Cramer’s rule in civilised form with Clifford algebra

To cite the clifford package in publications please use Hankin (2022). This short document shows a nice application of Clifford algebras to linear algebra. Suppose we have vectors a, b, c spanning ℝ3 and are given x ∈ ℝ3. We wish to write x = αa + βb + γc for some α, β, γ ∈ ℝ. The traditional Cramer’s rule for finding α, β, γ would be

$$ \alpha=\frac{ \det\begin{bmatrix} x_1&b_1&c_1\\ x_2&b_2&c_2\\ x_3&b_3&c_3 \end{bmatrix} }{ \det\begin{bmatrix} a_1&b_1&c_1\\ a_2&b_2&c_2\\ a_3&b_3&c_3 \end{bmatrix} } \qquad\beta=\frac{ \det\begin{bmatrix} a_1&x_1&c_1\\ a_2&x_2&c_2\\ a_3&x_3&c_3 \end{bmatrix} }{ \det\begin{bmatrix} a_1&b_1&c_1\\ a_2&b_2&c_2\\ a_3&b_3&c_3 \end{bmatrix} } \qquad \gamma=\frac{ \det\begin{bmatrix} a_1&b_1&x_1\\ a_2&b_2&x_2\\ a_3&b_3&x_3 \end{bmatrix} }{ \det\begin{bmatrix} a_1&b_1&c_1\\ a_2&b_2&c_2\\ a_3&b_3&c_3 \end{bmatrix} } $$

where x = (x1, x2, x3)T, a = (a1, a2, a3)T, b = (b1, b2, b3)T and c = (c1, c2, c3)T. However, observe that this solution, while accurate, requires one to take a coordinate basis; and offers little in the way of intuition.

Using Clifford algebra

Considering ℝ3 as a vector space and given vectors a, b, c spanning the space we can express any vector x ∈ ℝ3 as

$$\mathbf{x}= \left(\frac{{\mathbf x}\wedge{\mathbf b}\wedge{\mathbf c}}{{\mathbf a}\wedge{\mathbf b}\wedge{\mathbf c}}\right){\mathbf a}+ \left(\frac{{\mathbf a}\wedge{\mathbf x}\wedge{\mathbf c}}{{\mathbf a}\wedge{\mathbf b}\wedge{\mathbf c}}\right){\mathbf b}+ \left(\frac{{\mathbf a}\wedge{\mathbf b}\wedge{\mathbf x}}{{\mathbf a}\wedge{\mathbf b}\wedge{\mathbf c}}\right){\mathbf c} $$

which is Cramer’s rule expressed directly in vector form (rather than components). Observe that the numerator and denominator of each bracketed term is a pseudoscalar; the ratio of two pseudoscalars is an ordinary scalar. Package idiom is straightforward:

a <- as.1vector(runif(3))
b <- as.1vector(runif(3))
c <- as.1vector(runif(3))

(x <- as.1vector(1:3))
## Element of a Clifford algebra, equal to
## + 1e_1 + 2e_2 + 3e_3
options(maxdim = 3)  # needed to drop() pseudoscalars
abc <- drop(a ^ b ^ c)

alpha <- drop(x ^ b ^ c)/abc
beta  <- drop(a ^ x ^ c)/abc
gamma <- drop(a ^ b ^ x)/abc

c(alpha,beta,gamma)
## [1] -3.805997 -5.328439  8.309581
alpha*a + beta*b + gamma*c
## Element of a Clifford algebra, equal to
## + 1e_1 + 2e_2 + 3e_3
Mod(alpha*a + beta*b + gamma*c-x)
## [1] 0

Thus we have expressed x (except for possible roundoff error) as a linear combination of a, b, c, specifically x = αa + βb + γc. Conversely, we might know the coefficients and try to determine them using package idiom. Here we will use 1, 2, 3 and suppose that y = 1a + 2b + 3c:

y <- a*1 + b*2 + c*3
c(
    drop(y ^ b ^ c)/abc,
    drop(a ^ y ^ c)/abc,
    drop(a ^ b ^ y)/abc
)
## [1] 1 2 3

Higher dimensional space

To accomplish this in arbitrary-dimensional space is straightforward. Here we consider ℝ5:

n <- 5                                                 # dimensionality of space
options(maxdim=5)                                      # safety precaution
x <- as.1vector(seq_len(n))                            # target vector
x
## Element of a Clifford algebra, equal to
## + 1e_1 + 2e_2 + 3e_3 + 4e_4 + 5e_5
L <- replicate(n,as.1vector(rnorm(n)),simplify=FALSE)  # spanning vectors
subst <- function(L,n,x){L[[n]] <- x; return(L)}       # list substitution
coeff <- function(n,L,x){
      drop(Reduce(`^`,subst(L,n,x))/Reduce(`^`,L))
}

Then the coefficients are given by:

(alpha <- sapply(seq_len(n),coeff,L,x))
## [1]  21.610237  27.030973  11.866383 -22.222116   3.199427

and we can reconstitute vector x:

out <- as.clifford(0)
f <- function(i){alpha[i]*L[[i]]}
for(i in seq_len(n)){
      out <- out + f(i)
}
Mod(out-x)  # zero to numerical precision
## [1] 2.340649e-14

Or, somewhat slicker:

Reduce(`+`,sapply(seq_len(n),f,simplify=FALSE))
## Element of a Clifford algebra, equal to
## + 1e_1 + 2e_2 + 3e_3 + 4e_4 + 5e_5

Conversely, if we know the coefficients are, say, 15:11, then we would have

coeffs <- 15:11
x <- 0
for(i in seq_len(5)){x <- x + coeffs[i]*L[[i]]}
x
## Element of a Clifford algebra, equal to
## + 29.23618e_1 + 18.68409e_2 + 20.23065e_3 - 0.6361232e_4 - 40.82507e_5

And then to find the coefficients:

sapply(seq_len(n),coeff,L,x)
## [1] 15 14 13 12 11

Above we see that the original coefficients are recovered, up to numerical accuracy.

References

Hankin, R. K. S. 2022. “Clifford Algebra in R.” arXiv. https://doi.org/10.48550/ARXIV.2209.13659.