A^k for matrix multiplication in R? - r

Question

Suppose A is some square matrix. How can I easily exponentiate this matrix in R?
I tried two ways already: Trial 1 with a for-loop hack and Trial 2 a bit more elegantly but it is still a far cry from Ak simplicity.
Trial 1
set.seed(10)
t(matrix(rnorm(16),ncol=4,nrow=4)) -> a
for(i in 1:2){a <- a %*% a}
Trial 2
a <- t(matrix(c(0,1,0,0,0,0,1,0,0,0,0,1,0,0,0,0),nrow=4))
i <- diag(4)
(function(n) {if (n<=1) a else (i+a) %*% Recall(n-1)})(10)

Answer 1

If A is diagonizable, you could use eigenvalue decomposition:
matrix.power <- function(A, n) { # only works for diagonalizable matrices
e <- eigen(A)
M <- e$vectors # matrix for changing basis
d <- e$values # eigen values
return(M %*% diag(d^n) %*% solve(M))
}
When A is not diagonalizable, the matrix M (matrix of eigenvectors) is singular. Thus, using it with A = matrix(c(0,1,0,0),2,2) would give Error in solve.default(M) : system is computationally singular.

Answer 2

The expm package has an %^% operator:
library("sos")
findFn("{matrix power}")
install.packages("expm")
library("expm")
?matpow
set.seed(10);t(matrix(rnorm(16),ncol=4,nrow=4))->a
a%^%8

Answer 3

Although Reduce is more elegant, a for-loop solution is faster and seems to be as fast as expm::%^%
m1 <- matrix(1:9, 3)
m2 <- matrix(1:9, 3)
m3 <- matrix(1:9, 3)
system.time(replicate(1000, Reduce("%*%" , list(m1,m1,m1) ) ) )
# user system elapsed
# 0.026 0.000 0.037
mlist <- list(m1,m2,m3)
m0 <- diag(1, nrow=3,ncol=3)
system.time(replicate(1000, for (i in 1:3 ) {m0 <- m0 %*% m1 } ) )
# user system elapsed
# 0.013 0.000 0.014
library(expm) # and I think this may be imported with pkg:Matrix
system.time(replicate(1000, m0%^%3))
# user system elapsed
#0.011 0.000 0.017
On the other hand the matrix.power solution is much, much slower:
system.time(replicate(1000, matrix.power(m1, 4)) )
user system elapsed
0.677 0.013 1.037
#BenBolker is correct (yet again). The for-loop appears linear in time as the exponent rises whereas the expm::%^% function appears to be even better than log(exponent).
> m0 <- diag(1, nrow=3,ncol=3)
> system.time(replicate(1000, for (i in 1:400 ) {m0 <- m0 %*% m1 } ) )
user system elapsed
0.678 0.037 0.708
> system.time(replicate(1000, m0%^%400))
user system elapsed
0.006 0.000 0.006

Answer 4

Indeed the expm's package does use exponentiation by squaring.
In pure r, this can be done rather efficiently like so,
"%^%" <- function(mat,power){
base = mat
out = diag(nrow(mat))
while(power > 1){
if(power %% 2 == 1){
out = out %*% base
}
base = base %*% base
power = power %/% 2
}
out %*% base
}
Timing this,
m0 <- diag(1, nrow=3,ncol=3)
system.time(replicate(10000, m0%^%4000))#expm's %^% function
user system elapsed
0.31 0.00 0.31
system.time(replicate(10000, m0%^%4000))# my %^% function
user system elapsed
0.28 0.00 0.28
So, as expected, they are the same speed because they use the same algorithm. It looks like the overhead of the looping r code does not make a significant difference.
So, if you don't want to use expm, and need that performance, then you can just use this, if you don't mind looking at imperative code.

Answer 5

A shorter solution with eigenvalue decomposition:
"%^%" <- function(S, power)
with(eigen(S), vectors %*% (values^power * t(vectors)))

Answer 6

Simple solution
`%^%` <- function(A, n) {
A1 <- A
for(i in seq_len(n-1)){
A <- A %*% A1
}
return(A)
}