# How could an orthogonal transformation map an orthonormal basis to a non-orthogonal one?

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:

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?

• 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. Commented Feb 13, 2023 at 17:20

$$\newcommand\R{\mathbb R} \newcommand\trans[1]{#1^{\mathrm T}}$$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 $$\trans OO = 1$$.