,m,
j j i i
i=1
X(cid:10) (cid:11)
hence a′ = At(f∗),e . By Proposition3.1, it follows that
ij j i
(cid:10) (cid:11) a′ij = fj∗,A(ei) .
On the other hand, we may compute the right s(cid:10)ide of this(cid:11)last equation by using equation (1), which
yields
n n
f∗,A(e ) = f∗, f a = hf∗,f ia =a .
j i j k ki j k ki ji
(cid:10) (cid:11) D kX=1 E Xk=1
Thus, a′ =a for i=1, ... , n and j =1, ... , m.
ij ji
3.3. Proposition. 1. For A, B: E →F and α, β ∈K,
(αA+βB)t =αAt+βBt.
2. For A: E →F and B: F →G,
(B◦A)t =At◦Bt.
3. For A: E →F,
(At)t =A
(under the identifications E∗∗ =E and F∗∗ =F, since (At)t: E∗∗ →F∗∗).
Proof. These properties can be proved either by reduction to the corresponding properties of
matrices(byusingarbitrarybasesoftherelevantspaces),orbyusingthecharacterizationoftranspose
maps in Proposition 3.1.
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.
1.1. Lemma. For x, y ∈E, the equation x+V =y+V is equivalent to x−y ∈V.
Proof. Supposefirstx+V =y+V. Sincex=x+0∈x+V,wehavex∈y+V,hencex=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.
Itiseasilycheckedthattheseoperationsarewell-definedandendowthesetE/V withaK-vector
space structure.
In view of the definitions above, the map
ε : E →E/V
V
defined by
ε (x)=x+V
V
is a surjective linear map. It is calledthe canonical epimorphism of E onto E/V. Its kernelis 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
1.2. Proposition. Suppose E is finite-dimensional; then
dimE/V =dimE−dimV.
Moreover,
(a) if(e ,... ,e )isabasisofE whichextendsabasis(e ,... ,e )ofV,then(e +V,... ,e +V)
1 n 1 r r+1 n
is a basis of E/V;
(b) if(e ,... ,e )isabasisofV andu ,... ,u arevectorsofE suchthat(u +V,... ,u +V)
1 r r+1 n r+1 n
is a basis of E/V, then (e ,... ,e ,u ,... ,u ) is a basis of E.
1 r r+1 n
5
2. INDUCED LINEAR MAPS 6
Proof. Since Kerε = V and Imε =E/V, the relation between the dimensions of the kernel
V V
and the image of a linear map yields
dimE =dimV +dimE/V.
(a)SincedimE/V =dimE−dimV =n−r,itsufficestoprovethatthesequence(e +V,... ,e +V)
r+1 n
spans E/V. Every x∈E is a linear combination of e , ... , e . If
1 n
x=e x +···+e x ,
1 1 n n
then
x+V =ε (x)=ε (e )x +···+ε (e )x .
V V 1 1 V n n
Now, ε (e )=0 for i=1, ... , r since e ∈V, hence
V i i
x+V =(e +V)x +···+(e +V)x .
r+1 r+1 n n
(b) Since dimE = dimV +dimE/V, it suffices to prove that e , ... , e , u , ... , u are linearly
1 r r+1 n
independent. Suppose
(1) e α +···+e α +u β +···+u β =0.
1 1 r r r+1 r+1 n n
Taking the image of each side under ε , we obtain (u +V)β +···+(u +V)β = 0, hence
V r+1 r+1 n n
β =···=β =0. Therefore, by (1) it follows that α =···=α =0, since (e ,... ,e ) is a basis
r+1 n 1 r 1 r
of V.
2. Induced linear maps
LetA: E →F be alinear mapand letV ⊂E be a subspace. A linear mapA: E/V →F is said
to be induced by A if
A(x+V)=A(x) for all x∈E.
In other words,
A◦ε =A.
V
This condition is also expressed as follows: the diagram
E εV //E/V
?
?A??????(cid:31)(cid:31) }}{{{{{A{{{
F
is commutative or commutes.
2.1. Proposition. A linear map A: E →F induces a linear map
A: E/V →F
if and only if V ⊂KerA.
Proof. If A◦ε =A, then A(v)=0 for all v ∈V since V =Kerε . For the converse, suppose
V V
V ⊂KerA and consider x, x′ ∈E. If x+V =x′+V, then x−x′ ∈KerA, hence A(x−x′)=0 and
therefore
A(x)=A(x′).
We may then define a map A: E/V →F by
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
A: E/V →F/W
is said to be induced by A if
A(x+V)=A(x)+W for all x∈E
or,alternatively,ifA◦ε =ε ◦A. Thisconditioncanalsobe expressedbysayingthatthe following
V W
diagram is commutative:
A //
E F
εV εW
(cid:15)(cid:15) (cid:15)(cid:15)
A //
E/V F/W
or that A makes the above diagram commute.
2.2. 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 ◦A: E → F/W. The preceding proposition shows
W
that there exists a linear map A such that A′ = A◦ε if and only if V ⊂ KerA′. In view of the
V
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 →W.
V
We aim to compare the matrices of A, of A| and of A: E/V →F/W.
V
Consider a basis e′ = (e ,... ,e ) of V, a basis e = (e + V,... ,e + V) of E/V and the
1 r r+1 n
sequence e=(e ,... ,e ,e ,... ,e ) which, according to Proposition1.2, is a basis of E. Similarly,
1 r r+1 n
let f′ = (f ,... ,f ) be a basis of W, f =(f +W,... ,f +W) a basis of F/W and consider the
1 s s+1 m
basis f =(f ,... ,f ,f ,... ,f ) of F.
1 s s+1 m
2.3. Proposition.
f′(A|V)e′ ∗
(A) = .
f e
0 f(A)e !
Proof. Let A(e )= m f a for j =1, ... , n, so that
j i=1 i ij
(A) =(a ) .
P f e ij 16i6m
16j6n
For j = 1, ... , r we have e ∈ V, hence A(e ) ∈ W. Therefore, A(e ) is a linear combination of f ,
j j j 1
... , f , hence
s
a =0 for j =1, ... , r and i=s+1, ... , m.
ij
s
Moreover,A(e )=A| (e ) for j =1, ... , n, hence A| (e )= f a and therefore
j V j V j i=1 i ij
f′(A|V)e′ =(aij)16i6s. P
16j6r
On the other hand, for j =r+1, ... , n,
m
A(e +V)=A(e )+W = (f +W)a .
j j i ij
i=1
X
Fori=1,... ,s,wehavef ∈W,hence f +W =0. Theprecedingequationthus yieldsA(e +V)=
i i j
m
(f +W)a for j =r+1, ... , n, hence
i=s+1 i ij
P f(A)e =(aij)s+16i6m.
r+16j6n
3. TRIANGULATION 8
The proof is thus complete.
In particular, if F =E and W =V,
e′(A|V)e′ ∗
(A) =
e e
0 e(A)e !
andsincethedeterminantofablock-triangularmatrixistheproductofthedeterminantsoftheblocks
the following corollary follows:
2.4. Corollary. detA=detA| ·detA and Pc =Pc ·Pc .
V A A|V A
3. Triangulation
3.1. Proposition. Let A: E → E be a linear operator on a finite-dimensional vector space E.
ThereexistsabasiseofE suchthatthematrix (A) isuppertriangularifandonlyifthecharacteristic
e e
polynomial Pc decomposes into a product of factors of degree 1.
A
Proof. If the matrix (A) is upper triangular,let
e e
λ ∗
1
e(A)e = ... ,
0 λ
 n 
 
then Pc (X)=(X −λ )...(X −λ ).
A 1 n
Conversely,supposePc decomposesintoaproductoffactorsofdegree1. Wearguebyinduction
A
on the dimension of E. If dimE =1, there is nothing to prove since every matrix of order 1 is upper
triangular. Let dimE = n. By hypothesis, the characteristic polynomial of A has a root λ. Let
e ∈E be an eigenvector of A with eigenvalue λ and let V =e K be the vector space spanned by e .
1 1 1
We have
A(V)=A(e )K =e λK ⊂V,
1 1
hence A induces a linear operator
A: E/V →E/V.
ByCorollary2.4,wehavePc (X)=(X−λ)Pc (X),hencethecharacteristicpolynomialofAdecom-
A A
poses into a product of factors of degree 1. By induction, there exists a basis e =
(e +V,... ,e +V) of E/V such that the matrix (A) is upper triangular. Proposition 1.2 shows
2 n e e
that the sequence e=(e ,e ,... ,e ) is a basis of E. Moreover,by Proposition 2.3,
1 2 n
λ ∗
(A) = .
e e
0 e(A)e !
Since the matrix (A) is upper triangular, so is (A) .
e e e e
CHAPTER III
Tensor product
1. Tensor product of vector spaces
Let E , ... , E , F be vector spaces over a field K. Recall that a map
1 n
f: E ×···×E →F
1 n
is called n-linear or multilinear if
f(x ,... ,x ,x α+x′α′,x ,... ,x )=
1 i−1 i i i+1 n
f(x ,... ,x ,x ,x ,... ,x )α+f(x ,... ,x ,x′,x ,... ,x )α′
1 i−1 i i+1 n 1 i−1 i i+1 n
foreachi=1,... ,nandforallx ∈E ,... ,x ∈E ,x ,x′ ∈E ,x ∈E ,... ,x ∈E and
1 1 i−1 i−1 i i i i+1 i+1 n n
α, α′ ∈K. In other words, for each i=1, ... , n and for all x ∈E , ... , x ∈E , x ∈E ,
1 1 i−1 i−1 i+1 i+1
... , x ∈E fixed, the map
n n
f(x ,... ,x ,•,x ,... ,x ): E →F
1 i−1 i+1 n i
which carries x ∈ E to f(x ,... ,x ,x,x ,... ,x ) is linear. For example, all the products for
i 1 i−1 i+1 n
which the distributive law holds are bilinear maps.
Clearly,if f: E ×···×E →F is multilinear and A: F →G is linear,then the composite map
1 n
A◦f: E ×···×E → G is multilinear. It turns out that, for given E , ... , E , there exists a
1 n 1 n
“universal” linear map ⊗: E ×···×E →E ⊗···⊗E , called the tensor product of E , ... , E ,
1 n 1 n 1 n
from which all the multilinear maps originating in E ×···×E can be derived. The goal of this
1 n
section is to define this universal map.
1.1. Universal property
Let E , ... , E be vector spaces over a field K. A pair (f,F) consisting of a K-vector space F and
1 n
a multilinear map
f: E ×···×E →F
1 n
is called a universal product of E , ... , E if the following property (called the universal property)
1 n
holds: for every K-vector space G and every multilinear map
g: E ×···×E →G
1 n
there exists a unique linear map A: F →G such that g =A◦f or, in other words, which makes the
following diagram commute:
f //
E1×···×MMEMMngMMMMMMM&& ~~~~~~~A~~~F
G.
The existence of a universal product is far from obvious. By contrast, the uniqueness (up to isomor-
phism) easily follows from the definition by abstract nonsense.
1.1. Proposition. Suppose (f,F) and (f′,F′) are two universal products of E , ... , E . Then
1 n
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:
f //
E1×···×MMEMMnfM′MMMMMM&& ~~}}}}}ϕ}}}F
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:
f // f //
E1×···×LLELLnfLLLLLLL&& (cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127)(cid:127)I(cid:127)d(cid:127)FF E1×···×MMEMMnfMMMMMMM&& (cid:127)(cid:127)~~~~~ϕ~~′~◦ϕF
F F.
In view of the uniqueness condition in the universal property of (f,F), it follows that ϕ′◦ϕ = Id .
F
Similarly,theuniversalpropertyof(f′,F′)showsthatϕ◦ϕ′ =IdF′. Therefore,ϕandϕ′arereciprocal
bijections.
This proposition shows that a universal product of E , ... , E , if it exists, can be considered
1 n
as unique. We shall refer to this universal product as the tensor product of E , ... , E and denote
1 n
it as (⊗,E ⊗···⊗E ). The image of an n-tuple (x ,... ,x ) ∈ E ×···×E under ⊗ is denoted
1 n 1 n 1 n
x ⊗···⊗x .
1 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.
1.2. The tensor product of two vector spaces
1.2. Proposition. Let µ: Km×Kn →Km×n be defined by
µ (x ) ,(y ) =(x y ) .
i 16i6m j 16j6n i j 16i6m
16j6n
(cid:0) (cid:1)
The pair (µ,Km×n) is a universal product of Km and Kn.
Proof. Denoting the elements of Km and Kn as column vectors, we have
µ(x,y)=x·yt for x∈Km and y ∈Kn,
where · denotes the matrix multiplication Km×1×K1×n → Km×n. Therefore, the bilinearity of µ
follows from the properties of the matrix multiplication.
Let (c ) (resp. (c′) ) be the canonical basis of Km (resp. Kn). Let also
i 16i6m j 16j6n
e =µ(c ,c′) for i=1, ... , m and j =1, ... , n.
ij i j
Thus, e ∈ Km×n is the matrix whose only non-zero entry is a 1 at the i-th row and j-th column.
ij
The family (e ) is a basis of Km×n. Therefore, given any bilinear map
ij 16i6m
16j6n
B: Km×Kn →F,
we may use the construction principle to obtain a linear map ϕ: Km×n → F such that ϕ(e ) =
ij
x y
1 1
B(c ,c′). For all x = .. ∈ Km and y = .. ∈ Kn, we have x = m c x , y = n c′y ,
i j  .   .  i=1 i i j=1 j j
x y
 m  n P P
hence    
µ(x,y)= x y µ(c ,c′)= x y e
i j i j i j ij
16i6m 16i6m
X X
16j6n 16j6n