TITLE: Is dot product a kind of linear transformation
QUESTION [7 upvotes]: I wonder in the field of Linear Algebra, if the dot product also referred to as an inner product:
$$ \langle u,v\rangle =u\cdot v=u^Tv = u_1v_1 + u_2v_2+...+ u_nv_n, \quad \text{for} \quad u,v \in\mathbb{R}^n$$
can be categorized in a type of linear transformation.
I'm quite confused here, that the definition of $\langle \cdot,\cdot \rangle: V\times V\to F$, is a map of all vectors in a vector space, which is similar to Matrix Multiplication (a way to represent linear transformation), but the properties in the inner product is a bit different from the former.
It's kind of transforming from one vector space to another.
So the image or codomain of this linear transformation is the inner product space, while the domain of this linear transformation is the original vector space.

REPLY [4 votes]: The dot product isn't a linear transformation, but it gives you a lot of linear transformations: if you think of $\langle v,w \rangle$ as a function of $v$, with $w$ fixed, then it is a linear transformation $\mathbb R^n \rightarrow \mathbb R$, sending an $n$-dimensional vector $v$ to the one dimensional vector $\langle v,w\rangle$.  
You can also fix $v$, and think of $\langle v, w \rangle$ as a function of $w$. Then this also defines a linear transformation $\mathbb R^n \rightarrow \mathbb R$.
In other words, the dot product is linear in each variable.
Actually, there is a way in which you can think of the dot product as a linear transformation.  Consider the function (not a linear transformation, just a function) $f: \mathbb R^n \times \mathbb R^n \rightarrow \operatorname{Mat}_n(\mathbb R)$ given by the formula
$$f(a_1, ... , a_n, b_1, ... , b_n) = C$$
where $C$ is the $n$ by $n$ matrix whose $ij$th entry is $a_ib_j$.  Then there exists a unique linear transformation $T: \operatorname{Mat}_n(\mathbb R) \rightarrow \mathbb R$ such that 
$$T(f(v,w))$$
is equal to the dot product of $v$ and $w$ for all $v, w \in \mathbb R^n$ (can you see what $T$ is, and why it is unique?).  This is a special case of the more general principle that multilinear maps identify with linear transformations on the tensor product.