# Positive definite symmetric bilinear forms on $\mathbb{R}$

A Euclidean space is defined to be a vector space $V$ over $\mathbb{R}$ together with a positive definite symmetric bilinear form $\tau$.

My notes say that if we assume that $V$ is a Euclidean space and that we have chosen an orthonormal basis $e_1, \dots, e_n$. Then $\tau$ corresponds to the standard scalar product, so $\tau(v,w) = v\,.w$. So surely this mean that every positive definite symmetric bilinear form on $\mathbb{R}$ with an orthomormal basis is the same as it's just equal to the standard dot product?

-
No - the standard scalar product corresponds to $\tau$ if the orthonormal basis is exactly $e_1,\ldots,e_n$. For different orthonormal bases, you get a different scalar product. – Johannes Kloos Apr 2 '12 at 15:30

What is true is that if $V$ is a Euclidean space, and $\beta=\{\mathbf{e}_1,\ldots,\mathbf{e}_n\}$ is an orthonormal basis, then for any vectors $v$ and $w$ we will have $$\tau(v,w) = [v]_{\beta}\cdot [w]_{\beta},$$ where $[x]_{\beta}$ is the coordinate vector with respect to the basis $\beta$.
To see this, note that if $v=\alpha_1\mathbf{e}_1+\cdots+\alpha_n\mathbf{e}_n$ and $w = a_1\mathbf{e}_1+\cdots+a_n\mathbf{e}_n$, then \begin{align*} \tau(v,w) &= \tau(\alpha_1\mathbf{e}_1+\cdots+\alpha_n\mathbf{e}_n,a_1\mathbf{e}_1+\cdots+a_n\mathbf{e}_n)\\ &= \sum_{i=1}^n\sum_{j=1}^n\tau(\alpha_i\mathbf{e}_i,a_j\mathbf{e}_j\\ &= \sum_{i=1}^n\sum_{j=1}^n \alpha_ia_j\tau(\mathbf{e}_i,\mathbf{e}_j)\\ &= \sum_{i=1}^n\sum_{j=1}^n \alpha_ia_j\delta_{ij} &\text{(Kronecker's }\delta\text{)}\\ &= \alpha_1a_1+\cdots+\alpha_na_n\\ &= (\alpha_1,\ldots,\alpha_n)\cdot (a_1,\ldots,a_n)\\ &= [v]_{\beta}\cdot [w]_{\beta}. \end{align*}
However, in terms of the standard basis for $V$, the inner product may "look" different.
While it is not true that every positive definite symmetric bilinear form on $\mathbb{R}^n$ is equal to the standard dot product, it is true that $(\mathbb{R}^n,\tau)$ will be isomorphic to $\mathbb{R}^n$ with the standard dot product; that is, there exists a linear transformation $T\colon\mathbb{R}^n\to\mathbb{R}^n$ that is invertible, and such that for any $v,w\in\mathbb{R}^n$, $\tau(u,v) = T(u)\cdot T(v)$; namely, pick an orthonormal basis for $(\mathbb{R}^n,\tau)$ and let $T$ be the map that sends $u$ to its coordinate vector with respect to that basis.