Cramer's Rule
The roots of Cramer's rule have always seemed a little mysterious. If we have a system of equations\[ a_1 x_1 + b_1 x_2 = c_1 \]
\[ a_2 x_1 + b_2 x_2 = c_2 \]
Then Cramer says
\[ x_1 = \frac{\det \begin{bmatrix} c_1 & b_1 \\ c_2 & b_2 \end{bmatrix}}{\det \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}} \]
and
\[ x_2 = \frac{\det \begin{bmatrix} a_1 & c_1 \\ a_2 & c_2 \end{bmatrix}}{\det \begin{bmatrix} a_1 & b_1 \\ a_2 & b_2 \end{bmatrix}} \]
That determinants play any role in isolating linear parameters is not obvious. So it's delightful how geometric algebra shines light on this.
A note on notation: in this section, latin letters are vectors and greek letters are scalars, so that
\[ x = \alpha a + \beta b\]
is a vector equation relating vector $x$ to vectors $a$ and $b$. The gift we are given by geometric algebra is that vectors become invertible, so we can write, since $ a \wedge a = 0 $:
\[ a \wedge x = \beta a \wedge b \]
\[ \beta = \frac {a \wedge x}{a \wedge b} \]
And similarly,
\[ \alpha = \frac {x \wedge b}{a \wedge b} \]
so that
\[ x = \left( \frac {x \wedge b}{a \wedge b} \right) a + \left( \frac {a \wedge x}{a \wedge b} \right) b \]
Now, in $\mathbb{R_2}$, 2-vectors are pseudoscalars, and pseudoscalars are scalar multiples of $I$, the unique (up to a scalar) highest-grade multivector. The determinant is just that scalar (say, $\lambda$): $a \wedge b = \lambda I$. So we can write:
\[ x = \left(\frac {\det (x \wedge b) I}{\det (a \wedge b) I} \right) a + \left( \frac {\det (a \wedge x) I}{\det (a \wedge b) I} \right) b \]
\[ x = \frac {\det (x \wedge b) }{\det (a \wedge b) } a + \frac {\det (a \wedge x) }{\det (a \wedge b) } b \]
That's the fastest path to Cramer's Rule anywhere.
BAC-CAB
Consider the product of a vector and bivector: $ a (b \wedge c)$.\[ \frac{1}{2} a (bc - cb) \]
\[ \frac{1}{2} (abc - acb) \]
Since $ab =2 a \cdot b -ba$, we can write
\[ \frac{1}{2} (2 a \cdot b \; c - bac -2a \cdot c \; b -cab ) \]
\begin{equation} \label{eq1}
a \cdot b c - a \cdot c b -\frac{1}{2} (bac-cab)
\end{equation}
The last two terms can be expanded as before:
\[ - \frac12(bac-cab) = -b a \cdot c + \frac{bca}{2} + c a \cdot b - \frac{cba}{2} \]
So in the expanded \eqref{eq1}, the dot products combine to give us
\[ 2 a \cdot b c - 2 a \cdot c b + \frac{(bc-cb) a}{2} \]
So finally
\[ a (b \wedge c) = 2 ( a \cdot b c - a \cdot c b ) + (b \wedge c) a \]
\[ \frac{ a (b \wedge c) - (b \wedge c) a}{2} = a \cdot b c - a \cdot c b \]
\[ a \cdot (b \wedge c) = a \cdot b c - a \cdot c b \]
This is a handy result. And, not restricted to $\mathbb{R^3}$.
Using the Dual to Exchange Dot and Wedge Products
We will develop a nice result with the dual operation that lets us exchange dot and wedge products.For any multivector $A$, we can write $A = A^\perp I$, where $A^\bot$ is whatever multivector you need to multiply with $I$ to get $A$, called the dual of $A$. In other words, taking the dual is multiplying on right with $I^{-1}$. Now consider the dual of $AB$
\[ (AB)^\bot = AB^\bot\]
This can be confirmed by examining $(AB)I^{-1} = ABI^{-1}$.
Now compare grades of this expression, and conclude that the largest grades give us
\[ (A \cdot B)^{\bot} = A \wedge B^\bot \]
and the smallest grades give us
\[ (A \wedge B)^{\bot} = A \cdot B^\bot \]
Gibbs' Vector Product
The dual lets us move between Gibbs-style cross products $a \times b$ and geometric products:
$$ a \times b \equiv (a \wedge b)^{\bot} $$
Of course Gibbs' cross product is only defined in $ \mathbb{R^3}$.
Here's an example, which for some readers may complete the BAC-CAB story:
\begin{align}
a \times (b \times c) & = \{ a \wedge (b \times c) \} ^\bot \\
& = a \cdot (b \times c)^\bot \\
& = -a \cdot b \wedge c
\end{align}
Another:
\begin{align}
a \cdot b \times c & = a \cdot (b\wedge c)^\bot \\
& = (a \wedge b \wedge c)^\bot
\end{align}
Rotations
On our way to describing angular momentum under a geometric algebra framework, we need to establish a preliminary object known as a rotor. And on the way to the rotor, we need to work through reflections. Consider a reflection axis $n$ ($n$ is a unit vector) and a vector $v$.
$$ v = v n n^{-1} $$
$$ = [ v \cdot n + v \wedge n ] n^{-1} $$
The first term is the component of $v$ along $n$. So, the reflected $v$ is just
$$ v' = -v \cdot n n^{-1} + v \wedge n n^{-1} $$
A small manipulation lets us write
$$ v' = -n \cdot v n^{-1} - n \wedge v n^{-1} $$
$$ = - [nv] n^{-1} = - nvn^{-1} $$
Kinda cute. And, due to Hamilton, this gets us halfway to rotations, because he noticed that a rotation is simply two reflections. Say we reflect along $b$, followed by $a$:
$$ v_{rotated} = (ab)v(ab)^{-1} = RvR^{-1} $$
$R$ is known as a rotor: a product of two unit vectors.
$$ R = ab = a \cdot b + a \wedge b $$
We can find the magnitude of $ a \wedge b $ by writing
$$ ba = b \cdot a + b \wedge a = a \cdot b - a \wedge b$$
and then considering the following product:
$$ abba = a^2 b^2 = 1 = (a \cdot b)^2 - (a \wedge b)^2 $$
Since $ a \cdot b $ is $\cos \theta$ we can write
$$ 1 - \cos^2 \theta = \sin^2 \theta = - (a \wedge b)^2 $$
So $ a \wedge b $ is a bivector whose magnitude is $\sin \theta $ and square is negative. Let $I$ be the unit bivector
$$ I = \frac{a \wedge b}{\sin \theta} $$
and write
$$ ab = R = \cos \theta + I \sin \theta = \exp (I \theta) $$
Looks like the common exponential phase factor $ e^{i \theta}$. Aside: what are the complex numbers? In two dimensions, vector $(a,b)$ is related to complex $a + ib$ like so: $a \hat{x} + b \hat{y} = \hat{x} (a + \hat{x} \hat{y} b) = \hat{x} (a + I b)$. Just pre-multiply by $\hat{x}$, selecting the real axis. You always knew there was something like this going on behind the scenes.
Moving along: let's use the dagger operator to reverse the order of vector products: $ (ab)^\dagger = ba $. Since unit vectors are their own inverse, we can write $y$ as a rotated version of $x$ like so:
$$ y = R x R^{-1} = R x R^\dagger $$
If $R$ is in fact a function of time, we can consider $ \dot{y} $:
\begin{align}
\dot{y} &= \dot{R} x R^\dagger + R x \dot{R^\dagger} \\
&= \dot{R} R^\dagger y + y R \dot{R^\dagger} \\
&= (\dot{R} R^\dagger) y - y (\dot{R}R^\dagger)
\end{align}
For the last step, take the time derivative of $R R^\dagger = 1$ and show $\dot{R} R^\dagger = - R \dot{R}^\dagger$.
Let $ \Omega= - {1 \over 2} \dot{R} R^\dagger $. $\Omega$ is apparently of even grade, and has no scalar part. $\Omega$ must be a bivector, and we can write:
$$ \dot{y} = - \Omega\cdot y = y \cdot \Omega$$
$$ v = v n n^{-1} $$
$$ = [ v \cdot n + v \wedge n ] n^{-1} $$
The first term is the component of $v$ along $n$. So, the reflected $v$ is just
$$ v' = -v \cdot n n^{-1} + v \wedge n n^{-1} $$
A small manipulation lets us write
$$ v' = -n \cdot v n^{-1} - n \wedge v n^{-1} $$
$$ = - [nv] n^{-1} = - nvn^{-1} $$
Kinda cute. And, due to Hamilton, this gets us halfway to rotations, because he noticed that a rotation is simply two reflections. Say we reflect along $b$, followed by $a$:
$$ v_{rotated} = (ab)v(ab)^{-1} = RvR^{-1} $$
$R$ is known as a rotor: a product of two unit vectors.
$$ R = ab = a \cdot b + a \wedge b $$
We can find the magnitude of $ a \wedge b $ by writing
$$ ba = b \cdot a + b \wedge a = a \cdot b - a \wedge b$$
and then considering the following product:
$$ abba = a^2 b^2 = 1 = (a \cdot b)^2 - (a \wedge b)^2 $$
Since $ a \cdot b $ is $\cos \theta$ we can write
$$ 1 - \cos^2 \theta = \sin^2 \theta = - (a \wedge b)^2 $$
So $ a \wedge b $ is a bivector whose magnitude is $\sin \theta $ and square is negative. Let $I$ be the unit bivector
$$ I = \frac{a \wedge b}{\sin \theta} $$
and write
$$ ab = R = \cos \theta + I \sin \theta = \exp (I \theta) $$
Looks like the common exponential phase factor $ e^{i \theta}$. Aside: what are the complex numbers? In two dimensions, vector $(a,b)$ is related to complex $a + ib$ like so: $a \hat{x} + b \hat{y} = \hat{x} (a + \hat{x} \hat{y} b) = \hat{x} (a + I b)$. Just pre-multiply by $\hat{x}$, selecting the real axis. You always knew there was something like this going on behind the scenes.
Moving along: let's use the dagger operator to reverse the order of vector products: $ (ab)^\dagger = ba $. Since unit vectors are their own inverse, we can write $y$ as a rotated version of $x$ like so:
$$ y = R x R^{-1} = R x R^\dagger $$
If $R$ is in fact a function of time, we can consider $ \dot{y} $:
\begin{align}
\dot{y} &= \dot{R} x R^\dagger + R x \dot{R^\dagger} \\
&= \dot{R} R^\dagger y + y R \dot{R^\dagger} \\
&= (\dot{R} R^\dagger) y - y (\dot{R}R^\dagger)
\end{align}
For the last step, take the time derivative of $R R^\dagger = 1$ and show $\dot{R} R^\dagger = - R \dot{R}^\dagger$.
Let $ \Omega= - {1 \over 2} \dot{R} R^\dagger $. $\Omega$ is apparently of even grade, and has no scalar part. $\Omega$ must be a bivector, and we can write:
$$ \dot{y} = - \Omega\cdot y = y \cdot \Omega$$
Orbital Dynamics
With rotational motion in our toolkit, let's develop the first few results in central-force dynamics. A nice trick for describing this is to shift the dynamics to a rotor. We accomplish this by considering a vector $x$ as a rotation of some basis vector, plus a stretch.
$$ x= U e_1 U^{\dagger} $$
We are now allowing a stretch as well as rotation, so U is a little more than a rotor: we are now considering a general even element. In three dimensions this is a scalar plus bivector [side note: the spinor and quaternion concepts fall out of this]. This presents us with a slight issue right away, in that U has four degrees of freedom, but we only need three for $x$. This allows us to apply a constraint on $U.$ But before deciding on that constraint, consider the rate of change of $x$:
$$ \dot{x} = \dot{U} e_1 U^\dagger + U e_1 \dot{U}^\dagger $$
Under the special case where $\dot{U} e_1 U^\dagger = U e_1 \dot{U}^\dagger $ (this is our constraint) we can write
$$ \dot{x} = 2 \dot{U} e_1 U^\dagger $$
Introduce a new variable s, such that $ {dt \over ds} =r= U U^\dagger$
$$ 2 {dU \over ds} = \dot{x} U e_1 $$
$$ 2 {d^2 U \over {ds}^2} = r \ddot{x} U e_1+ \dot{x} {dU \over ds} e_1 = (\ddot{x} x+{1 \over 2} \dot{x}^2) U $$
The last step follows because $r U e_1 = x U$. If we have inverse square central force motion
$$ m \ddot{x} = -{k x\over r^3} $$
then we have
$$ 2 {d^2 U \over {ds}^2} = (\ddot{x} x+{1 \over 2} \dot{x}^2) U $$
$$= ( -{k \over m} {x^2 \over r^3} + {1 \over 2} \dot{x}^2) U $$
$$= {1 \over m} ( -{k \over r} + {1 \over 2}m \dot{x}^2) U $$
$$ {d^2 U \over {ds}^2} = \left( {E \over 2m} \right) U $$
Harmonic motion. Nice.
Orbit Shape
A mass $m$, at $r$, moving with respect to some origin, defines a plane described by the bivector $L$, the angular momentum:\begin{align*}
L &\equiv r \wedge p \\
&= m r \wedge \dot{r}\\
&= m r \wedge (r \dot{\hat{r}} + \dot{r}\hat{r})\\
&= m r^2 (\hat{r} \wedge \dot{\hat{r}}) = m r^2 \hat{r} \dot{\hat{r}}
\end{align*}
For motion in a potential $V= -{k \over r}$ we have
$$ \dot{v} = -{k \over m r^2} \hat{r} $$
Form the product of this with $L$:
$$ L\dot{v} = -mr^2\dot{\hat{r}}\hat{r} \left( {-k \over mr^2} \right) \hat{r} = k\dot{\hat{r}} $$
$$ {d \over {dt}} (Lv - k \hat{r}) = 0 $$
And we see that $Lv- k \hat{r}$ is a constant of the motion. This vector, which sits in the plane of motion $(L \wedge v = 0)$, is known as the Runge-Lenz vector. Let's be explicit about the conserved vector and call it $e$:
$$ Lv = k \hat{r} + ke$$
With this definition, $e$ is dimensionless. If we multiply through with $r$ we have
\begin{equation} \label{eq2}
Lvr = k \hat{r} r + ker
\end{equation}
The left hand side is
\begin{align}
Lvr &= L (v \cdot r + v \wedge r)\\
&= L \left(v \cdot r + \frac{L^\dagger}{m} \right) \\
&= L (v \cdot r) + \frac{l^2}{m} \label{eq3}
\end{align}
The right hand side is
\begin{align*}
k \hat{r} r + ker &= k \abs{r} + k(e \cdot r + e \wedge r) \\
&= k \abs{r} + \abs{e} \abs{r} \cos \theta + e \wedge r
\end{align*}
So we have, finally
$$ l^2/m + (r \cdot v) L = k \abs{r} + k \abs{e} \abs{r} \cos \theta + e \wedge r $$
The scalar part is
$$l^2/m = k \abs{r} + k \abs{e} \abs{r} \cos \theta $$
or
$$ \abs{r} = \frac{l^2/mk}{1+\abs{e} \cos \theta}$$
and the bivector part is
$$ (r \cdot v) L = e \wedge r $$
or
$$ L = \frac{e \wedge r }{r \cdot v} $$
Returning to $\eqref{eq2}$ we can develop a relation between $E$, and $l$ and $e$
\begin{align*}
(Lv - k \hat{r})^2 &= k^2 e ^2 \\
(Lv)^2 - 2k(Lv) \cdot \hat{r} + k^2 &= k^2 e^2 \\
l^2 v^2 -2k \frac{\left< Lvr \right>}{r} &= k^2 (e^2 -1)
\end{align*}
Using $\eqref{eq3}$
\begin{align*}
l^2 v^2 - \frac{2kl^2}{mr} &= k^2 (e^2 -1)\\
\frac{2 l^2}{m} \left( \frac12 m v^2 - \frac kr \right) &= k^2 (e^2-1) \\
E &= \frac{m k^2}{2 l^2} (e^2 -1)
\end{align*}
