Infer that basis is orthogonal

Question

$x^n = \sum_{i=1}^D \alpha_i^nu_i = \sum_i^D(x^{nT}u_i)u_i$

$x^n$ is the $nth$ data point in $D$ dimensional space.

I read a claim that I cannot make sense of:

"We can infer that the basis $u_i,\dots, u_d$ must be orthogonal. Otherwise, it would not be possible to express $x^n$ as it was."

Why is this true?

Answer 1

I am guessing that $x^n$ can be any vector. So we have for all vectors $v$ $$v=\sum<v,u_i>u_i.$$ In particular, taking $v=u_j$ we have $u_j=\sum <u_j,u_i>u_i$ and since $(u_i)$ is a basis we obtain (equating coefficients) $1=<u_j,u_j>$ and $0=<u_j,u_i>$ (for $i\neq j$).