∀i∈N+,∀L1,L2∈L(V,W):(L1+L2)(vi)=L1(vi)+L2(vi) by the definition of L(V,W).
So, [(L1+L2)(vi)]Bw=[L1(vi)+L2(vi)]Bw⇒[(L1+L2)(vi)]Bw=[L1(vi)]BW+[L2(vi)]BW by the definition of f(L)⇔[(L1+L2)(v)]BW=[L1(v)]BW+[L2(v)]BW∴f(L1+L2)=f(L1)+f(L2)
∀k∈F,∀i∈N+,L∈L(V,W):(kL)(vi)=kL(vi) by the definition of L(V,W).
So, [(kL)(vi)]Bw=[kL(vi)]Bw⇒[(kL)(vi)]Bw=k[L(vi)]Bw by the definition of f(L)⇔[(kL)(v)]Bw=k[L(v)]Bw∴f(kL)=kf(L)
∀i∈N+,∀L1,L2∈L(V,W):f(L1)=f(L2)⇔[L1(vi)]BW=[L2(vi)]BW⇔L1(vi)=L1(vi)
Let {v1,…,vn}=BV, then ∀v∈V:∃k1,…,kn∈F s.t. v=∑i=1nkivi
Since L1(vi)=L1(vi), ∑i=1nkiL1(vi)=∑i=1nkiL2(vi)⇔∑i=1nL1(kivi)=∑i=1nL2(kivi)⇔L1(∑i=1nkivi)=L2(∑i=1nkivi) by the definition of L(V,W).
∴L1(v)=L2(v)⇔L1=L=2
∀M1,M2∈Mm×n(F),∀v∈V:[LM1+M2(v)]BW=(M1+M2)[v]BV by the definition of the operation g(M)(M1+M2)[v]BV=M1[v]BV+M2[v]BV=[LM1(v)]BW+[LM2(v)]BW by the definition of matrix operation
∴g(M1+M2)=g(M1)+g(M2)