(3) Let V denote a vector space over the field F and let v,..., Un E...

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(3) Let V denote a vector space over the field F and let v,..., Un E V. (a) Show that span(vn, 2,. , Un) (b) Show that span (

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Answer 1

Answer #1

(a) . Suppose x € span ( v₁ , v₂ , v₃,....,v_n) .

Then there exists scalars c₁ , c₂ , c₃ , ...., c_n € F such that

x = c₁v₁ + c₂v₂ + .....+ c_nv_n

Now , c₁v₁ € span (v₁) ; c₂v₂ € span (v₂) ,......., c_nv_n € span(v_n)

$\Rightarrow$ c₁v₁ + c₂v₂ + .....+ c_nv_{n €} span (v₁) + span(v₂) +....+ span(v_n)

$\Rightarrow$ x € span(v₁) + span (v₂) +....+span(v_n)

So x € span ( v_{1 ,} v₂ , ...., v_n) $\Rightarrow$ x€ span (v₁) + span (v₂) +....+span(v_n)

Hence ,

span (v₁ , v₂,....,v_n) $\subset$ span(v₁) + span(v₂) +....+ span(v_n) ..........(1)

Conversely suppose , y € span(v₁) + span(v₂) +....+ span(v_n)

$\Rightarrow$ there exist d₁ , d₂ , ....d_n € F such that y = d₁ v₁ + d₂ v₂ + ...+ d_n v_n

$\Rightarrow$ y € span (v₁ , v₂,....,v_n)

So y € span(v₁) + span(v₂) +....+ span(v_n) $\Rightarrow$ y€ span (v₁ , v₂,....,v_n)

Hence ,

span (v₁) +span(v₂) +....+span(v_n) $\subset$ span ( v₁ , , v₂,....,v_n) ..........(2)

From (1) and (2) we get that ,

span ( v₁ , , v₂,....,v_n) = span (v₁) +span(v₂) +....+span(v_n) .

(b) we have already prove that span ( v₁ , , v₂,....,v_n) = span (v₁) +span(v₂) +....+span(v_n) . To prove that span ( v₁ , , v₂,....,v_n) = span (v₁) $\oplus$ span(v₂) $\oplus$ ... $\oplus$ span(v_n) we only need to prove that ,

span (v₁) $\cap$ span (v₂) $\cap$ ..... $\cap$ span (v_n) = {0}

Let x € span (v₁) $\cap$ span (v₂) $\cap$ ..... $\cap$ span (v_n)

$\Leftrightarrow$ x € span (v₁) , x € span (v₂) ,.....,x€ span (v_n)

$\Leftrightarrow$ x = a₁v₁ , x=a₂v₂ , .....,x = a_nv_n.

$\Leftrightarrow$ a₁v₁ = a₂v₂ = ......= a_nv_n

( v₁ , v₂ , .....,v_n) is linearly dependent , a contradiction .

Hence , span (v₁) $\cap$ span(v₂) $\cap$ ..... $\cap$ span (v_n) ={0}

.

If you have any doubt or need more clarification at any step please comment.

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Answer 2

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