A Clifford Algebra $C_k$ is a Real Algebra of dimension $2^k$ with its algebra generators being $\{e_1,\ldots,e_k\}$, satisfying the following relations: $$ e_i^2 = -1 ~~\& ~~e_je_i = -e_ie_j ~~\text{if} ~~i \neq j. $$ The vector space basis of $C_k$ is $\{e_{i_1}\cdots e_{i_r} : i_1 < i_2 \ldots < i_r, ~0 \le r \le k\}$
Let $R^k$ denote the k-space in $C_k$ spanned by $e_1,\ldots,e_k$. Now let $\{u_1,\ldots, u_k\}$ be some other basis of $R^k$. Then my question is following:
Do we still have the following relations?
$$ u_i^2 = -1 ~~\& ~~u_ju_i = -u_iu_j ~~\text{if} ~~i \neq j. $$
If no, then under the restriction of $\{u_1,\ldots,u_k\}$ being an orthonormal basis, do we have the above-mentioned relations on $ u_i$s?
What I have tried is given below:
Let $u_i = \sum_l a_l e_l$ and $u_j = \sum_n b_n e_n$. Then we have $$ u_i u_j = \sum_i^k a_ib_i e_i^2 + \sum_{~m,n \\ m \neq n} a_m b_n e_me_n $$ $$ \& $$ $$ u_j u_i = \sum_i^k b_ia_i e_i^2 + \sum_{~n,m \\ m \neq n} b_n a_m e_ne_m = \sum_i^k a_ib_i e_i^2 + (-1)\sum_{~n,m \\ m \neq n} b_n a_m e_me_n $$ From here I cannot establish the fact $u_ju_i = - u_iu_j$
$\newcommand\Cl{\mathrm{Cl}}$
Given any $K$-vector space $V$ with quadratic form $Q$, we may define the associated Clifford algebra $\Cl(V,Q)$. Canonically identifying $K$ and $V$ as subsets of $\Cl(V,Q)$, its fundamental property is that $v^2 = Q(v)$ for all $v \in V$. By definition, $Q$ has an associated symmetric bilinear form $B$, where when the characteristic of $K$ is not 2 we have $$ B(v,w) = \frac12(Q(v+w) - Q(v) - Q(w)) = \frac12((v+w)^2 - v^2 - w^2) = \frac12(vw + wv) $$ for $v,w \in V$. It follows that if $B(v,w) = 0$, i.e. $v$ and $w$ are orthogonal, then $$ 0 = vw + wv \implies vw = -wv, $$ and it is clear that $v$ and $w$ anticommute iff they are orthogonal. This is still all true when the characteristic of $K$ is 2, but we instead have to define $B$ by $$ B(v,w) = Q(v+w) - Q(v) - Q(w). $$
For more on defining Clifford algebras, see chapter 14 of Clifford Algebras and Spinors (2001) by Pertti Lounesto.