How to understand this equation by linear algebra?

Question

Assume that $$ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} u & Dv \\ v & -u \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} $$ with $D>0$, then we have $$ x'^2+Dy'^2 = (ux+Dvy)^2 + D(vx-uy)^2= (u^2+Dv^2)(x^2+Dy^2) $$ But I don't know how to understand this equation from the point of view of linear algebra. I try to define an inner product $\langle(a,b),(c,d)\rangle= ac+Dbd$, then $x^2+Dy^2=\langle(x,y), (x,y)\rangle$. Let $$ A = \begin{pmatrix} u & Dv \\ v & -u \end{pmatrix}, b' = \begin{pmatrix} x' \\ y' \end{pmatrix},b = \begin{pmatrix} x \\ y \end{pmatrix} $$ We have $b'= Ab$, $\langle b',b'\rangle=-|A|\langle b,b\rangle$. How this happen? I am stuck.

Answer 1

Inner products are not the way to go. The identity you're trying to understand is called Brahmagupta's identity, and note that in that identity $(u, v)$ and $(x, y)$ have symmetric roles but you've given them asymmetric roles in your matrix equation, with one being a matrix and one being a vector.

Instead the way to proceed is to promote everything to a matrix, as follows. Consider the matrix $X = \left[ \begin{array}{cc} 0 & -d \\ 1 & 0 \end{array} \right]$. We have $X^2 = \left[ \begin{array}{cc} -d & 0 \\ 0 & -d \end{array} \right]$ (so $X$ satisfies $X^2 + d = 0$), which is a multiple of the identity, so it follows that $X$ generates a subalgebra of matrices of the form

$$a + bX = \left[ \begin{array}{cc} a & -db \\ b & a \end{array} \right]$$

(here when I write $a$ I mean $aI$ but I don't like including the $I$'s explicitly) and in particular that this subalgebra of matrices is closed under multiplication. Note that this matrix has determinant $a^2 + db^2$ and that the signs are different from the ones you chose. Now closure under multiplication explicitly looks like

$$(a + bX)(p + qX) = (ap - dbq) + (aq + bp) X$$

or, in matrix form,

$$\left[ \begin{array}{cc} a & -db \\ b & a \end{array} \right] \left[ \begin{array}{cc} p & -dq \\ q & p \end{array} \right] = \left[ \begin{array}{cc} ap-dbq & -d(aq+bp) \\ aq+bp & ap-dbq \end{array} \right]$$

and taking determinants of both sides gives Brahmpagupta's identity

$$\boxed{ (a^2 + db^2)(p^2 + dq^2) = (ap - dbq)^2 + d(aq + bp)^2 }$$

as desired. Abstractly this matrix $X$ comes from considering the action of left multiplication by $\sqrt{-d}$ on the basis $\{ 1, \sqrt{-d} \}$ of the quadratic field $\mathbb{Q}(\sqrt{-d})$. Left multiplication embeds $\mathbb{Q}(\sqrt{-d})$ into $M_2(\mathbb{Q})$ and taking the determinant produces what is called the norm $N(a + b \sqrt{-d}) = a^2 + db^2$. The multiplicativity of the determinant then translates into the multiplicativity of the norm.