Question

*Please show work and explain*

Prove that a vector space V over a field F is isomorphic to the vector space L(F,V) of all linear maps from F to V.

Answer 1

Let V(F) be a vector space and V' be set of all linear transformations T : VR. We have to show V is sioorphic to V'.

Let {b1,b2,…,bn}. Define linear maps h_i:V→R such that

h_i(b_i)=1 and h_i(b_j)=0 when i≠j.

We will show that {h1,h2,…,hn} is a basis for V'. Let us assume that the dimension of V is m (if m=0, it is trivial , so let us assume n>0). Consider {s1,s2,…,sn} as basis of V . let us consider m linear maps defined as

h_i(s_j)=1 if i=j and h_i(s_j)=0 if i≠j.

Our aim is to prove that these maps form basis of V'. We first show that they span V' Let v be any arbitary vector in V such that

v=c₁s₁+c₂s₂+⋯+c_ms_m

Now h_i(v)=h_i(c₁s₁+c₂s₂⋯+c_ms_m)=c₁h_i(s₁)+c2h_i(s₂)+⋯+c_mh_i(s_m)

Now, applying an arbitrary linear map L to v . Then

L(v)=L(c₁s₁+c₂s₂⋯+c_ms_m)=c₁L(s₁)+c₂L(s₂)+⋯+c_mL(s_m) .

But L(s_i)  just a real number, and we have

L(v)=h₁(v)L(s₁)+h₂(v)L(s₂)+⋯+h_m(v)L(s_m)

Hence L is a linear combination of the h_i' s, so the h_i' s span V'.

Now It remains to show that the h_i's are linearly independent. For this suppose that

r₁h₁+r₂h₂+⋯+r_mh_m=0

where r_i∈R. Hence

r₁h₁(v)+r₂h₂(v)+⋯+r_mh_m(v)=0

for every v∈V.

However, choosing v=s_i  gives us

0=r₁h₁(v_i)+r₂h₂(v_i)+⋯+r_mh_m(v_i)=r_i   

and so

r₁=r₂=⋯=r_m=0

hence h_i's are linearly independent and therefore form a basis (of m elements), so V and V' are isomorphic. Hence the result.