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 \({\mathbf a}, {\mathbf b}, {\mathbf c}\) spanning \(\mathbb{R}^3\) and are given \({\mathbf x}\in\mathbb{R}^3\). We wish to write \({\mathbf x}=\alpha {\mathbf a}+\beta {\mathbf b}+\gamma {\mathbf c}\) for some \(\alpha,\beta,\gamma\in\mathbb{R}\). The traditional Cramer’s rule for finding \(\alpha,\beta,\gamma\) 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 \({\mathbf x}=(x_1,x_2,x_3)^T\), \({\mathbf a}=(a_1,a_2,a_3)^T\), \({\mathbf b}=(b_1,b_2,b_3)^T\) and \({\mathbf c}=(c_1,c_2,c_3)^T\). 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 \(\mathbb{R}^3\) as a vector space and given vectors \({\mathbf a}, {\mathbf b}, {\mathbf c}\) spanning the space we can express any vector \({\mathbf x}\in\mathbb{R}^3\) 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 \({\mathbf x}\) (except for possible roundoff error) as a linear combination of \({\mathbf a},{\mathbf b},{\mathbf c}\), specifically \({\mathbf x}=\alpha{\mathbf a}+\beta{\mathbf b}+\gamma{\mathbf c}\). Conversely, we might know the coefficients and try to determine them using package idiom. Here we will use \(1,2,3\) and suppose that \({\mathbf y}=1{\mathbf a}+2{\mathbf b}+3{\mathbf c}\):

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 \(\mathbb{R}^{5}\):

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