Homomorphisms of a Vector Space forms a Basis Assume V is a vector space with finite dimensions over F with the basis T. 
Let us define $\phi_t : T \to F$ for all $t \in T$ by the following:
\begin{cases} 
      1 & x=t \\
      0 & x \neq t 
   \end{cases}
Now extend $\phi_t$ linearly to V s.t. $∀t \in T$ $ \phi_t$ : V → F is a linear homomorphism.
Show that the homomorphisms, which is $\phi_t$ for t ∈ T, form a basis for $Hom_F(V,F)$ over F. Show that if V is not finite-dimensional, then $\phi_t$, t ∈ T, could not form a basis for $Hom_F (V, F )$ over F.
How would i show that the homomorphisms form a basis? I am a little confused on what exactly is $Hom_F (V, F )$ over F. I know that by definition of basis, for the vectors to span the entire vector space, all the vectors need to be linearly independent and therefore must form a basis. 
 A: Note that $\operatorname{Hom}(V,F)$ is the set of linear maps $\phi:V \to F$.  To show that the $\phi_t$ form a basis, we need to show that they span $\operatorname{Hom}(V,F)$ and that they are linearly independent.
To show that they are linearly independent, consider any sum $\phi = \sum_{t \in T} a_t \phi_t = 0$ with $a_t \in F$.  We want to show that if $\phi = 0$ (which is to say that $\phi(v) = 0$ for all $v \in V$), then we must have $a_t = 0$ for all $t \in T$. With that in mind, suppose that $\phi$ as defined above satisfies $\phi = 0$.  First, verify that $\phi(t) = a_t$.  Since $\phi = 0$, we have $\phi(t) = 0$, which means that each $a_t$ is $0$ as desired.
To show that the maps $\phi_t$ spans $\operatorname{Hom}(V,F)$, we have to consider an arbitrary linear $\phi: V \to F$ and show that there exist coefficients $a_t$ such that $\phi = \sum_{t \in T} a_t \phi_t$. Verify that if we define $\psi = \sum_{t \in T}\phi(t) \phi_t$, then we will have $\psi(v) = \phi(v)$ for all $v \in V$.  In other words, setting $a_t = \phi(t)$ gives us the desired linear combination.

As for the case of an infinite-dimensional space $V$, the key is to note that only finite sums are defined over an algebraic vector space.  Thus, the set $\{\phi_t : t \in T\}$ fails to span $\operatorname{Hom}(V,F)$.
