Is the $\mathbb{Z}$-grading of a Clifford algebra basis independent? Let $V$ be a finite dimensional vector space over a field $K$ of characteristic $\neq 2$, and let $q \colon V \to K$ be a quadratic form. One of the first things to show when learning the theory of Clifford algebras is that any orthogonal basis $\{ e_1, \dots, e_n \}$ of $V$ gives rise to a basis $\{ e_{i_1} \dots e_{i_k} : 1 \leq i_1 < i_2 < \dots < i_k \leq n \}$ of $\operatorname{C\ell}(V,q)$ as a $K$-vector space. In that way, we obtain a direct sum decomposition
$$
  \operatorname{C\ell}(V,q) = C_0 \oplus \dots \oplus C_n,
$$
where each $C_k$ is generated by the $k$-products $e_{i_1} \dots e_{i_k}$. It is often regarded as a grading, although it is not a grading of a $K$-algebra in the strict sense.
Question 1:
By looking at simple examples, it appears to me that this grading is independent of the orthogonal basis of $V$ chosen. Is this true?
Question 2:
The Wikipedia article on Clifford algebras suggests that the map $\operatorname{C\ell}(V,q) \to K$ sending each element to its $C_0$-part is basis independent (of course this would be implied by a positive answer to the first question). Is this true? Is there some kind of basis independent formula?
Thank you in advance!

This is an answer to the second question (can't believe I haven't seen this immediately). David's answer already implies that the map $\chi \colon \operatorname{C \ell}(V,q) \to K$ sending an element to its scalar part is well defined (i.e., independent of the orthogonal basis of $V$ chosen). But in fact, this map is even independent of the embedding $V \hookrightarrow \operatorname{C \ell}(V,q)$:
By choosing a standard basis of $\operatorname{C \ell}(V,q)$ given by an orthogonal basis of $V$, you can easily check that we have $\chi(x) = 2^{- \dim(V)} \cdot \operatorname{Tr}(x)$ for all $x \in \operatorname{C \ell}(V,q)$, where the trace is taken with respect to the standard action of $\operatorname{C \ell}(V,q)$ on itself by left multiplications (this is well defined since we are assuming $\operatorname{char}(K) \neq 2$).
