TITLE: Some aspects of inner products in $\mathbb R^3$
QUESTION [5 upvotes]: I read several descriptions about inner products recently due to an exercise (here it is no longer active so I like to ask again). I am still very confused about this concept. Any clarification will be greatly appreciated.

The very first definition of inner product I encountered is $\langle
    v, w\rangle =v_1w_1+v_2w_2+v_3w_3$. Or in general, it is a function
defined on $\mathbb R^3\times\mathbb R^3$ into $\mathbb R$ which
satisfies certain positivity and linearity conditions. I have no
problem in understanding this definition.
Then people talk about the set of inner products in $\mathbb R^3$.
It seems that an inner product can be represented by a matrix. Why?
And how? This does not make sense to me because if we write inner
product in vector notation, then we have $\langle v, w\rangle=v^Tw$.
There is no matrix in this notation, is there? Moreover, what does
it mean by a set of inner products? Does that mean a set of
functions (since inner product can be treated as a function as
indicated in 1)?
What is the natural topology on the set of inner products? Can I say
that the topology is the sub product topology of $\mathbb R^3\times
\mathbb R^3$?

REPLY [3 votes]: Abstractly, inner products are special cases of something from geometry called a symmetric bilinear form on a vector space. A bilinear form on and $F$ vector space $V$ is just a function $B:V\times V\to F$ satisfying 

$B(x,y)=B(y,x)$
$B(x+y,z)=B(x,z)+B(y,z)$ 
$B(\lambda x,y)=B( x,\lambda y)=\lambda B(x,y)$

You'll notice the positive definiteness criterion ($B(x,x)>0$ for all $x\neq 0$) is missing. 
Another thing that's missing is any mention of coordinates or components of vectors! If you pick any basis at all, then you can bulid a matrix $X$, necessarily a symmetric matrix, such that $\vec{x}^\top X \vec{y}=B(x,y)$, where $\vec{x}$ is the coordinate representation of $x$ in that particular basis.
When you are working in the simple basis with $(1,0,0)$, $(0,1,0)$ and $(0,0,1)$ and the usual inner product, then $X$ is just the identity matrix. But keep in mind $X$ doesn't have to be the identity matrix.
Now a natural question is this: 

If I'm willing to undergo a change of basis, is there any way I can simplify $X$?

For the case of $F=\Bbb R$, you can always change basis so that the matrix for the bilinear form is zero everywhere except on the diagonal, where it can have $0$'s, $1$'s and/or $-1$'s. If you additionally require that $B(x,x)>0$ for all $x\neq 0$, then $X$ can be changed into the identity matrix. You can learn a little about this from the wiki article on metric signatures.
Also keep in mind that here are some fields $F$ that are not as nice as $\Bbb R$, and won't admit such nice simplifications.

So how can you view the set of inner products (for, say, $\Bbb R$)? For a fixed basis, you can view the set of possible symmetric bilinear forms on $\Bbb R^n$ as the set of $n\times n$ symmetric matrices. If you want to focus on positive definite forms only, then you have to further restrict this to the set of positive definite symmetric matrices.
Any positive definite symmetric matrix at all will define an inner product on the vectors in the basis you chose, and you can see by playing around with some examples that they give different forms. For example, you could take the matrix $X=\begin{bmatrix}1&0\\0&2\end{bmatrix}$ and examine $\vec{v}^\top=(0,1)$. Notice that $\vec{v}^\top X\vec{v}=2 $, but $\vec{v}^\top I_2 \vec{v}=1$, so the identity matrix $I_2$ and $X$ are definitely giving two different inner products on $\Bbb R^2$ for the same input vector.

The set of symmetric matrices $S$, and the set of positive definite symmetric matrices $S^+$ could be given the subspace topology from whatever topology you choose on the set of $n\times n$ matrices. Perhaps one final thing to note is that while $S$ is a linear subspace of the $n\times n$ matrices, $S^+$ isn't.