Reversion in Geometric Algebra- how is it computed? In all of the sources that I have encountered, the GA reversion for $A=nm$ is defined as $\widetilde{A}= \widetilde{m}\widetilde{n}$, with $\widetilde{v}=v$ for a vector $v$. I have two issues with this:

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*Is it necessary to decompose a multivector into a multiplication of two other multivectors? If I wish to compute a reverse, can I write $A$ using basis vectors and reverse them?

*If there is more than one way to decompose $A$, we need to prove that the outcome is the same. All the sources I've seen seem to let this point slide for some reason. For example: This article about GA in physics
 A: *

*"Is it necessary to decompose a multivector into a multiplication of two other multivectors?"

No, and as you suggest, decomposition of a multivector in terms of basis vectors, is a straightforward way to compute the reverse.
Take for example
$$   A = a + b \mathbf{e}_1 + c \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_4,$$
where $ a, b, c$ are scalars, and $\mathbf{e}_k$ are all mutually orthogonal.  Reversion can be performed term by term, resulting in a sign flip for the trivector term.
$$   \tilde{A} = a + b \mathbf{e}_1 - c \mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_4.$$


*For your bivector example $ A = \mathbf{n} \mathbf{m} $,
first note that such a factorization may not
generally be possible.  For example,
an $\mathbb{R}^{4}$ bivector such as $ A = \mathbf{e}_{12} + \mathbf{e}_{34} $ cannot be decomposed into a product of two vectors (this is an example of a bivector that
is not a 2-blade.)

If the bivector has such a factorization, that decomposition amounts to a choice of basis.
For example:
$$A = \mathbf{e}_{12} + \mathbf{e}_{23} + \mathbf{e}_{31},$$
may be factored as
$$A = \left( { \mathbf{e}_1 + \mathbf{e}_2 - 2 \mathbf{e}_3 } \right) \frac{ \mathbf{e}_2 - \mathbf{e}_1 }{2},$$
or
$$A = \frac{ \mathbf{e}_3 - \mathbf{e}_2 }{2} \left( { 2 \mathbf{e}_1 - \mathbf{e}_2 - \mathbf{e}_3 } \right).$$
We may compute the reverse from the original representation
$$   \tilde{A} = \mathbf{e}_{21} + \mathbf{e}_{32} + \mathbf{e}_{13},$$
or using the first factorization
$$\begin{aligned}\tilde{A} &= \frac{ \mathbf{e}_2 - \mathbf{e}_1 }{2}\left( { \mathbf{e}_1 + \mathbf{e}_2 - 2 \mathbf{e}_3 } \right)  \\ &=\frac{1}{{2}} \left( { \mathbf{e}_{21} - \mathbf{e}_{11} + \mathbf{e}_{22} - \mathbf{e}_{12} - 2 \mathbf{e}_{23} + 2 \mathbf{e}_{13} } \right) \\ &= \mathbf{e}_{21} + \mathbf{e}_{32} + \mathbf{e}_{13},\end{aligned}$$
or using the second factorization
$$\begin{aligned}   \tilde{A}    &= \left( { 2 \mathbf{e}_1 - \mathbf{e}_2 - \mathbf{e}_3 } \right)\frac{ \mathbf{e}_3 - \mathbf{e}_2 }{2}  \\    &= \frac{1}{{2}} \left( {   2 \mathbf{e}_{13} - 2 \mathbf{e}_{12}   - \mathbf{e}_{23} + \mathbf{e}_{22}    - \mathbf{e}_{33} + \mathbf{e}_{32}} \right) \\ &= \mathbf{e}_{21} + \mathbf{e}_{32} + \mathbf{e}_{13}.\end{aligned}$$
One should get the same answer regardless (and we do.)
Also observe that such a 2-blade factorization depends on the orthogonality of the factors.  If that were not the case, then there would be a scalar term in the product, and the result would not be a 2-blade.  To see that, consider the following product of vectors
$$A = \left( { \mathbf{e}_1 e^{i \theta} } \right) \mathbf{e}_1 = \cos\theta - i \sin\theta$$
where $\{\mathbf{e}_{1}, \mathbf{e}_{2}\} $ is assumed to be an orthonormal Euclidean basis, and $ i = \mathbf{e}_{1} \mathbf{e}_{2} $ is the pseudoscalar for the plane that $ A $ lies in (i.e. $ A \wedge i = 0 $).  For this to be a bivector, we require $ \theta \in \pi/2 + n \pi $.  That is, the two factors must be orthogonal.
If one wanted to prove that the reverse of any bivector $ A = \mathbf{n} \mathbf{m} $ is independent of the factorization (if such a factorization is possible), the task is essentially to show that the reverse is independent of any change of basis.
