Doubts about the scalar product

Question

I've a doubt regarding the proof of this very basic property of the scalar product using matrix notation. The property is

$\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}$

If $A$ $\in \mathbb{R^{n}}$ and $B$ $\in \mathbb{R^{n}}$ are two matrices containing the coordinates with respect to the same orthonormal base $\mathscr{C}$ of two vectors $\vec{a}$ and $\vec{b}$, than the property can be rewritten as

$A^{T} B=B^{T} A$

Can I say that $A^{T} B=(A^{T} B)^{T}$ since $A$ $\in \mathbb{R^{n}}$ and $B$ $\in \mathbb{R^{n}}$ ? (Then I would conclude easily)

By the way I have a doubt on the scalar product in general. There is a theorem that states that for any base $\mathscr{C}$ in a vector space $V$ there is always a scalar product for which $\mathscr{C}$ is orthonormal.

This is true because

$\mathscr{C}$ is orthonormal $\iff$ $\vec{a} \cdot \vec{b}=A^{T} B$

And it is always possible to do the product $A^{T} B$ and I will hopfeully manage to demonstrate that $A^{T} B$ satisfies all the four properties of the scalar product, therefore it is always a scalar product.

Is this right?

Answer 1

For your first question:

You can see that $a^Tb$ and $b^Ta$ are transposes of each other. But they are also just 1x1 matrices, so they are unchanged under transposition. Hence they are equal to each other.

For your second question:

The standard inner product is not the only possible one. In a space where the vectors are column vectors, you can also define an inner product as $a^TMb$ where $M$ is any symmetric, positive-definite matrix.

The map from geometric Euclidean vectors (with inner product given by the standard dot product $\vec a \cdot \vec b$) to linear algebraic column vectors (with inner product given by $a^TMb$ for a given $M$) depends on a choice of basis for the Euclidean vectors, and $M$ has to be chosen appropriately depending on the basis for the map to be an inner product space isomorphism and not just a vector space isomorphism. The basis in Euclidean space is almost universally chosen to be orthonormal, so that $M=I$ and the inner product is just $a^Tb$ for the column vectors.

But if you don't care about keeping the same inner product structure, you can choose $M=I$ for any basis, and then that basis is orthonormal by definition of the inner product. This shows that there is an inner product for any given basis where that basis is orthonormal. (If you then change to a new basis, $M$ transforms so it is no longer equal to $I$. The new basis vectors are not orthonormal, but the original ones still are after this change of basis.)