I am working on a piece of code to find the QR factorization of a matrix in R.
X <- structure(c(0.8147, 0.9058, 0.127, 0.9134, 0.6324, 0.0975, 0.2785,
0.5469, 0.9575, 0.9649, 0.1576, 0.9706, 0.9572, 0.4854, 0.8003
), .Dim = c(5L, 3L))
myqr <- function(A) {
n <- nrow(A)
p <- ncol(A)
Q <- diag(n)
Inp <- diag(nrow = n, ncol = p)
for(k in c(1:ncol(A))) {
# extract the kth column of the matrix
col<-A[k:n,k]
# calculation of the norm of the column in order to create the vector
norm1<-sqrt(sum(col^2))
# Define the sign positive if a1 > 0 (-) else a1 < 0(+)
sign <- ifelse(col[1] >= 0, -1, +1)
# Calculate of the vector a_r
a_r <- col - sign * Inp[k:n,k] * norm1
# beta = 2 / ||a-r||^2
beta <- 2 / sum(t(a_r) %*% a_r)
# the next line of code calculates the matrix Q in every step
Q <- Q - beta *Q %*% c(rep(0,k-1),a_r) %*% t(c(rep(0,k-1),a_r))
# calculates the matrix R in each step
A[k:n,k:p] <- A[k:n,k:p] - beta * a_r %*% t(a_r) %*% A[k:n,k:p]
}
list(Q=Q,R=A)
}
But, Here I have not calculated in every step the matrix H that represents the householder reflection, also I have not calculated the matrix A in every step.
As H = I - 2 v v', if I multiply by Q I obtain
QH = Q - 2 (Qv) v' // multiplication on the left
HQ = Q - 2 v (Q'v)' // multiplication on the right
Now, this operations should be work in every step. However if I consider the first matrix H and he the second matrix H1.... these matrices will be of smaller that the first one. In order to avoid that I have used the next line of code:
Q <- Q - beta * Q %*% c(rep(0,k-1),a_r) %*% t(c(rep(0,k-1),a_r))
but, I am not sure why the code is working well, when I generate the new vector a_r with the first k entries of zeros at every step.
I thought you want exactly the same output as returned by qr.default, which uses compact QR storage. But then I realized that you are storing Q and R factors separately.
Normally, QR factorization only forms R but not Q. In the following, I will describe QR factorization where both are formed. For those who lack basic understanding of QR factorization, please read this first: lm(): What is qraux returned by QR decomposition in LINPACK / LAPACK, where there are neat math formulae arranged in LaTeX. In the following, I will assume that one knows what a Householder reflection is and how it is computed.
First of all, a Householder refection vector is H = I - beta * v v' (where beta is computed as in your code), not H = I - 2 * v v'.
Then, QR factorization A = Q R proceeds as (Hp ... H2 H1) A = R, where Q = H1 H2 ... Hp. To compute Q, we initialize Q = I (identity matrix), then multiply Hk on the right iteratively in the loop. To compute R, we initialize R = A and multiply Hk on the left iteratively in the loop.
Now, at k-th iteration, we have a rank-1 matrix update on Q and A:
Q := Q Hk = Q (I - beta v * v') = Q - (Q v) (beta v)'
A := Hk A = (I - beta v * v') A = A - (beta v) (A' v)'
v = c(rep(0, k-1), a_r), where a_r is the reduced, non-zero part of a full reflection vector.
The code you have is doing such update in a brutal force:
Q <- Q - beta * Q %*% c(rep(0,k-1), a_r) %*% t(c(rep(0,k-1),a_r))
It first pads a_r to get the full reflection vector and performs the rank-1 update on the whole matrix. But actually we can drop off those zeros and write (do some matrix algebra if unclear):
Q[,k:n] <- Q[,k:n] - tcrossprod(Q[, k:n] %*% a_r, beta * a_r)
A[k:n,k:p] <- A[k:n,k:p] - tcrossprod(beta * a_r, crossprod(A[k:n,k:p], a_r))
so that only a fraction of Q and A are updated.
t() and "%*%" a lot! But almost all of them can be replaced by crossprod() or tcrossprod(). This eliminates the explicit transpose t() and is more memory efficient;You initialize another diagonal matrix Inp which is not necessary. To get householder reflection vector a_r, you can replace
sign <- ifelse(col[1] >= 0, -1, +1)
a_r <- col - sign * Inp[k:n,k] * norm1
by
a_r <- col; a_r[1] <- a_r[1] + sign(a_r[1]) * norm1
where sign is an R base function.
## QR factorization: A = Q %*% R
## if `complete = FALSE` (default), return thin `Q`, `R` factor
## if `complete = TRUE`, return full `Q`, `R` factor
myqr <- function (A, complete = FALSE) {
n <- nrow(A)
p <- ncol(A)
Q <- diag(n)
for(k in 1:p) {
# extract the kth column of the matrix
col <- A[k:n,k]
# calculation of the norm of the column in order to create the vector r
norm1 <- sqrt(drop(crossprod(col)))
# Calculate of the reflection vector a-r
a_r <- col; a_r[1] <- a_r[1] + sign(a_r[1]) * norm1
# beta = 2 / ||a-r||^2
beta <- 2 / drop(crossprod(a_r))
# update matrix Q (trailing matrix only) by Householder reflection
Q[,k:n] <- Q[,k:n] - tcrossprod(Q[, k:n] %*% a_r, beta * a_r)
# update matrix A (trailing matrix only) by Householder reflection
A[k:n, k:p] <- A[k:n, k:p] - tcrossprod(beta * a_r, crossprod(A[k:n,k:p], a_r))
}
if (complete) {
A[lower.tri(A)] <- 0
return(list(Q = Q, R = A))
}
else {
R <- A[1:p, ]; R[lower.tri(R)] <- 0
return(list(Q = Q[,1:p], R = R))
}
}
Now let's have a test:
X <- structure(c(0.8147, 0.9058, 0.127, 0.9134, 0.6324, 0.0975, 0.2785,
0.5469, 0.9575, 0.9649, 0.1576, 0.9706, 0.9572, 0.4854, 0.8003
), .Dim = c(5L, 3L))
# [,1] [,2] [,3]
#[1,] 0.8147 0.0975 0.1576
#[2,] 0.9058 0.2785 0.9706
#[3,] 0.1270 0.5469 0.9572
#[4,] 0.9134 0.9575 0.4854
#[5,] 0.6324 0.9649 0.8003
First for thin-QR version:
## thin QR factorization
myqr(X)
#$Q
# [,1] [,2] [,3]
#[1,] -0.49266686 -0.4806678 0.17795345
#[2,] -0.54775702 -0.3583492 -0.57774357
#[3,] -0.07679967 0.4754320 -0.63432053
#[4,] -0.55235290 0.3390549 0.48084552
#[5,] -0.38242607 0.5473120 0.03114461
#
#$R
# [,1] [,2] [,3]
#[1,] -1.653653 -1.1404679 -1.2569776
#[2,] 0.000000 0.9660949 0.6341076
#[3,] 0.000000 0.0000000 -0.8815566
Now the full QR version:
## full QR factorization
myqr(X, complete = TRUE)
#$Q
# [,1] [,2] [,3] [,4] [,5]
#[1,] -0.49266686 -0.4806678 0.17795345 -0.6014653 -0.3644308
#[2,] -0.54775702 -0.3583492 -0.57774357 0.3760348 0.3104164
#[3,] -0.07679967 0.4754320 -0.63432053 -0.1497075 -0.5859107
#[4,] -0.55235290 0.3390549 0.48084552 0.5071050 -0.3026221
#[5,] -0.38242607 0.5473120 0.03114461 -0.4661217 0.5796209
#
#$R
# [,1] [,2] [,3]
#[1,] -1.653653 -1.1404679 -1.2569776
#[2,] 0.000000 0.9660949 0.6341076
#[3,] 0.000000 0.0000000 -0.8815566
#[4,] 0.000000 0.0000000 0.0000000
#[5,] 0.000000 0.0000000 0.0000000
Now let's check standard result returned by qr.default:
QR <- qr.default(X)
## thin R factor
qr.R(QR)
# [,1] [,2] [,3]
#[1,] -1.653653 -1.1404679 -1.2569776
#[2,] 0.000000 0.9660949 0.6341076
#[3,] 0.000000 0.0000000 -0.8815566
## thin Q factor
qr.Q(QR)
# [,1] [,2] [,3]
#[1,] -0.49266686 -0.4806678 0.17795345
#[2,] -0.54775702 -0.3583492 -0.57774357
#[3,] -0.07679967 0.4754320 -0.63432053
#[4,] -0.55235290 0.3390549 0.48084552
#[5,] -0.38242607 0.5473120 0.03114461
## full Q factor
qr.Q(QR, complete = TRUE)
# [,1] [,2] [,3] [,4] [,5]
#[1,] -0.49266686 -0.4806678 0.17795345 -0.6014653 -0.3644308
#[2,] -0.54775702 -0.3583492 -0.57774357 0.3760348 0.3104164
#[3,] -0.07679967 0.4754320 -0.63432053 -0.1497075 -0.5859107
#[4,] -0.55235290 0.3390549 0.48084552 0.5071050 -0.3026221
#[5,] -0.38242607 0.5473120 0.03114461 -0.4661217 0.5796209
So our results are correct!