Linear algebra/Generating a vector space

< Linear algebra

HomePage | Recent changes | View source | Discuss this page | Page history | Log in |

Printable version | Disclaimers | Privacy policy

Definition: Given a vector space V over a field K, the subset of W={v1,v2,…,vn} of V, is said to generate V, if for every element of V, say, X=(x1,x2,...,xn), there exists a1,a2,...an in K such that a1v1+a2v2+…+anvn=X.
Alternatively, the subset W={v1,v2,…,vn} is said to be a span of V.

Example I:. Let U and W be subspaces of the vector space V over K. Define U+W={u+v: u is a member of U and w is a member of W}. Then U+W is a vector space and a subspace of V called the sum of U and W. And, U and W generate or span U+W.

Proof:

1. Property 1: Let v1 and v2 belong to U+W. Then v1=u1+w1 and v2=u2+w2. Then v1+v2=(u1+w1)+(u1+w2)=(u1+u2)+(w1+w2). Since u1+u2 is a member of U and w1+w2 is a member of W, v1+v2 is a member of U+W, by definition of U+W.

2. Property 2: Let v be a member of U+W and c be a member of K. Consider c*v. Since v is a member of U+W, v=u+w for some u in U and w in W. Then c*v=c*(u+w)=cu+cw. Since U and W are vector spaces, themselves, cu is a member of U and cw is a member of W. Then cv is a member of U+W.

3. Property 3: Since U and W are vector spaces, 0 is a member of both U and W. Then 0+0 =0 is a member of U+W.

It should be obvious that, given an element v=u+w of U+W, it is generated by U and W, that is, it is a linear combination of an element from U and an element from W.

Example 2: Let S be a subspace of R3, defined by S= {(s1,s2,s,3) such that s2=0}. Show that S is generated by E1=(1,0,0) and E3=(0,0,1).

Proof:

Let v belong to S. Then v=(v1,0,v3). To show that V is generated by E1 and E3, it is necessary to prove v is a linear combination of E1 and E2. But, v1*E1+v3*E3=v1(1,0,0,)+v3*(0,0,1)=(v1,0,0)+(0,0,v3)=(v1,0,v3)=v. So v=v1*E1+v3*3E3, a linear combination of E1 and E3.