9 Pasivos Financieros
CAPITULO 1.- GASTOS DE PERSONAL
A concept of critical importance in vector space theory is linear independence. Let
V be a vector space over a field F and let X be a non-empty subset of V . Then X is called linearly dependent if there exist distinct vectors x1, x2, . . . , xkin X and scalars
a1, a2, . . . , ak∈ F , not all zero, such that
a1x1+ a2x2+ · · · + akxk= 0.
This amounts to saying that one of the xican be expressed as a linear combination of the others. For if, say, ai= 0, we can solve for xi, obtaining
xi= k j=1 j=i (−ai−1)ajvj.
A subset which is not linearly dependent is said to be linearly independent. For example, the elementary vectors E1, E2, . . . , En form a linearly independent subset of Fnfor any field F .
Homogeneous linear systems. To make any significant progress with linear inde- pendence, a basic result about systems of linear equations is needed. Let F be a field and consider a system of m linear equations over F
a11x1+ · · · + a1nxn= 0 a21x1+ · · · + a2nxn= 0 .. . am1x1+ · · · + amnxn = 0
Here aij ∈ F and x1, . . . , xnare the unknowns. This system has the trivial solution
x1= x2= · · · = xn= 0. The interesting question is whether there are any non-trivial solutions. A detailed account of the theory of systems of linear equations can be found in any book on linear algebra, for example [9]. For our purposes the following result will be sufficient.
8.2 Linear independence, basis and dimension 139
(8.2.1) If the number of equations m is less than the number of unknowns n, then the
homogeneous linear system has a non-trivial solution.
Proof. We adopt a method of systematic elimination which is often called Gaussian elimination. We may assume that a11 = 0 – if this is not true, replace equation 1 by
the first equation in which x1 appears. Since equation 1 can be multiplied by a−111,
we may also assume that a11= 1. Then, by subtracting multiples of equation 1 from
equations 2 through m, eliminate x1from these equations.
Next find the first of equations 2 through m which contains an unknown with smallest subscript > 1, say xi2. Move this equation up to second position. Now make
the coefficient of xi2equal to 1 and subtract multiples of equation 2 from equations 3
through m. Repeat this procedure until the remaining equations involve no more unknowns, i.e., they are of the form 0 = 0. Say this happens after r steps. At this point the matrix of coefficients is in row echelon form and has r linearly independent rows.
Unknowns other than x1= xi1, xi2, . . . , xircan be given arbitrary values. The non-
trivial equations may then be used to solve back for xi1, . . . , xir. Since r≤ m < n, at
least one unknown can be given an arbitrary value, so the linear system has a non-trivial
solution.
This result will now be used to establish the fundamental theorem about linear dependence in vector spaces.
(8.2.2) Let v1, v2, . . . , vk be vectors in a vector space V over a field F . Then any
subset of Fv1, v2, . . . , vk containing k + 1 or more vectors is linearly dependent.
Proof. Let u1, u2, . . . , uk+1 ∈ S = F v1, . . . , vk. It is enough to show that {u1, u2, . . . , uk+1} is a linearly dependent set. This amounts to finding field elements
a1, a2, . . . , ak+1, not all zero, such that a1u1+ · · · + ak+1uk+1= 0.
Since ui ∈ S, there is an expression ui = d1iv1+ · · · + dkivkwhere dj i ∈ F . On substituting for the ui, we obtain
a1u1+ · · · + ak+1uk+1 = k+1 i=1 ai k j=1 dj ivj = k j=1 k +1 i=1 aidj i vj.
Therefore a1u1+ · · · + ak+1uk+1will equal 0 if the aisatisfy the equations k+1
i=1
aidj i = 0, j = 1, . . . , k.
But this is a system of k linear homogeneous equations in the k+ 1 unknowns ai. By (8.2.1) there is a non-trivial solution a1, . . . , ak+1. Therefore{u1, . . . , uk+1} is
Corollary (8.2.3) If a vector space V can be generated by k elements, then every
subset of V with k+ 1 or more elements is linearly dependent.
Bases. A basis of a vector space V is a non-empty subset X such that: (i) X is linearly independent;
(ii) X generates V .
Notice that these are contrasting properties since (i) implies that X is not too large and (ii) that X is not too small.
For example, the elementary vectors E1, E2, . . . , En form a basis of the vector space Fn – called the standard basis – for any field F . More generally a basis of
Mm,n(F )is obtained by taking the m×n matrices over F with just one non-zero entry which is required to be 1F.
A important property of bases is unique expressibility.
(8.2.4) If{v1, v2, . . . , vn} is a basis of a vector space V over a field F , then every
vector v in V is uniquely expressible in the form v= a1v1+ · · · + anvnwith ai ∈ F .
Proof. In the first place such expressions for v exist. If v in V had two such expressions v=ni=1aivi=
n
i=1bivi, then we would have n
i=1(ai− bi)vi= 0, from which it follows that ai = biby linear independence of the vi. This result shows that a basis may be used to introduce coordinates in a vector space. Suppose that V is a vector space over field F and thatB = {v1, v2, . . . , vn} is a basis of V with its elements written in a specific order, i.e., an ordered basis. Then by (8.2.4) each v ∈ V has a unique expression v =ni=1civiwith ci ∈ F . So v is determined by the column vector in Fnwhose entries are c1, c2, . . . , cn; this is called the coordinate column vector of v with respect toB and it is written
[v]B.
Coordinate vectors provide a concrete representation of the vectors of an abstract vector space.
The existence of bases. There is nothing in the definition of a basis to make us certain that bases exist. Our first task will be to show that this is true for finitely generated non-zero vector spaces. Notice that the zero space does not have a basis since it has no linearly independent subsets.
(8.2.5) Let V be a finitely generated vector space and suppose that X0is a linearly
8.2 Linear independence, basis and dimension 141
Proof. Suppose that V can be generated by m vectors. Then by (8.2.3) no linearly
independent subset of V can contain more than m vectors. It follows that X0 is
contained in a largest linearly independent subset X; for otherwise it would be possible to form ever larger finite linearly independent subsets containing X0.
We complete the proof by showing that X generates V . If this is false, there is a vector u in V − F X . Then u ∈ X, so X = X ∪ {u} and X ∪ {u} must be linearly independent by maximality of X. Writing X = {v1, . . . , vk}, we deduce that there is a relation of the type
a1v1+ · · · + akvk+ bu = 0,
where a1, . . . , ak, b∈ F and not all of these scalars are 0. Now b cannot equal 0: for otherwise a1v1+· · ·+akvk= 0 and a1= · · · = ak= 0 since v1, . . . , vkare known to be linearly independent. Therefore u= −b−1a1v1− · · · − b−1akvk∈ F X, which
is a contradiction.
Corollary (8.2.6) Every finitely generated non-zero vector space V has a basis.
Indeed, since V = 0, we can choose a non-zero vector v from V and apply (8.2.5) with X0= {v}.
In fact every infinitely generated vector space has a basis, but advanced methods are needed to prove this – see (12.1.1) below.
Dimension. A vector space usually has many bases, but it is a fact that all of them have the same number of elements.
(8.2.7) Let V be a finitely generated non-zero vector space. Then any two bases of V
have the same number of elements.
Proof. In the first place a basis of V is necessarily finite by (8.2.3). Next let {u1, . . . , um} and {v1, . . . , vn} be two bases. Then V = v1, . . . , vn and so by (8.2.2) there cannot be a linearly independent subset of V with more than n elements. Therefore m≤ n. By the same reasoning n ≤ m, so we obtain m = n, as required.
This result enables us to define the dimension dim(V )
of a finitely generated vector space V . If V = 0, define dim(V ) to be 0, and if V = 0, let dim(V ) be the number of elements in a basis of V . By (8.2.7) this definition is unambiguous. In the future we shall speak of finite dimensional vector spaces instead of finitely generated ones.
(8.2.8) Let X1, X2, . . . , Xk be vectors in Fn, F a field. Let A = [X1, X2, . . . , Xk]
be the n× k matrix which has the Xi as columns. Then dim(FX1, . . . , Xk) = r
Proof. We shall need to use some elementary facts about matrices here. In the first
place, S = F X1, . . . , Xk is the column space of A, and it is unaffected when column operations are applied to A. Now by applying column operations to A, just as we did for row operations during Gaussian elimination in the proof of (8.2.1), we can replace
Aby a matrix with the same column space S which has the so-called column echelon
form with r non-zero columns. Here r is the rank of A. These r columns are linearly
independent, so they form a basis of S (if r > 0). Hence dim(S)= r. Next we consider the relation between the dimension of a vector space and that of a subspace.
(8.2.9) If V is a vector space with finite dimension n and U is a subspace of V , then
dim(U )≤ dim(V ). Furthermore dim(U) = dim(V ) if and only if U = V .
Proof. If U = 0, then dim(U) = 0 ≤ dim(V ). Assume that U = 0 and let X be
a basis of U . By (8.2.5) X is contained in a basis Y of V . Hence dim(U ) = |X| ≤ |Y | = dim(V ).
Finally, suppose that 0 < dim(U )= dim(V ), but U = V . Then U = 0. Using the notation of the previous paragraph, we see that Y must have at least one more element than X, so dim(U ) < dim(V ), a contradiction. The next result can simplify the task of showing that a subset of a finite dimensional vector space is a basis.
(8.2.10) Let V be a finite dimensional vector space with dimension n and let X be a
subset of V with n elements. Then the following statements about X are equivalent:
(a) X is a basis of V ;
(b) X is linearly independent; (c) X generates V .
Proof. Of course (a) implies (b) and (c). Assume that (b) holds. Then X is a basis
of the subspace it generates, namely FX; hence dim(F X) = n = dim(V ) and (8.2.9) shows that FX = V . Thus (b) implies (a).
Finally, assume that (c) holds. If X is not a basis of V , it must be linearly dependent, so one of its elements can be written as a linear combination of the others. Hence
n= dim(V ) = dim(F X) < |X| = n, a contradiction.
Change of basis. As previously remarked, vector spaces usually have many bases, and a vector may be represented with respect to each basis by a coordinate column vector. A natural question is: how are these coordinate vectors related?
LetB = {v1, . . . , vn} and B = {v1, . . . , vn} be two ordered bases of a finite dimensional vector space V over a field F . Then each vican be expressed as a linear
8.2 Linear independence, basis and dimension 143 combination of v1, . . . , vn, say vi= n j=1 sj ivj,
where sj i ∈ F . The change of basis B → B is described by the transition matrix
S = [sij]. Observe that S is n × n and its i-th column is the coordinate vector [vi]B. To understand how S determines the change of basisB→ B, choose an arbitrary vector v from V and write v = ni=1civi where c1, . . . , cn are the entries of the coordinate vector[v]B. Replace vibynj=1sj ivj to get
v= n i=1 ci n j=1 sj ivj = n j=1 n i=1 sj ici vj.
Therefore the entries of the coordinate vector[v]Bareni=1sj icifor j = 1, 2, . . . , n. This shows that
[v]B = S[v]B,
i.e., left multiplication by the transition matrix S transforms coordinate vectors with respect toBinto those with respect toB.
Notice that the transition matrix S must be non-singular. For otherwise, by standard matrix theory, there would exist a non-zero X ∈ Fn such that SX = 0; however, if
u ∈ V is defined by [u]B = X, then [u]B = SX = 0, which can only mean that
u= 0 and X = 0. From [v]B= S[v]Bwe deduce that S−1[v]B= [v]B. Thus S−1 is the transition matrix for the change of basisB → B. We sum up these conclusions in:
(8.2.11) LetB and B be two ordered bases of an n-dimensional vector space V . Define S to be the n× n matrix whose i-th column is the coordinate vector of the i-th vector ofB with respect to B. Then S is non-singular: also [v]B = S[v]B and
[v]B= S−1[v]Bfor all v in V .
Example (8.2.1) Let V be the vector space of all real polynomials in t with degree at
most 2. ThenB = {1, t, t2} is clearly a basis of V , and so is B= {1 + t, 2t, 4t2− 2} since this set can be verified to be linearly independent. Write down the coordinate vectors of 1+ t, 2t, 4t2− 2 with respect to B as columns of the matrix
S = 11 02 −20 0 0 4 .
This is the transition matrix for the change of basisB → B. The transition matrix forB → Bis S−1= 1 0 1 2 −1 2 1 2 − 1 4 0 0 14 .
For example, to express f = a + bt + ct2in terms of the basisB, we compute [f ]B= S−1[f ]B= S−1 ab c = a+ c 2 −1 2a+ 1 2b− 1 4c 1 4c .
Thus f = (a + c2)(1+ t) + (−12a+ 12b− 14c)(2t)+14c(4t2− 2), which is clearly correct.
Dimension of the sum and intersection of subspaces. Since a vector space V is an additively written abelian group, one can form the sum of two subspaces U , W ; thus
U + W = {u + w | u ∈ U, w ∈ W}. It is easily verified that U + W is a subspace of V. There is a useful formula connecting the dimensions of the subspaces U+ W and
U ∩ W.
(8.2.12) Let U and W be subspaces of a finite dimensional vector space V . Then
dim(U+ W) + dim(U ∩ W) = dim(U) + dim(W).
Proof. If U = 0, then U + W = W and U ∩ W = 0; in this case the formula is
certainly true. Thus we can assume that U = 0 and W = 0.
Choose a basis for U∩ W, say z1, . . . , zr, if U∩ W = 0; should U ∩ W be 0, just ignore the zi. By (8.2.5) we can extend{z1, . . . , zr} to bases of U and of W, say
{z1, . . . , zr, ur+1, . . . , um} and {z1, . . . , zr, wr+1, . . . , wn} .
Now the vectors z1, . . . , zr, ur+1, . . . , un vr+1, . . . , vn surely generate U + W: for any vector of U + W is expressible in terms of them. We claim that these elements are linearly independent.
To establish the claim suppose there is a linear relation r i=1 eizi+ m j=r+1 cjuj + n k=r+1 dkwk = 0
where ei, cj, dkare scalars. Then n k=r+1 dkwk= r i=1 (−ei)zi+ m j=r+1 (−cj)uj,
which belongs to U and to W , and so to U ∩ W. Hencenk=r+1dkwk is a linear combination of the zi. But z1, . . . , zr, wr+1, . . . , wn are linearly independent, which means that dk= 0 for all k. Our linear relation now reduces to
r i=1 eizi+ m j=r+1 cjuj = 0.
8.2 Linear independence, basis and dimension 145 But z1, . . . , zr, ur+1, . . . , um are linearly independent. Hence cj = 0 and ei = 0, which establishes the claim of linear independence.
Finally, dim(U+ W) equals the number of the vectors z1, . . . , zr, ur+1, . . . , um,
vr+1, . . . , vn, which is
r+ (m − r) + (n − r) = m + n − r = dim(U) + dim(W) − dim(U ∩ W),
and the required formula follows.
Direct sums of vector spaces. Since a vector space V is an additive abelian group, we can form the direct sum of subspaces U1, . . . , Ukof V – see 4.2. Thus
U = U1⊕ · · · ⊕ Uk where U = {u1+· · ·+uk| ui ∈ Ui} and Ui∩
j=iUj = 0. Clearly U is a subspace of V . Note that by (8.2.12) and induction on k
dim(U1⊕ · · · ⊕ Uk)= dim(U1)+ · · · + dim(Uk).
Next let{v1, . . . , vn} be a basis of V . Then clearly V = F v1 ⊕ · · · ⊕ F vn, so that we have established:
(8.2.13) An n-dimensional vector space is the direct sum of n 1-dimensional sub-
spaces.
This is also true when n= 0 if the direct sum is interpreted as 0.
Quotient spaces. Suppose that V is a vector space over a field F and U is a subspace of V . Since V is an abelian group and U is a subgroup, the quotient
V /U = {v + U | v ∈ V }
is already defined as an abelian group. Now make V /U into a vector space over F by defining scalar multiplication in the natural way:
a(v+ U) = av + U (a ∈ F ).
This is evidently a well-defined operation. After an easy check of the vector space axioms, we conclude that V /U is a vector space over F , the quotient space of U in V . The dimension of a quotient space is easily computed.
(8.2.14) Let U be a subspace of a finite dimensional space V . Then
Proof. If U = 0, the statement is obviously true because V /0 V . Assuming U = 0, we choose a basis {v1, . . . , vm} of U and extend it to a basis of V , say {v1, . . . , vm, vm+1, . . . , vn}. We will argue that {vm+1+ U, . . . , vn+ U} is a basis of
V /U.
Assume thatni=m+1ai(vi+ U) = U where ai ∈ F . Then n
i=m+1aivi ∈ U, so this element is a linear combination of v1, . . . , vm. It follows by linear independence that each ai= 0, and thus {vm+1+U, . . . , vn+U} is linearly independent. Next, if v ∈
V, write v=ni=1aivi, with scalars ai, and observe that v+U = n
i=m+1ai(vi+U) since v1, . . . , vm ∈ U. It follows that vm+1+ U, . . . , vn+ U form a basis of V /U and dim(V /U )= n − m = dim(V ) − dim(U), as required.
Two applications. To conclude this section let us show that the mere existence of a basis in a finite dimensional vector space is enough to prove two important results about abelian groups and finite fields.
Let p be a prime. An additively written abelian group A is called an elementary
abelian p-group if pa = 0 for all a in A. For example, the Klein 4-group is an
elementary abelian 2-group. The structure of finite elementary abelian p-groups is given in the next result.
(8.2.15) Let A be a finite abelian group. Then A is an elementary abelian p-group if
and only if A is a direct sum of copies ofZp.
Proof. The essential idea of the proof is to view A as a vector space over the field
Zp. Here the scalar multiplication is the natural one, namely (i+ pZ)a = ia where
i ∈ Z, a ∈ A. One has to verify that this operation is well-defined, which is true since (i+ pm)a = ia + mpa = ia for all a ∈ A. Since A is finite, it is a finite dimensional
vector space over Zp. By (8.2.13) A = A1⊕ A2⊕ · · · ⊕ An where each Ai is a 1-dimensional subspace, i.e., |Ai| = p and Ai Zp. Conversely, any direct sum of copies ofZp certainly satisfies pa = 0 for every element a and so is elementary
abelian p.
The second result to be proved states that the number of elements in a finite field is always a prime power. This is in marked contrast to the behavior of groups and rings, of which examples exist with any finite order.
(8.2.16) Let F be a finite field. Then|F | is a power of a prime.
Proof. By (6.3.9) the field F has characteristic a prime p and pa = 0 for all a ∈ F .
Thus, as an additive group, F is elementary abelian p. It now follows from (8.2.15)
8.3 Linear mappings 147