Geometric Algebra: How to prove that the Grade Projection Operator is well defined The Grade Projection Operator $<\cdot>_r$ is widely used in Geometric Algebra to prove numerous relations and results. I'm looking for a proof that Grade Projection is a well-defined operation.
To do as such it should be sufficient to show that an $r$-blade $f_1f_2...f_r$ (geometric product of $r$ normal vectors) becomes an $r$-vector when each $f_{i}$ is expressed in terms of a basis $\{e_1,e_2,...,e_n\}$, meaning:
$$f_1f_2...f_r = \sum_{i,j,...,k} (a_{1i}a_{2j}...a_{rk})e_ie_j...e_k = \sum_{i \neq j \neq ... \neq k} (a_{1i}a_{2j}...a_{rk})e_ie_j...e_k$$
Unfortunately, most texts I've found do not bother proving this fact, as they present Geometric Algebra axiomatically. Due to the fact that it remains to be proven that such an axiomatic definition produces an actually existing entity, a constructive approach (be it as a quotient of the tensor algebra or through direct construction from a canonical orthonormal basis) seems to still be necessary.
In An elementary construction of the geometric algebra, Alan Macdonald provides such a construction, including a short "proof" of what I'm looking for. The proof seems to me more of an example than an actual demonstration and moreover, it relies on the principle that you can transform a certain orthonormal basis to another one through a series of reflections operations.
This fact seems to me to be reliant on the fact Macdonald considers a vector space with a positive definite inner product (meaning $a\cdot b \ge 0$ or Euclidean space with Euclidean norm). Geometric Algebra is also used with the Minkowski metric $\{1,-1,-1,-1\}$ for example, for which I do not know if the "consecutive reflections method" for basis vectors transformation still holds.
I'm then looking for a general proof for a Geometric Algebra with signature $\{n,m\}$ (therefore, for any non-degenerate inner product).

Summary:
Prove that the $r$-blade $f_1f_2...f_r$ is expressed as an $r$-vector in another normal basis $\{e_1,e_2,...,e_n\}$ (all components of grade $\neq r$ are null)
N.B. I'm working with the definition of $r$-blade as geometric product of $r$ normal vectors, NOT with the definition using the wedge product. The definition of the wedge product should if possible come AFTER the verification that the Grade Projection Operator is well-defined.

EDIT 1:
I'd like to thank @Nullius in Verba for their answer. I report their answer with expanded algebraic passages (hoping no error has been made and eventually asking for corrections). This is mostly to help myself visualize better their explanation, but I hope this will be of help to others too. The reasoning is/should be the following:

*

*Consider $r$-blade $f_1f_2...f_j...f_k...f_r$


*Express it in another basis as:
$$f_1f_2...f_j...f_k...f_r=\sum_{i_1,i_2,...,i_j,..,i_k,...,i_r} (a_{i_1}a_{i_2}...a_{i_j}...a_{i_k}...a_{i_r})(e_{i_1}e_{i_2}...e_{i_j}...e_{i_k}...e_{i_r})$$


*Due to the anti-symmetry property of the product of normal vectors (the property can be shown to hold even if working with the $f_i$ basis vectors already expressed in the $e_i$ basis, as shown in An elementary construction of the geometric algebra by Alan Macdonald):

$$f_1f_2...f_j...f_k...f_r+f_1f_2...f_k...f_j...f_r=0$$


*By expressing both in the $e_i$ basis:


*

*$f_1f_2...f_j...f_k...f_r=
\sum_{i_1 \neq ... \neq i_j \neq i_k \neq ... \neq i_r} (a_{i_1}...a_{i_j}...a_{i_k}...a_{i_r})(e_{i_1}...e_{i_j}...e_{i_k}...e_{i_r}) +\\
\sum_{\exists i_j,i_k:i_j=i_k} (a_{i_1}...a_{i_j}...a_{i_k}...a_{i_r})(e_{i_1}...e_{i_j}...e_{i_k}...e_{i_r})$


*$f_1f_2...f_k...f_j...f_r=\sum_{i_1 \neq ... \neq i_k \neq i_j \neq ... \neq i_r} (a_{i_1}...a_{i_k}...a_{i_j}...a_{i_r})(e_{i_1}...e_{i_k}...e_{i_j}...e_{i_r}) +\\
\sum_{\exists i_j,i_k:i_j=i_k} (a_{i_1}...a_{i_k}...a_{i_j}...a_{i_r})(e_{i_1}...e_{i_j}...e_{i_k}...e_{i_r})$


*The second expression can be rewritten as:

$$f_1f_2...f_k...f_j...f_r=-\sum_{i_1 \neq ... \neq i_j \neq i_k \neq ... \neq i_r} (a_{i_1}...a_{i_j}...a_{i_k}...a_{i_r})(e_{i_1}...e_{i_j}...e_{i_k}...e_{i_r}) +\\
\sum_{\exists i_j,i_k:i_j=i_k} (a_{i_1}...a_{i_j}...a_{i_k}...a_{i_r})(e_{i_1}...e_{i_j}...e_{i_k}...e_{i_r})$$

*

*First summation: scalars reordered via commutativity and $e_{i_j}$ and $e_{i_k}$ reordered with the addition of a minus sign in front due to antisymmetry (in the first summation, the condition $i_1 \neq ... \neq i_j \neq i_k \neq ... \neq i_r$ holds)

*Second summation: indices re-labelled, due to the second summary containing ONLY and ALL members with at least one pair $e_{i_j}$ and $e_{i_k}$ for which $i_j=i_k$ holds (guaranteeing the possibility of re-labelling elements accordingly)



*By summing the two expressions, the following is obtained:

$$f_1f_2...f_j...f_k...f_r+f_1f_2...f_k...f_j...f_r=\\
2\sum_{\exists i_j,i_k:i_j=i_k} (a_{i_1}...a_{i_j}...a_{i_k}...a_{i_r})(e_{i_1}...e_{i_j}...e_{i_k}...e_{i_r})=0$$
Then $\sum_{\exists i_j,i_k:i_j=i_k} (a_{i_1}...a_{i_j}...a_{i_k}...a_{i_r})(e_{i_1}...e_{i_j}...e_{i_k}...e_{i_r})=0$, proving that even if the basis is changed, the only elements which have the possibility of "surviving" the geometric product are precisely those of grade $r$.
Proven this, the Grade Projection Operator $<\cdot>_r$ can be said to be well-defined, as no strange phenomena (appearance of terms of different grades) should happen even if the basis is changed.
N.B. If there exists also a "coordinate-free proof" (which seems to be the motto of Geometric Algebra), which does not require expansion in an orthonormal basis but which follows directly from the properties of Geometric Algebra defined as the quotient of the corresponding tensor algebra, it would be welcome. Nevertheless, I believe that the provided answer is more than satisfactory.
 A: "it relies on the principle that you can transform a certain orthonormal basis to another one through a series of reflections operations"
That's the Cartan-Dieudonné Theorem: For an n-dimensional, non-degenerate symmetric bilinear space over a field with characteristic not equal to 2, every element of the orthogonal group is a composition of at most n reflections. It works fine on the Minkowski metric.
"N.B. I'm working with the definition of r-blade as geometric product of r normal vectors, NOT with the definition using the wedge product."
The concept of 'normal' vectors is probably best defined after defining the inner product, at the same time as the wedge product. For the axiomatic approach, you want to start with algebraic properties of the geometric product and derive the geometric interpretation of thoe properties later. Define r-blades by specifying the vector basis to be pairwise antisymmetric: $f_if_j=-f_jf_i$ for $i\neq j$.
Hopefully with the full Cartan-Dieudonné Theorem in hand, the McDonald proof may be deemed more acceptable. But I don't think it's too hard to prove using the antisymmetry property. Expand the product of $f_i$s writing each as a linear combination of $e_i$s.
$$f_1f_2\ldots f_r=(a_{11}e_1+a_{12}e_2+\ldots +a_{1n}e_n)\ldots (a_{r1}e_1+a_{r2}e_2+\ldots +a_{rn}e_n)$$
Multiply out all the $r$ brackets to get terms with $r$ vectors $e_{i_1}e_{i_2}\ldots e_{i_r}$ in each. If the vectors in a term are all distinct, then the product is an r-blade by the antisymmetry of the $e_i$. If any are repeated, then we can permute the order (with appropriate sign-changes because the $e_i$ are antisymmetric) to put them next to each other and collapse them to a scalar, which commutes under multiplication with any multivector. Now we use this and the antisymmetry of the $f_i$ to show that these must sum to zero. Swapping any pair of the brackets above swaps the order of the corresponding $e_{i_j}$, $e_{i_k}$ in the products, and by antisymmetry of the $f_i$ this must reverse the sign of the sum of all such terms. But if $i_j=i_k$ then $e_{i_j}e_{i_k}=e_{i_k}e_{i_j}$ for all such terms, and the only way this can reverse sign is if they sum to zero.
I am inclined to agree that elementary introductions to Geometric Algebra are lacking. Those directed at physicists do usually try to motivate them geometrically at the start, but at some point the leap to abstraction is made, and from that point on only the most minimal sketch is provided - a scaffold on which, with a lot of work, a full derivation could be constructed. Usually this seems to be because authors want to demonstrate the utility of Geometric Algebra to other professionals by getting on to the more interesting applications quickly. Teaching vector algebra ramps up gradually, from drawing arrows on graph paper at the age of 10 to maybe starting vector calculus (grad, div, curl and all that) at the age of 20. Introductory texts on Geometric Algebra do a couple of pages of drawing arrows to show willing, but only about 10 pages later are discussing Dirac spinors and general relativity and differential geometry! I exaggerate, but not by much. There are about 10 years worth of intermediate detail missing. But such student-oriented texts are unlikely to be written until Geometric Algebra is widely enough accepted and established to be taught at the same elementary stage as vectors are. (Hestenes made the attempt, but again, in my view, tried to do too much at once.) The pedagogy is still a work in progress. I expect it will be another 20 years or more before we get that far.
A: $
\newcommand\K{\mathbb K}
\newcommand\G{\mathcal G}
\newcommand\Ext{{\bigwedge}}
\newcommand\End{\mathrm{End}}
\newcommand\proj[1]{\langle#1\rangle}
\newcommand\form[1]{\langle#1\rangle}
\newcommand\lintr{\mathbin\lrcorner}
\newcommand\rintr{\mathbin\llcorner}
\newcommand\lcontr{\mathbin{{\rfloor}}}
\newcommand\rcontr{\mathbin{{\lfloor}}}
\newcommand\tr{\mathrm{tr}}
$
Let $\K$ be a field with characteristic $\not= 2$,
$V$ an $n$-dimensional $\K$-vector space,
$Q : V \to \K$ a quadratic form,
$\G(V, Q)$ the corresponding geometric algebra,
and $\Ext V$ the exterior algebra of $V$.
Once we have an isomorphism $\phi : \G(V, Q) \cong \Ext V$,
the grade projection is defined by
$$
  \proj{X}_r := \phi^{-1}\!\left(\proj{\phi(X)}^\wedge_r\right)
$$
where $\proj\cdot^\wedge_r$ is the grade projection on $\Ext V$.
I will give two ways of constructing
the canonical $\phi : \G(V, Q) \cong \Ext V$
in an essentially basis-free manner.
The first way I think is not too difficult to grasp,
but the second way feels much more involved to me.
If simplicity is what you're after,
I think that Nullius' answer is perfect for your question,
and is much more accessible that what I am going to get into.
Both of the constructions I will demonstrate
are special cases of Chevalley's construction;
see chapter 22 of Clifford Algebras and Spinors (2001)
by Pertti Lounesto constructing $\G(V, Q)$ in $\Ext V$,
and see these notes
for the full generality giving isomorphisms $\G(V, Q) \cong \G(V, Q+Q')$.
The key in both cases is to exploit the universal properties of $\Ext V$ and $\G(V , Q)$, which define them uniquely up to algebra isomorphism:

*

*We assume $V \subseteq \Ext V$ in the canonical fashion. Let $A$ be any associative algebra. Then every $f : V \to A$ such that $f(v)^2 = 0$ for all $v \in V$ extends uniquely to an algebra homomorphism $f_\wedge : \Ext V \to A$; i.e. $f_\wedge$ is linear, $f_\wedge(v) = f(v)$ for all $v \in V$, and
$$
  f_\wedge(v_1\wedge\cdots\wedge v_k) = f(v_1)f(v_2)\cdots f(v_k).
$$

*We assume $V \subseteq \G(V, Q)$ in the canonical fashion.
Let $A$ be any associative algebra.
Then every $f : V \to A$ such that $f(v)^2 = Q(v)1_A$
for all $v \in V$ extends uniquely to an algebra homomorphism
$f_Q : \G(V, Q) \to A$;
i.e. $f_A$ is linear, $f_Q(v) = f(v)$ for all $v \in V$, and
$$
  f_Q(v_1v_2\cdots v_k) = f(v_1)f(v_2)\cdots f(v_k).
$$
It should be apparent from these universal properties that $\Ext V$ is just $\G(V, 0)$.
Within $\G(V, Q)$
We construct an exterior product within $\G(V, Q)$
which gives us the isomorphism with $\Ext V$.
While we can construct the isomorphism completely basis-free,
we will appeal to the existence of an orthogonal basis
to show that it is indeed an isomorphism.
We assume $\K, V \subseteq \G(V, Q)$ canonically. Then for any $v \in V$ and $X \in \G(V, Q)$, define
$$
  v\wedge X := \frac12(vX + \hat Xv)
$$
where $\hat X$ is the main or grade involution of $X$; this is defined by extending the map $v \mapsto -v$ using the universal property of $\G(V, Q)$. The $\wedge$ operation gives us a map from $V$ into the linear endomorphisms $\End(\G(V, Q))$ by
$$
  v \mapsto (X \mapsto v\wedge X).
\tag{$*$}
$$
We then confirm for any $v, w \in V$ that
$$
  v\wedge (v\wedge X)
= \frac12(v(v\wedge X) + \widehat{(v\wedge X)}v)
= \frac14(vvX + v\hat Xv + \widehat{vX}v + \widehat{\hat Xv}v)
= \frac14(v^2X + v\hat Xv - v\hat Xv - Xv^2)
= 0,
$$
where we've used the fact that $X \mapsto \hat X$ is an algebra involution, that $\hat v = -v$, and that $v^2$ is a scalar. Hence the map ($*$) extends to an algebra homomorphism $\phi : \Ext V \to \End(\G(V, Q))$ (the product on $\End(\G(V, Q))$ being function composition) where
$$
  \phi(v_1\wedge\cdots\wedge v_k)(X) = v_1\wedge(v_2\wedge(\cdots\wedge(v_k\wedge X)\cdots)),\quad X \in \G(V, Q).
$$
Then we get a linear map $\phi_1 : \Ext V \to \G(V, Q)$ by $\phi_1(X) = \phi(X)(1)$ for $X \in \Ext V$. Now let $\{e_i\}_{i=1}^n$ be a basis; we know that
$$
  \{1\}\cup\{e_i\}_{1\leq i\leq n}\cup\{e_ie_j\}_{1\leq i<j\leq n}\cup\{e_ie_je_k\}_{1\leq i<j<k\leq n}\cup\cdots
$$
is a basis for $\G(V, Q)$. Then it is easy to confirm that when $\{e_i\}_{i=1}^n$ is an orthogonal basis that
$$
  e_i\wedge(e_{j_1}e_{j_2}\cdots e_{j_k}) = e_ie_{j_1}e_{j_2}\cdots e_{j_k}
$$
if $i \not\in \{j_1, j_2,\dotsc, j_k\}$; hence
$$
  \phi_1(e_{j_1}\wedge e_{j_2}\wedge\cdots\wedge e_{j_k}) = e_{j_1}e_{j_2}\cdots e_{j_k}
$$
and $\phi_1$ is a linear isomorphism. In this way, we've also explicitly constructed the exterior product on $X, Y \in \G(V, Q)$:
$$
  X\wedge Y
= \phi_1\bigl(\phi_1^{-1}(X)\wedge\phi_1^{-1}(Y)\bigr)
= \phi\bigl(\phi_1^{-1}(X)\bigr)(Y).
$$
We can easily verify that starting analogously with a product $X\wedge v$ in $\G(V, Q)$ and following the same procedure would again result in the same $\phi_1$.
Within $\Ext V$
We construct a geometric product within $\Ext V$
which gives us the isomorphism with $\G(V, Q)$.
While we can construct the isomorphism completely basis-free,
we will appeal to the existence of an orthogonal basis
to show that it is indeed an isomorphism.
To present this in full generality,
we will make use of the dual space $V^*$
(though we can avoid this in the case that $Q$ is non-degenerate).
Let $\form{\cdot,\cdot} : V^*\times V \to \K$
denote the bilinear natural pairing;
that is $\form{v^*, v} = v^*(v)$ for all $v^* \in V^*$ and all $v \in V$.
This extends (naturally and in a basis-free manner that I will not go into here)
to a bilinear pairing $\Ext V^*\times\Ext V \to \K$ given by
$$
  \form{v^*_k\wedge v^*_{k-1}\wedge\cdots v^*_1,\;
    v_1\wedge v_2\wedge\cdots\wedge\cdots v_l}
= \delta_{kl}\det\bigl(\form{v^*_i, v_j}\bigr)_{i,j=1}^n.
$$
This pairing is non-degenerate, so the exterior product has well-defined adjoints:
$$
  \form{X^*\wedge Y^*, Z} = \form{X^*, Y^*\lintr Z} = \form{Y^*, Z\rintr X^*},
$$
where ${\lintr} : \Ext V^*\times\Ext V \to \Ext V$ is the left interior product on $\Ext V$ and ${\rintr} : \Ext V\times\Ext V^* \to \Ext V$ is the right interior product on $\Ext V$. There are other conventions in defining these, but the above is the most natural from the perspective of $\G(V, Q)$. Now $Q$ gives us a symmetric bilinear form $B : V\times V \to \K$ by
$$
  \form{v, w}_Q := \frac12(Q(v+w) - Q(v) - Q(w))
$$
and this gives us a map $\flat : V \to V^*$ by
$$
  v^\flat := (w \mapsto \form{v, w}_Q),
$$
and note that in fact
$$
  \form{v, w}_Q = \form{v^\flat, w}.
$$
$\flat$ extends to an algebra homomorphism $\flat : \Ext V \to \Ext V^*$
by the universal property of $\Ext V$.x
Then $\form{\cdot,\cdot}_Q$ extends to $\Ext V\times\Ext V$
in the same way as $\form{\cdot,\cdot}$.
We may now define the left and right contractions
${\lcontr}, {\rcontr} : \Ext V\times\Ext V \to \K$ by
$$
  X\lcontr Y = X^\flat\lintr Y,\quad X\rcontr Y = X\rintr Y^\flat,
$$
where in fact we have
$$
  \form{X\wedge Y, Z}_Q = \form{X, Y\lcontr Z}_Q = \form{Y, Z\rcontr X}_Q,
$$$$
  \form{X, Y\wedge Z}_Q = \form{X\rcontr Y, Z}_Q = \form{Z\lcontr X, Y}_Q.
$$
(We cannot use this to define the contractions since $Q$ may be degenerate.) This definition of the contractions immediately gives us the properties
$$
  v\lcontr w = \form{v, w}_Q,\quad
  v\lcontr(w\wedge X) = (v\lcontr w)X - w\wedge(v\lcontr X),\quad
  v\lcontr(w\lcontr X) = (v\wedge w)\lcontr X
$$
for $v, w \in V$ and $X \in \Ext V$. We now define the geoemtric product between $v \in V$ and $X \in \Ext V$ by
$$
  v\bullet X = v\lcontr X + v\wedge X.
$$
This gives us a map $V \to \End(\Ext V)$ by
$$
  v \mapsto (X \mapsto v\bullet X),
$$
and we see that
$$\begin{aligned}
  v\bullet(v\bullet X)
&= v\lcontr(v\bullet X) + v\wedge(v\bullet X)
\\
&= v\lcontr(v\lcontr X) + v\lcontr(v\wedge X)
  + v\wedge(v\lcontr X) + v\wedge(v\wedge X)
\\
&= (v\wedge v)\lcontr X + (v\lcontr v)X - v\wedge(v\lcontr X)
  + v\wedge(v\lcontr X) + (v\wedge v)\wedge X
\\
&= Q(v)X.
\end{aligned}$$
Thus by the univeral property of $\G(V, Q)$ we get an algebra homomorphism $\phi : \G(V, Q) \to \End(\Ext V)$ where
$$
  \phi(v_1v_2\cdots v_k)(X) = v_1\bullet(v_2\bullet(\cdots\bullet(v_k\bullet X)\cdots)).
$$
Then we get a linear map $\phi_1 : \G(V, Q) \to \Ext V$ by $\phi_1(X) = \phi(X)(1)$ for $X \in \G(V, Q)$. Now, if $\{e_i\}_{i=1}^n$ is an orthogonal basis for $V$, it is straightforward to show (say, by induction) that
$$
  e_i\bullet(e_{j_1}\wedge\cdots\wedge e_{j_k})
= e_i\wedge e_{j_1}\wedge\cdots\wedge e_{j_k}
$$
when $i, j_1,\dotsc, j_k$ are all distinct. Hence
$$
  \phi_1(e_{j_1}e_{j_2}\cdots e_{j_k}) = e_{j_1}\wedge e_{j_2}\wedge\cdots e_{j_k}
$$
and so $\phi_1$ is an isomorphism $\G(V, Q) \to \Ext V$.
The Particular Case of the Scalar Projection
There is a actually a nice way of defining just the scalar projection
that is fairly simple.
For any $X \in \G(V, Q)$,
the left multiplication map $Y \mapsto XY : \G(V, Q) \to \G(V, Q)$ is linear.
So we define the trace $\tr(X)$ of $X$ as the trace of this map.
We can show that right multiplication
actually gives the same definition of trace.
The dimension of $\G(V, Q)$ is $2^n$, so $\tr(1) = 2^n$,
and we define the scalar projection as
$$
  \proj{X}_0 = \frac1{2^n}\tr(X).
$$
It is easy to show that $\proj{X}_0$ is the scalar component of $X$
expressed in any orthogonal basis.
