9 Pasivos Financieros
ARTICULO 22.- MATERIAL, SUMINISTROS Y OTROS
1. Show that X1 = 42 1 , X2= −52 −3 , X3= 13 0
form a basis of R3, and express
the elementary vectors E1, E2, E3in terms of X1, X2, X3.
2. Find a basis for the null space of the matrix A=
−3 1 4 −72 3 1 1
1 2 1 0
.
3. Find the dimension of the vector space Mm,n(F )where F is a arbitrary field. 4. Let v1, v2, . . . , vnbe vectors in a vector space V . Assume that each element of V is uniquely expressible as a linear combination of v1, v2, . . . , vn. Prove that the vi’s form a basis of V .
5. LetB = {E1, E2, E3} be the standard ordered basis of R3and let
B = 20 0 , −12 0 , 11 1 .
Show thatBis a basis ofR3and find the transition matrices for the changes of bases B → BandB→ B.
6. Let V be a vector space of dimension n and let i be an integer such that 0≤ i ≤ n. Show that V has at least one subspace of dimension i.
7. The same as Exercise 6 with “subspace” replaced by “quotient space”.
8. Let U be a subspace of a finite dimensional vector space V . Prove that there is a subspace W such that V = U ⊕ W. Show also that W V /U.
9. Let V be a vector space of dimension 2n and assume that U and W are subspaces of dimensions n and n+ 1 respectively. Prove that U ∩ W = 0.
10. Let vectors v1, v2, . . . , vmgenerate a vector space V . Prove that some subset of {v1, v2, . . . , vm} is a basis of V .
8.3
Linear mappings
Just as there are homomorphisms of groups and of rings, there are homomorphisms of vector spaces. Traditionally these are called linear mappings or transformations. Let V and W be two vector spaces over the same field F . Then a function
α: V → W
is called a linear mapping from V to W if the following rules are valid for all v1,
(i) α(v1+ v2)= α(v1)+ α(v2);
(ii) α(av1)= aα(v1).
If α is also bijective, it is called an isomorphism of vector spaces. Should there exist an isomorphism between vector spaces V and W over a field F , we will write V WF or V W. Notice that a linear mapping is automatically a homomorphism of additive groups by (i). So all results established for group homomorphisms may be carried over to linear mappings. A linear mapping α : V → V is called a linear operator on V .
Example (8.3.1) Let A be an m × n matrix over a field F and define a function
α : Fn → Fm by the rule α(X) = AX where X ∈ Fn. Then simple properties of matrices reveal that α is a linear mapping.
Example (8.3.2) Let V be an n-dimensional vector space over a field F and let
B = {v1, v2, . . . , vn} be an ordered basis of V . Recall that to each vector v in V there corresponds a unique coordinate vector[v]Bwith respect toB. Use this corre- spondence to define a function α : V → Fnby α(v)= [v]B. By simple calculations we see that[u + v]B = [u]B+ [v]B and[av]B = a[vB] where u, v ∈ V , a ∈ F . Hence α is a linear mapping. Clearly[v]B = 0 implies that v = 0; thus α is injective, and it is obviously surjective. Our conclusion is that α is an isomorphism and V FF n.
We state this result formally as:
(8.3.1) If V is a vector space with dimension n over a field F , then V FF n. Thus any two vector spaces with the same finite dimension over F are isomorphic.
(The second statement follows from the fact that isomorphism of vector spaces is an equivalence relation).
An important way of defining a linear mapping is by specifying its effect on a basis.
(8.3.2) Let{v1, . . . , vn} be a basis of a vector space V over a field F and let w1, . . . , wn
be any n vectors in another F -vector space W . Then there is a unique linear mapping α: V → W such that α(vi)= wifor i = 1, 2, . . . , n.
Proof. Let v ∈ V and write v = ni=1aivi, where ai in F are unique. Define a function α : V → W by the rule
α(v)=
n i=1
aiwi.
Then an easy check shows that α is a linear mapping, and of course α(vi)= wi. If
α: V → W is another such linear mapping, then α= α; for α(v)=in=1aiα(vi)= n
8.3 Linear mappings 149 Our experience with groups and rings suggests it may be worthwhile to examine the kernel and image of a linear mapping.
(8.3.3) Let α: V → W be a linear mapping. Then Ker(α) and Im(α) are subspaces
of V and W respectively.
Proof. Since α is a group homomorphism, it follows from (4.3.2) that Ker(α) and
Im(α) are additive subgroups. We leave the reader to complete the proof by showing that these subgroups are also closed under scalar multiplication. Just as for groups and rings, there are Isomorphism Theorems for vector spaces.
(8.3.4) If α: V → W is a linear mapping between vector spaces over a field F , then
V /Ker(α) Im(α).F
In the next two results U and W are subspaces of a vector space V over a field F .
(8.3.5) (U+ W)/W U/(U ∩ W).F
(8.3.6) If U ⊆ W, then (V /U)/(W/U) V /W.F
Since the Isomorphism Theorems for groups are applicable, all one has to prove here is that the functions introduced in the proofs of (4.3.4), (4.3.5) and (4.3.6) are linear mappings, i.e., they act appropriately on scalar multiples.
For example, in (8.3.4) the function in question is θ : V / Ker(α) → Im(α) where
θ (v+ Ker(α)) = α(v). Then
θ (a(v+ Ker(α)) = θ(av + Ker(α)) = α(av) = aα(v) = aθ(v + Ker(α)),
and it follows that θ is a linear mapping.
There is an important dimension formula connecting kernel and image.
(8.3.7) If α: V → W is a linear mapping between finite dimensional vector spaces,
then dim(Ker(α))+ dim(Im(α)) = dim(V ).
This follows directly from (8.3.4) and (8.2.14). An immediate application is to the null space of a matrix.
Corollary (8.3.8) Let A be an m× n matrix with rank r over a field F . Then the
dimension of the null space of A is n− r.
Proof. Let α be the linear mapping from Fn to Fm defined in Example (8.3.1) by
α(X) = AX. Now Ker(α) is the null space of A, and it is readily seen that Im(α)
is the column space. By (8.2.8) dim(Im(α)) = r, the rank of A, and by (8.3.7)
dim(Ker(α))= n − r, so the result follows.
As another application of (8.3.7) we give a different proof of the dimension formula for sum and intersection of subspaces – see (8.2.12).
(8.3.9) If U and W are subspaces of a finite dimensional vector space, then
dim(U+ W) + dim(U ∩ W) = dim(U) + dim(W).
Proof. By (8.3.5) (U+W)/W U/(U ∩W). Hence, taking dimensions and applying
(8.2.14), we find that dim(U + W) − dim(W) = dim(U) − dim(U ∩ W).
Vector spaces of linear mappings. It is useful to endow sets of linear mappings with the structure of a vector space. Suppose that V and W are two vector spaces over the same field F . We will write
L(V , W )
for the set of all linear mappings from V to W . Define addition and scalar multiplication in L(V , W ) by the natural rules
α+ β(v) = α(v) + β(v), a · α(v) = a(α(v)),
where α, β ∈ L(V, W), v ∈ V , a ∈ F . It is simple to verify that α + β and a · α are linear mappings. The basic result about L(V , W ) is:
(8.3.10) Let V and W be vector spaces over a field F . Then:
(i) L(V , W ) is a vector space over F ;
(ii) if V and W are finite dimensional, then L(V , W ) has finite dimension equal to dim(V )· dim(W).
Proof. We omit the routine proof of (i) and concentrate and (ii). Let{v1, . . . , vm} and {w1, . . . , wn} be bases of V and W respectively. By (8.3.2), for i = 1, 2, . . . , m and
j = 1, 2, . . . , n, there is a unique linear mapping αij : V → W such that
αij(vk)=
wj if k= i 0 if k= i .
Thus αij sends basis element vito basis element wj and all other vk’s to 0. First we show that the αij are linearly independent in the vector space L(V , W ).
Let aij ∈ F ; then by definition of αij we have for each k m i=1 n j=1 aijαij (vk)= n j=1 m i=1 aij(αij(vk))= n j=1 akjwj. (∗)
Thereforemi=1nj=1aijαij = 0 if and only if akj = 0 for all j, k. It follows that the αij are linearly independent.
Finally, we claim that αij actually generate L(V , W ). To prove this let α ∈
L(V , W ) and write α(vk) = n
j=1akjwj where akj ∈ F . Then from the equation
(∗) above we see that α = mi=1jn=1aijαij. Therefore the αij’s form a basis of
8.3 Linear mappings 151
The dual space. If V is a vector space over a field F , the vector space
V∗= L(V, F )
is called the dual space of V ; here F is to be regarded as a 1-dimensional vector space. Elements of V∗are linear mappings from V to F : these are called linear functionals on V .
Example (8.3.3) Let Y ∈ Fn and define α : Fn → F by the rule α(X) = YTX. Then α is a linear functional on Fn.
If V has finite dimension n, then
dim(V∗)= dim(L(V, F )) = dim(V )
by (8.3.10). Thus V , V∗and the double dual V∗∗ = (V∗)∗all have the same dimen- sion. Therefore these vector spaces are isomorphic by (8.3.1).
In fact there is a canonical linear mapping θ : V → V∗∗. Let v∈ V and define
θ (v)∈ V∗∗by the rule
θ (v)(α)= α(v)
where α ∈ V∗. So θ (v) simply evaluates each linear functional on V at v. Concerning the function θ , we will prove:
(8.3.11) The function θ : V → V∗∗ is an injective linear mapping. If V has finite dimension, then θ is an isomorphism.
Proof. In the first place θ (v)∈ V∗∗: for
θ (v)(α+ β) = α + β(v) = α(v) + β(v) = θ(v)(α) + θ(v)(β).
Also θ (v)(a· α) = a · α(v) = a(α(v)) = a(θ(v)(α)). Next we have
θ (v1+ v2)(α)= α(v1+ v2)= α(v1)+ α(v2)
= θ(v1)(α)+ θ(v2)(α)= (θ(v1)+ θ(v2))(α).
Hence θ (v1+ v2)= θ(v1)+ θ(v2). We leave the reader to verify in a similar way that
θ (a· v) = a(θ(v)). So far it what we have shown that θ is a linear mapping from V
to V∗∗.
Now suppose that θ (v)= 0. Then 0 = θ(v)(α) = α(v) for all α ∈ V∗. This can only mean that v = 0: for if v = 0, we could include v in a basis of V and construct a linear functional α such that α(v)= 1F, by (8.3.2). It follows that θ is injective.
Finally, assume that V has finite dimension; then dim(V )= dim(V∗)= dim(V∗∗). Also dim(V∗∗) = dim(Im(θ)) since θ is injective. It follows from (8.2.9) that
Representing linear mappings by matrices. A linear mapping between finite di- mensional vector spaces can be described by matrix multiplication, which provides us with a concrete way of representing linear mappings.
Let V and W be vector spaces over the same field F with finite dimensions m and
nrespectively. Choose ordered bases for V and W , sayB = {v1, v2, . . . , vm} and C = {w1, w2, . . . , wn} respectively. Now let α ∈ L(V, W); then
α(vi)= n j=1
aj iwj
for some aj i ∈ F . This enables us to form the n × m matrix over F
A= [aij] ,
which is to represent α. Notice that the i-th column of A is precisely the coordinate column vector of α(vi)with respect to the basisC. Thus we have a function
θ: L(V, W) → Mn,m(F ) defined by the rule that column i of θ (α) is[α(vi)]C
To understand how the matrix A= θ(α) reproduces the effect of α on an arbitrary vector v=mi=1biviof V , we compute α(v)= m i=1 bi(α(vi))= m i=1 bi n j=1 aj iwj = n j=1 m i=1 aj ibi ωj.
Hence the coordinate column vector of α(v) with respect toC has entriesmi=1aj ibi, for j = 1, . . . , n, i.e., it is just A
b1 .. . bm
= A[v]B. Therefore we have the basic formula
[α(v)]C = A[v]B = θ(α)[v]B.
Concerning the function θ we shall prove:
(8.3.12) The function θ: L(V, W) → Mn,m(F ) is an isomorphism of vector spaces. Proof. We show first that θ is a linear mapping. Let α, β ∈ L(V, W) and v ∈ V . Then
the basic formula above shows that
θ (α+ β)[v]B= [α + β(v)]C = [α(v) + β(v)]C= [α(v)]C+ [β(v)]C,
which equals
8.3 Linear mappings 153 Hence θ (α + β) = θ(α) + θ(β), and in a similar fashion it may be shown that
θ (a· α) = a(θ(α)), where a ∈ F .
Next, if θ (α)= 0, then [α(v)]C = 0 for all v ∈ V , so α(v) = 0 and α = 0. Hence
θ is injective. Both the vector spaces concerned have dimension mn, so Im(θ ) =
Mm,n(F )and θ is an isomorphism.
Example (8.3.4) Consider the dual space V∗ = L(V, F ), where V is an n-dimen-
sional vector space over a field F . Choose an ordered basisB of V and use the basis {1F} for V . Then a linear functional α ∈ V∗is represented by a row vector, i.e., by
XT where X ∈ Fn, according to the rule α(v) = XT[v]B. So the effect of a linear functional is produced by left multiplication of coordinate vectors by a row vector, (cf. Example (8.3.3)).
The effect of a change of basis. We have seen that any linear mapping between finite dimensional vector spaces can be represented by multiplication by a matrix. However the matrix depends on the choice of ordered bases of the vector spaces. The precise nature of this dependence will now be investigated.
LetB and C be ordered bases of finite dimensional vector spaces V and W over a field F , and let α : V → W be a linear mapping. Then α is represented by a matrix A over F where[α(v)]C = A[v]B. Now suppose now that two different ordered bases BandCare chosen for V and W . Then α will be represented by another matrix A.
The question is: how are A and Arelated?
To answer the question we introduce the transition matrices S and T for the re- spective changes of basesB → BandC → C(see (8.2.11)). Thus for any v∈ V and w∈ W we have
[v]B= S[v]B and [w]C = T [w]C.
Therefore
[α(v)]C= T [α(v)]C = T A[v]B = T AS−1[v]B,
and it follows that A = T AS−1. We record this conclusion in:
(8.3.13) Let V and W be non-zero finite dimensional vector spaces over the same field.
LetB, B be two ordered bases of V andC, C two ordered bases of W . Suppose further that S and T are the transition matrices for the changes of basesB → B and C → C respectively. If the linear mapping α : V → W is represented by matrices A and Awith respect to the respective pairs of bases (B, C) and (B,C), then A= T AS−1.
The case where α is a linear operator on V is especially important. Here V = W and we can takeB = C and B = C. Thus S= T and A= SAS−1, i.e., A and A are similar matrices. Consequently, the matrices which can represent a given linear
The algebra of linear operators. Let V be a vector space over a field F and suppose also that V is a ring with respect to some multiplication operation. Then V is called an F -algebra if, in addition to the vector space and ring axioms, we have the law
a(uv)= (au)v = u(av)
for all a ∈ F , u, v ∈ V . For example, the set of all n × n matrices Mn(F ) is an
F-algebra with respect to the usual matrix operations.
Now let V be any vector space over a field F ; we shall write
L(V )
for the vector space L(V , V ) of all linear operators on V . Our aim is to make L(V ) into an F -algebra. Now there is a natural product operation on L(V ), namely functional composition. Indeed, if α1, α2 ∈ L(V ), then α1α2 ∈ L(V ) by an easy check. We
claim that with this product operation L(V ) becomes an F -algebra.
The first step is to verify that all the ring axioms hold for L(V ). This is fairly routine. For example,
α1(α2+ α3)(v)= α1(α2(v)+ α3(v))= α1α2(v)+ α1α3(v)= (α1α2+ α1α3)(v).
Hence α1(α2+ α3)= α1α2+ α1α3. Once the ring axioms have been dealt with, we
have to check that a(α1α2)= (aα1)α2 = α1(aα2). This too is not hard; indeed all
three mappings send v to a(α1(α2(v))). Therefore L(V ) is an F -algebra.
A function α: A1→ A2between two F -algebras is called an algebra isomorphism
if it is bijective and it is both a linear mapping of vector spaces and a homomorphism of rings.
(8.3.14) Let V be a vector space with finite dimension n over a field F . Then L(V )
and Mn(F ) are isomorphic as F -algebras.
Proof. Choose an ordered basisB of V and let : L(V ) → Mn(F )be the function which associates with a linear operator α the n× n matrix that represents α with respect toB. Thus [α(v)]B = (α)[v]Bfor all v ∈ V . Clearly is bijective, so to prove that it is an F -algebra isomorphism we need to establish
(α+ β) = (α) + (β), (a · α) = a · (α) and (αβ) = (α)(β).
For example, take the third statement. If v∈ V , then
[αβ(v)]B = (α)[β(v)]B= (α)((β)[v]B)= (α)(β)[v]B.
Therefore (αβ)= (α)φ(β).
Thus (8.3.14) tells us is in a precise way that linear operators on an n-dimensional vector space over F behave in very much the same manner as n× n matrices over F .