Linear algebra/Linear combination

Definition: Let V be an arbitrary vector space over a field K. Let {v1,…,vn} belong to V. Then, given a1,a2,a3,…,an, elements of K, an expression of the type:

             a1v1+a2v2+a3v3+…+anvn is called a linear combination of the vectors {v1,v2,…vn}.

This definition, alone, is not too useful. The next question is, what can we say about a set of vectors and all their linear combinations. Do {v1,v2,…,vn} form a subspace? of V?

First, there are many examples of linear combinations of vectors from a vector space.

Example I:

Given the vector space R³, consider the elements of R³, {E1,E2,E3} where E1=(1,0,0), E2=(0,1,0) and E3=(0,0,1). Given any vector in R³, say (a1,a2,a3), it is a linear combination of E1,E2, and E3.

Proof: Given (a1,a2,a3) in R³, we can find {x1,x2,x3} in R such that (a1,a2,a3)= x1E1+x2E2+x3E3.
Since a1=a1E1=a1(1,0,0)=a1 and, similarly, with a2 and a3, then (a1,a2,a3)=a1*E1+a2*E2+a3*E3. So x1=a1,x2=a2, and x3=a3.

Example 2:

Given the vector space of functions from R into R, F:R->R. Denote this vector space F(R). Consider two functions f, g that are elements of F(R), f(t)=e^t and f(t)=e^2t. What are some of the linear combinations of f and g?
3e^t+5e^2t, 2e^t+πe^2t and in fact all functions h(t)=ae^t+be^t2 where a,b are elements of R.
This includes the zero function, 0, since 0*et+0*et2 is a linear combination of f and g.

Example 3: Consider the vector space P₃=all polynomials of degree less than equal to 3 with real coefficients.
(Prove yourself that P₃ is a vector space.)
Given the polynomial, –1+x², in P₃, show that it is a linear combination of the polynomials, p=1+x+x³ and q=-x-x²-x³, elements of P₃.

Proof:

Find a,b be real numbers such that a*(1+x+x³)+b(-x-x²-x³)=-1+x².
Then, a+ax+ax³-bx-bx²-bx³=-1+x².
And, a+(a-b)x-bx²+(a-b)x³=-1+x2.
Equating coefficients of the same powers,

  a=-1, a-b=0, -b=1, a-b=0.

Then a=b=-1.
Hence, -1*p-1*q=-1+x².

Now for the proof that given a subset of a vector space and all of the linear combinations of the elements of this subset, one has a subspace?.

Theorem: Let V be an arbitrary vector space over a field K, and let v1,v2,…,vn be elements of V. Then the set of all linear combinations of v1,v2,…,vn is a subspace? W of V.

Proof:

Property I:

Let x1,x2,…,xn and y1,y2,…,yn be elements of K. Then S=(x1v1,x2v2,…xnvn) and T=(y1v1,y2v2,…ynvn) are elements of W.
S+T=(x1v1+x2v2+…xnvn)+(y1v1+y2v2+…ynvn)= ((x1+y1)v1,(x2+y2)v2,…(xn+yn)vn).
Since K is a field, xi+yi is an element of K, for all i={1,2,…n}, so S+T=((x1+y1)v1,(x2+y2)v2,…,(xn+yn)vn) is a linear combination of v1,v2,…vn and is an element of W.

Property 2:

Let c be an element of K and S=(x1v1,x2v2,…,xnvn).
Then c*S=c*(x1v1,x2v2,…,vn)=(c*x1v1,c*x2v2,…c*xnvn).
Since K is a field, cxi is an element of K for all i={1,2,…n}.
Then c*T= (cx1v1,cx2v2,…cxnvn) is a linear combination of v1,v2,…vn and is an element of W.

Property 3:

Since K is a field, K has an additive identity, 0. Let T=(v1,v2,…vn).
Consider 0*T= 0*(v1,v2,…,vn)=(0v1,0v2,…0vn).
Since K is a field, 0*vi=0 for all I+{1,2,…n}.
Then 0*T=(0, 0,…0)= 0. So 0 is an element of W.