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I'm reading a text on geometric algebra, and in proving that the wedge product $a_1 \wedge a_2 \wedge \dots \wedge a_r$ for general $a_i$ (so not necessarily mutually orthogonal $a_i$) gives an $r$-blade, the author asserts the following:

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The matrix $R$ is an orthogonal matrix, so, according to equation 34, it's the matrix representation, relative to the $\{ e_1, e_2, \dots, e_r \}$ basis, of an orthogonal transformation that should map an orthonormal basis $\{ e_1, e_2, \dots, e_r \}$ to a possibly non-orthogonal basis $\{ a_1, a_2, \dots, a_r \}$. But how? How can an orthonormal basis be mapped to one that isn't orthogonal by an orthogonal transformation? How does that follow from $M$ being symmetric? Shouldn't the transformation preserve inner products? Or am I misunderstanding something?

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  • $\begingroup$ It may be worth computing $M_{ij}$ using equation (34), to see how $M$ relates to $R$. It looks to me like $M=R^T R$, which would render their claim about $R$ being orthogonal rather strange. $\endgroup$ Commented Feb 13, 2023 at 17:20

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$ \newcommand\R{\mathbb R} \newcommand\trans[1]{#1^{\mathrm T}} $I will assume our field is $\R$, but this is unimportant here.

Let $V$ be our abstract vector space with symmetric bilinear form $B$ which generates the geometric algebra. It suffices to assume that vectors $a_i \in V$ are linearly independent, otherwise their wedge product is zero which is trivially an $r$-blade. Let $A = \mathrm{span}\{a_i\}_{i=1}^r \subseteq V$ be their span. By forming the matrix $M$ and applying this to vectors, what the author is doing is implicitly identifying $A$ with $\R^r$ via $a_i$ coordinates, i.e. $$ x = x_1a_1 + x_2a_2 + \dotsb + x_ra_r \mapsto [x] = \trans{(x_1, x_2, \dotsc, x_r)}. $$ When the author applies $M$ to a vector $x \in A$ really what they're doing is $M[x]$ where $[x] \in \R^r$, and then maybe also applying the inverse transformation back into $A$.

Now let $E(\cdot,\cdot)$ be the inner product on $\R^r$ such that the standard basis is orthonormal. A matrix $O \in \R^{r\times r}$ is orthogonal if its columns $N_i$ are $E$-orthonormal: $E(N_i, N_j) = \delta_{i,j}$. But back in $A$, using $E$ corresponds to declaring the $a_i$ basis to be orthonormal. This doesn't necessarily have anything to do with the form $B$ on $V$. So while $M$ factors into $M = OD\trans O$ with $D$ diagonal and $O$ orthogonal, this only means that $O$ preserves $E$ and says nothing about whether $O$ preserves $B$, and indeed it will only preserve $B$ when $B$ is positive-definite and the $a_i$ basis is already $B$-orthonormal, in which case $$ B(x, y) = E([x], [y]),\quad $x, y \in A$. $$

Put another way, $B$ restricts to $A$ and is represented in $a_i$ coordinates by a matrix $[B]_{ij} = B(a_i, a_j)$. Vectors $x, y \in A$ are then $B$-orthogonal iff $\trans{[x]}[B][y] = 0$, and a matrix $Q \in \R^{r\times r}$ represents a $B$-orthogonal transformation on $A$ iff $\trans Q[B]Q = 1$ where $1$ is the identity matrix. This does not apply to the matrix $O$ above; $O$ satisfies $\trans OO = 1$.

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