Is dot product a kind of linear transformation

Question

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.

Answer 1

There are a couple ways to view a dot product as a linear map by changing your view slightly.

The map $\langle \cdot,\cdot \rangle : V\times V\to F$ is not linear, it is what we call bilinear, which means that it is linear in each variable. I.e. a map $B : V\times V' \to W$ is bilinear if for all fixed $v\in V$, and all fixed $v'\in V'$ the maps $u'\mapsto B(v, u')$ and $u\mapsto B(u, v')$ are linear. (Though these maps are equal in the case of a dot product, since it is symmetric, so you just need to check that one is linear. Symmetric meaning that $\langle v, w\rangle= \langle w, v\rangle$)

The other view is perhaps a little more faithful to the idea of viewing the dot product as a linear map, but essentially equivalent. Though perhaps a little more abstract (though such judgments are inherently subjective).

The idea is that we can take a bilinear map $B: V\times V' \to W$ and turn it into a linear map $\tilde{B}: V\to \newcommand\Hom{\operatorname{Hom}}\Hom_F(V',W)$. Where $\Hom_F(V',W)$ denotes the vector space of $F$-linear maps from $V'$ to $W$. We define $\tilde{B}(v) = v'\mapsto B(v,v')$. Then one can use the defining property of bilinear maps given above to show that $\tilde{B}$ is linear, and for any $v\in V$, $\tilde{B}(v)$ is a linear map from $V'$ to $W$. This process is called currying. Then $\tilde{B}$ is basically the same as $B$, since we can recover $B$ from $\tilde{B}$ from the fact that $B(v,v')=(\tilde{B}v)v'$ (sorry for changing notation to parenthesis-less function application, I just think it's much more readable here).

Thus one can curry the dot product to get a linear map, call it $D$ from $V$ to $\Hom_F(V,F)$. In general, $\Hom_F(V,F)$ is a vector space called $V$-dual, often written $V^*$, so we can say $D$ is a linear map from $V$ to $V^*$. I.e., we can view the dot product as being equivalent to a particular nice linear map from $V$ to $V^*$.

Answer 2

This answer considers not just the dot product but in general bilinear mappings to a different vector space: $U\times V\to W$. Formally their definition goes along the lines given in the answer of jgon. Only the domain and codomain are allowed to be more general. There is a construction called tensor product of two vector spaces $U$ and $V$. This tensor product, denoted by $U\otimes V$ is a device to "convert" bilinear mappings to linear mappings. So every bilinear mapping above will be equivalently represented by a linear transformation $U\otimes V\to W$.

This is akin to quotient construction, say in groups, any homomorphism $f\colon G\to H$ is equivalent to an injective homomorphism $\tilde f \colon G/\ker f\to H$. Please wikipedia pages for construction of tensor products.

Answer 3

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.