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 $R^{3}$ and are given $x \in R^{3}$ . We wish to write $x = α a + β b + γ c$ for some $α, β, γ \in R$ . The traditional Cramer’s rule for finding $α, β, γ$ would be

$α = \frac{det [\begin{matrix} x_{1} & b_{1} & c_{1} \\ x_{2} & b_{2} & c_{2} \\ x_{3} & b_{3} & c_{3} \end{matrix}]}{det [\begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix}]} β = \frac{det [\begin{matrix} a_{1} & x_{1} & c_{1} \\ a_{2} & x_{2} & c_{2} \\ a_{3} & x_{3} & c_{3} \end{matrix}]}{det [\begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix}]} γ = \frac{det [\begin{matrix} a_{1} & b_{1} & x_{1} \\ a_{2} & b_{2} & x_{2} \\ a_{3} & b_{3} & x_{3} \end{matrix}]}{det [\begin{matrix} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{matrix}]}$

where $x = (x_{1}, x_{2}, x_{3})^{T}$ , $a = (a_{1}, a_{2}, a_{3})^{T}$ , $b = (b_{1}, b_{2}, b_{3})^{T}$ and $c = (c_{1}, c_{2}, c_{3})^{T}$ . 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 $R^{3}$ as a vector space and given vectors $a, b, c$ spanning the space we can express any vector $x \in R^{3}$ as

$x = (\frac{x \land b \land c}{a \land b \land c}) a + (\frac{a \land x \land c}{a \land b \land c}) b + (\frac{a \land b \land x}{a \land b \land c}) c$

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 $x = α a + β b + γ c$ . Conversely, we might know the coefficients and try to determine them using package idiom. Here we will use $1, 2, 3$ and suppose that $y = 1 a + 2 b + 3 c$ :

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 $R^{5}$ :

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.

Robin K. S. Hankin

Using Clifford algebra

Higher dimensional space

References