If $u$ and $v$ are any two vectors in $\mathbb{R}^n$ and $\mathcal{B}$ is any base for $\mathbb{R}^n$, then is it true or false that
$[u \cdot v]_{\mathcal{B}}=[u ]_{\mathcal{B}} \cdot [v ]_{\mathcal{B}}$ ?
I believe it is true right, because dot product preserves the angles between the vectors, right?
We can see $u\cdot v$ as the product $u\cdot v=u^{\top}v$ if we consider the matrices of the components of vectors as columns: $u= \left(\begin{array}{c} u^1\\ \vdots\\ u^n \end{array} \right)$ and similarly for $v$.
But it is known that under a change of basis the new components of a vector $u$ will be $B^{-1}u$ where $B$ is the matrix of that change, which is not singular to asure the linear independence of the new basis.
So \begin{eqnarray*} u\cdot v&=&u^{\top}v,\\ &=&(BB^{-1}u)^{\top}(BB^{-1}v),\\ &=&(B^{-1}u)^{\top}\ (B^{\top}B)\ B^{-1}v. \end{eqnarray*} and it is here that we can see how the matrix $B^{\top}B$ is "pairing" the new versions of both $u$ and $v$ without altering the number $u\cdot v$.