The dot produt (inner product in the context of Euclidean space) of two vectors $\mathbf{a}=\left [ a_{1},a_{2},…,a_{n} \right]$ and $\mathbf{b}=\left [ b_{1},b_{2},…,b_{n} \right ]$ is defined as:
$\left ( \textrm{0.1}\right )$ $\mathbf{a}\cdot \mathbf{b}=\sum_{i=1}^{n}a_{i}b_{i}$
in Wikipedia that quotes S. Lipschutz, M. Lipson (2009). Linear Algebra. McGraw Hill.
Given that the basis of the subspace is $\left \{ \mathbf{e}_{1}, \mathbf{e}_{2}, … , \mathbf{e}_{n},\right \}$ then $\mathbf{a}=a_{1}\mathbf{e}_{1}+a_{2}\mathbf{e}_{2}+ … + a_{n}\mathbf{e}_{n}$ and $\mathbf{b}=b_{1}\mathbf{e}_{1}+b_{2}\mathbf{e}_{2}+ … + b_{n}\mathbf{e}_{n}$, then:
$\left ( \textrm{0.2}\right )$ $\mathbf{a}\cdot \mathbf{b}= a_{1}b_{1}\mathbf{e}_{1}\cdot\mathbf{e}_{1}+a_{1}b_{2}\mathbf{e}_{1}\cdot\mathbf{e}_{2}+… +a_{1}b_{n}\mathbf{e}_{1}\cdot\mathbf{e}_{n} + … + a_{n}b_{n}\mathbf{e}_{n}\cdot\mathbf{e}_{n}$
so:
- if the basis is composed of pairwise orthogonal vectors then $\mathbf{e}_{i}\cdot\mathbf{e}_{i}=1$ and $\mathbf{e}_{i}\cdot\mathbf{e}_{j}=0$ with $i\neq j$ then $\left ( \textrm{0.2}\right )$ is reduced to $\left ( \textrm{0.1}\right )$
- if the basis is composed of not pairwise orthogonal vectors, then $\left ( \textrm{0.2}\right )$ can be written as:
$\left ( \textrm{0.3}\right )$
$\mathbf{a}\cdot \mathbf{b}=\sum_{j=1}^{n}\sum_{i=1}^{n}a_{j}b_{i}\mathbf{e_{j}}\cdot \mathbf{e_{i}}$
and actually $\left ( \textrm{0.3}\right )$ is not equivalent to the general definition $\left ( \textrm{0.1}\right )$.
What am I missing?
Best Answer
You aren't missing anything, (0.1) is the expression of the dot product with respect to an orthonormal basis. If the basis is not orthonormal you get (0.3). Notice that you do not just need the vectors to be pairwise orthogonal, but also of unit norm in order for (0.1) to hold.