I am trying to repeat the following proof of $O(2)$ invariance, but for a multivector. First, with a vector:
Let $\mathbf{v}$ be a vector and $A$ a linear transformation:
$$ \mathbf{v}:=\pmatrix{x\\y}\implies \mathbf{v}^T =\pmatrix{x& y} $$
Now I define a distance function as follows:
$$ f(\mathbf{v}):=x^2+y^2=\mathbf{v}^T\mathbf{v} $$
The linear transformations are:
$$ f(A\mathbf{v})=(A\mathbf{v})^T(A\mathbf{v})=\mathbf{v}^TA^TA\mathbf{v} \implies A^TA=1 $$
Now let us try with multi-vectors. Let $\mathbf{u}$ be a multivector of $Cl_{2,0}(\mathbb{R})$:
$$ \mathbf{u}:=xe_1+ye_2 $$
I am interested in the linear transformations that keep the geometric product $\mathbf{u}\mathbf{u}=x^2+y^2$ invariant.
To define $f(\mathbf{u})$ I am already running into problems. As far as I know, there are no such things as a poly-vector valued function, only vector-valued function. Consequently, I am assuming that defining my linear transformation as $f(A\mathbf{u})=(A\mathbf{u})(A\mathbf{u})$ is inadmissible.
So is f then a function of $x,y$? Assuming this is the case, let me define it as follows:
$$ f(x,y)=x^2+y^2= \mathbf{u}\mathbf{u} $$
And with a linear transformation:
$$ \begin{align} f(Ax,Ay)&=\mathbf{u}\mathbf{u} = (Axe_1+Aye_2)(Axe_1+Aye_2)\\ &=x^2(Ae_1)^2+xyAe_1Ae_2+xyAe_2Ae_1+y^2(Ae_2)^2 \end{align} $$
How do I simplify this further to show that $A^TA=1$? It feels like I am missing some identities. Does $[A,e_1]=0$ and $[A,e_2]=0$. What kind of object is $A$. I doubt its a matrix; is it a multivector?