Definition 2.2.12 LetW1andW2be vector subspaces of the real vector spaceV. The sumW1+W2is calleddirectifW1∩W2 = {0V}. A direct sum is denotedW1⊕W2. Proposition 2.2.13 Let W1, W2 be vector subspaces of the real vector space V . Their sum W =W1+W2 is direct if and only if any elementv ∈ W1+W2has a uniquedecomposition asv=w1+w2withwi ∈Wi,i=1,2.
Proof We first suppose that the sumW1+W2is direct, that isW1∩W2= {0V}. If there exists an elementv∈W1+W2withv=w1+w2=w1 +w2, andwi, wi ∈Wi, thenw1−w1 =w2−w2 and such an element would belong to both W1 andW2. This would then be zero, sinceW1∩W2= {0V}, and thenw1=w1andw2=w2.
Suppose now that any element v∈W1+W2 has a unique decomposition v=w1+w2 withwi∈Wi,i =1,2. Letv∈W1∩W2; thenv∈W1 andv∈W2
which gives 0V =v−v∈W1+W2, so the zero vector has a unique decomposition.
But clearly also 0V =0V +0Vand being the decomposition for 0V unique, this gives
v=0V.
These proposition gives a natural way to generalise the notion of direct sum to an arbitrary number of vector subspaces of a given vector space.
Definition 2.2.14 LetW1, . . . ,Wn be vector subspaces of the real vector spaceV. The sumW1+ · · · +Wn is calleddirectif any of its element has a unique decom- position asv=w1+ · · · +wn withwi ∈Wi,i =1, . . . ,n. The direct sum vector subspace is denotedW1⊕ · · · ⊕Wn.
2.3 Linear Combinations 25 The setL(I)is also called thelinear spanof I.
Proposition 2.3.2 The spaceL(v1, . . . , vn)is a vector subspace of V , called the spacegeneratedbyv1, . . . , vnor thelinear spanof the vectorsv1, . . . , vn.
Proof After Proposition 2.2.2, it is enough to show that L(v1, . . . , vn)is closed for the sum and the product by a scalar. Let v, w∈L(v1, . . . , vn); it is then v=λ1v1+ · · · +λnvn and w=μ1v1+ · · · +μnvn, for scalars λ1, . . . ,λn and μ1, . . . ,μn. Recalling point(2)in the Definition2.1.1, one has
v+w=(λ1+μ1)v1+ · · · +(λn+μn)vn ∈L(v1, . . . , vn).
Next, letα∈R. Again from the Definition2.1.1(point 4)), one hasαv=(αλ1)v1+
· · · +(αλn)vn, which givesαv∈L(v1, . . . , vn). Exercise 2.3.3 The following are two examples for the notion just introduced.
(1) Clearly one hasVO2 =L(i,j)andVO3 =L(i,j,k).
(2) Letv=(1,0,−1)andw=(2,0,0)be two vectors inR3; it is easy to see that L(v, w) is a proper subset of R3. For example, the vector u=(0,1,0) /∈L(v, w). Ifuwere inL(v, w), there should beα,β ∈Rsuch that
(0,1,0)=α(1,0,−1)+β(2,0,0)=(α+2β,0,−α).
No choice ofα,β ∈ Rcan satisfy this vector identity, since the second com- ponent equality would give 1=0, independently ofα,β.
It is interesting to explore which subsets I ⊆ V yieldL(I)=V. Clearly, one has V = L(V). The example (1) above shows that there are proper subsets I ⊂ V whose linear span coincides withV itself. We already know thatV2O=L(i,j)and that V3O=L(i,j,k): bothVO3 andVO2 are generated by afinitenumber of (their) vectors. This is not always the case, as the following exercise shows.
Exercise 2.3.4 The real vector space R[x] is not generated by a finite num- ber of vectors. Indeed, let f1(x), . . . , fn(x)∈R[x]be arbitrary polynomials. Any
p(x)∈L(f1, . . . , fn)is written as
p(x)=λ1f1(x)+ · · · +λnfn(x)
with suitableλ1, . . . ,λn ∈ R. If one writesdi =deg(fi)andd=max{d1, . . . ,dn}, from Remark A.3.5 one has that
deg(p(x))=deg(λ1f1(x)+ · · · +λnfn(x))≤max{d1, . . . ,dn} =d.
This means that any polynomial of degree d+1 or higher is not contained in L(f1, . . . , fn). This is the case for anyfinite n, giving afinite d; we conclude that, if nis finite, anyL(I)withI = (f1(x), . . . ,fn(x))is a proper subset ofR[x]which can thennotbe generated by afinitenumber of polynomials.
On the other hand,R[x]is indeed the linear span of the infinite set {1,x,x2, . . . ,xi, . . .}.
Definition 2.3.5 A vector space V over R is said to be finitely generated if there exists a finite number of elements v1, . . . , vn in V which are such that V =L(v1, . . . , vn). In such a case, the set{v1, . . . , vn}is called asystem of gen- eratorsforV.
Proposition 2.3.6 Let I ⊆ V andv∈V . It holds that L({v} ∪I)=L(I) ⇐⇒ v∈L(I).
Proof “⇒” Let us assume that L({v} ∪I)=L(I). Since v∈L({v} ∪I), then v∈L(I).
“⇐” We shall prove the claim under the hypothesis that we have a finite system {v1, . . . , vn}. The inclusion L(I)⊆L({v} ∪I) is obvious. To prove the inclusionL({v} ∪I)⊆L(I), consider an arbitrary elementw∈L({v} ∪I), so that w=αv+μ1v1+ · · · +μnvn. By the hypothesis, v∈L(I) so it is v=λ1v1+ · · · +λnvn. We can then write
w=α(λ1v1+ · · · +λnvn)+μ1v1+ · · · +μnvn.
From the properties of the sum of vectors inV, one concludes thatw∈L(v1, . . . , vn)
=L(I).
Remark 2.3.7 From the previous proposition one has also the identity L(v1, . . . , vn,0V)=L(v1, . . . , vn)
for anyv1, . . . , vn ∈ V.
If I is a system of generators forV, the next question to address is whether I contains aminimalset of generators forV, that is whether there exists a setJ ⊂I (withJ =I) such thatL(J)=L(I)=V. The answer to this question leads to the notion oflinear independencefor a set of vectors.
Definition 2.3.8 Given a collectionI = {v1, . . . , vn}of vectors in a real vector space V, the elements of Iare calledlinearly independentonR, and the systemI is said to befree, if the following implication holds,
λ1v1+ · · · +λnvn =0V =⇒ λ1= · · · =λn =0R.
That is, if theonlylinear combination of elements ofI giving the zero vector is the one whose coefficients are all zero.
Analogously, an infinite systemI ⊆V is said to befreeif any of its finite subsets is free.
2.3 Linear Combinations 27 The vectors v1, . . . , vn ∈ V are said to be linearly dependent if they are not linearly independent, that is if there are scalars(λ1, . . . ,λn)=(0, . . . ,0)such that λ1v1+ · · · +λnvn=0V.
Exercise 2.3.9 It is clear thati,j,kare linearly independent inV3O, while the vec- torsv1 =i+j,v2 =j−kandv3=2i−j+3kare linearly dependent, since one computes that 2v1−3v2−v3=0.
Proposition 2.3.10 Let V be a real vector space and I = {v1, . . . , vn}be a collec- tion of vectors in V . The following properties hold true:
(i) if0V ∈ I , then I is not free,
(ii) I is not free if and only if one of the elementsvi is a linear combination of the other elementsv1, . . . , vi−1, vi+1, . . . , vn,
(iii) if I is not free, then any J⊇I is not free,
(iv) if I is free, then any J such that J ⊆I is free; that is any subsystem of a free system is free.
Proof i) Without loss of generality we suppose thatv1=0V. Then, one has 1Rv1+0Rv2+ · · · +0Rvn=0V,
which amounts to say that the zero vector can be written as a linear combination of elements inI with a non zero coefficients.
(ii) SupposeIis not free. Then, there exists scalars(λ1, . . . ,λn) = (0, . . . ,0)giv- ing the combinationλ1v1+ · · · +λnvn =0V. Without loss of generality take λ1 =0; soλ1is invertible and we can write
v1=λ−11(−λ2v2− · · · −λnvn)∈L(v2, . . . , vn).
In order to prove the converse, we start by assuming that a vectorvi is a linear combination
vi =λ1v1+ · · · +λi−1vi−1+λi+1vi+1+ · · · +λnvn.
This identity can be written in the form
λ1v1+ · · · +λi−1vi−1−vi+λi+1vi+1+ · · · +λnvn=0V.
The zero vector is then written as a linear combination with coefficients not all identically zero, since the coefficient ofvi is−1. This amounts to say that the systemI is not free.
We leave the reader to show the obvious points (iii) and (iv).