Let dimK=m and let {β1,β2,…,βm} be a basis of K . Then dimR=n−m .
Since K⊆V , it follows that vectors {β1,β2,…,βm} can be complemented to a basis of V .
So, let β={β1,β2,…,βn} be a basis of V .
Let γ={γ1,γ2,…,γn−m} be a basis of R .
Now we define a linear transformation L:V→W for basis vectors:
Lβ1=…=Lβm=0 and Lβm+1=γ1 , … , Lβn=γn−m .
For v=a1β1+a2β2+…+anβn∈V we have Lv=am+1γ1+…+anγn−m
Im L={Lv ∣ v∈V}={am+1γ1+…+anγn−m∣ai∈F}=R
Lv=am+1γ1+…+anγn−m=0 if and only if am+1=…=an=0 .
Therefore, Lv=0 only for v=a1β1+…+amβm , where ai∈F . It means that Ker L=K .
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