Okay, so it really isn't that strange but I haven't come into contact with anything like it before, so I want to make sure I am interpreting it correctly.
Let $V$ be a finite dimensional real inner product space. Define a function $\Phi: V\rightarrow V^*$ from the vector space $V$ to its dual space $V^*$ by: $$\Phi(v) = \{w\mapsto\langle w,v \rangle\}$$
Now I know that the dual space of $V$ is defined to be the set of linear transformations from the vector space to its field of scalars; so in this case $V^* = \{T \: \vert \: T:V\rightarrow \mathbb{R}\}$
Moreover, we just did a theorem in class that said for any linear transformation $T:V\rightarrow \mathbb{F}$ there exists a unique vector $y\in V$ such that $T(x) = \langle x,y\rangle$. Initially I read the above as $w$ is representing the unique vector $y$ for the generic linear transforms. That is to say, if $T:V\rightarrow \mathbb{R}$, then $w \equiv T(x) = \langle w,x\rangle$.
This all seems quite clumsy to me and I wanted to make sure I have definition of the function right in my head before I continue with the question. Thanks for your help!
First of all, $V^*=\{T\colon V\longrightarrow\mathbb{R}\,|\,T\text{ is linear}\}$.
Besides, it is not correct to write that $w\equiv T(x)$. What happens here is that, for every linear map $T\colon V\longrightarrow$, there is one and only one vector $w$ such that $T=\langle w,\cdot\rangle$, which means that$$(\forall v\in V):T(v)=\langle w,v\rangle.$$