Inner product respect on a non-canonical base

Question

Let a,b be vectors, on the standard base we use the dot product by simply doing a.b. But when we consider an other base we put a symmetric matrix between them. Why? How does that work? Thanks

Answer 1

Let $V = \mathbb{R}^n$ and $v_1,\dots.v_n \in V$ some basis of $V$. Then for any $a,b \in V$ we have $a = \sum_{i = 1}^n a_i v_i$ and $b = \sum_{i = 1}^n b_i v_i$ for some real numbers $a_1,\dots,a_n, b_1,\dots, b_n \in \mathbb{R}$. (The tuples $(a_i)_i$ and $(b_i)_i$ are the coordinate vectors of $a$ and $b$ with respect ot the basis $v_1,\dots,v_n$). As the dot product is linear in both components we can calculate $$ a.b = \left(\sum_{i = 1}^n a_i v_i\right).\left(\sum_{j = 1}^n b_i v_i\right) = \sum_{i = 1}^n \sum_{j = 1}^n a_i b_j \;(v_i.\!v_j) = (a_i)\: G\: (b_i) $$ where $G := (v_i.\!v_j)_{i,j}$ is the matrix consisting of the dot products of the basis elements. Note that because the dot product is symmetric, so is $G$, and in the special case that our chosen basis is the standard basis we have $v_i.\!v_j = 0$ if $i \neq j$ and $v_i.\!v_i = 1$ for $1 \leq i \leq n$, so $G$ becomes the identity matrix in this case and we can calculate the dot product in the usual way.