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.