(a) . Suppose x € span ( v1 , v2 , v3,....,vn) .
Then there exists scalars c1 , c2 , c3 , ...., cn € F such that
x = c1v1 + c2v2 + .....+ cnvn
Now , c1v1 € span (v1) ; c2v2 € span (v2) ,......., cnvn € span(vn)
c1v1 + c2v2 + .....+ cnvn € span (v1) + span(v2) +....+ span(vn)
x € span(v1) + span (v2) +....+span(vn)
So x € span ( v1 , v2 , ...., vn) x€ span (v1) + span (v2) +....+span(vn)
Hence ,
span (v1 , v2,....,vn) span(v1) + span(v2) +....+ span(vn) ..........(1)
Conversely suppose , y € span(v1) + span(v2) +....+ span(vn)
there exist d1 , d2 , ....dn € F such that y = d1 v1 + d2 v2 + ...+ dn vn
y € span (v1 , v2,....,vn)
So y € span(v1) + span(v2) +....+ span(vn) y€ span (v1 , v2,....,vn)
Hence ,
span (v1) +span(v2) +....+span(vn) span ( v1 , , v2,....,vn) ..........(2)
From (1) and (2) we get that ,
span ( v1 , , v2,....,vn) = span (v1) +span(v2) +....+span(vn) .
(b) we have already prove that span ( v1 , , v2,....,vn) = span (v1) +span(v2) +....+span(vn) . To prove that span ( v1 , , v2,....,vn) = span (v1) span(v2) ...span(vn) we only need to prove that ,
span (v1) span (v2) ..... span (vn) = {0}
Let x € span (v1) span (v2) ..... span (vn)
x € span (v1) , x € span (v2) ,.....,x€ span (vn)
x = a1v1 , x=a2v2 , .....,x = anvn.
a1v1 = a2v2 = ......= anvn
( v1 , v2 , .....,vn) is linearly dependent , a contradiction .
Hence , span (v1) span(v2) ..... span (vn) ={0}
.
If you have any doubt or need more clarification at any step please comment.
(3) Let V denote a vector space over the field F and let v,..., Un E...
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