R has a qr()
function, which performs QR decomposition using either LINPACK or LAPACK (in my experience, the latter is 5% faster). The main object returned is a matrix "qr" that contains in the upper triangular matrix R (i.e. R=qr[upper.tri(qr)]
). So far so good. The lower triangular part of qr contains Q "in compact form". One can extract Q from the qr decomposition by using qr.Q()
. I would like to find the inverse of qr.Q()
. In other word, I do have Q and R, and would like to put them in a "qr" object. R is trivial but Q is not. The goal is to apply to it qr.solve()
, which is much faster than solve()
on large systems.
R uses the LINPACK dqrdc
routine, by default, or the LAPACK DGEQP3
routine, when specified, for computing the QR decomposition. Both routines compute the decomposition using Householder reflections. An m x n matrix A is decomposed into an m x n economy-size orthogonal matrix (Q) and an n x n upper triangular matrix (R) as A = QR, where Q can be computed by the product of t Householder reflection matrices, with t being the lesser of m-1 and n: Q = H1H2...Ht.
Each reflection matrix Hi can be represented by a length-(m-i+1) vector. For example, H1 requires a length-m vector for compact storage. All but one entry of this vector is placed in the first column of the lower triangle of the input matrix (the diagonal is used by the R factor). Therefore, each reflection needs one more scalar of storage, and this is provided by an auxiliary vector (called $qraux
in the result from R's qr
).
The compact representation used is different between the LINPACK and LAPACK routines.
A Householder reflection is computed as Hi = I - viviT/pi, where I is the identity matrix, pi is the corresponding entry in $qraux
, and vi is as follows:
qr
)Let's work through the example from the QR decomposition article at Wikipedia in R.
The matrix being decomposed is
> A <- matrix(c(12, 6, -4, -51, 167, 24, 4, -68, -41), nrow=3)
> A
[,1] [,2] [,3]
[1,] 12 -51 4
[2,] 6 167 -68
[3,] -4 24 -41
We do the decomposition, and the most relevant portions of the result is shown below:
> Aqr = qr(A)
> Aqr
$qr
[,1] [,2] [,3]
[1,] -14.0000000 -21.0000000 14
[2,] 0.4285714 -175.0000000 70
[3,] -0.2857143 0.1107692 -35
[snip...]
$qraux
[1] 1.857143 1.993846 35.000000
[snip...]
This decomposition was done (under the covers) by computing two Householder reflections and multiplying them by A to get R. We will now recreate the reflections from the information in $qr
.
> p = Aqr$qraux # for convenience
> v1 <- matrix(c(p[1], Aqr$qr[2:3,1]))
> v1
[,1]
[1,] 1.8571429
[2,] 0.4285714
[3,] -0.2857143
> v2 <- matrix(c(0, p[2], Aqr$qr[3,2]))
> v2
[,1]
[1,] 0.0000000
[2,] 1.9938462
[3,] 0.1107692
> I = diag(3) # identity matrix
> H1 = I - v1 %*% t(v1)/p[1] # I - v1*v1^T/p[1]
> H2 = I - v2 %*% t(v2)/p[2] # I - v2*v2^T/p[2]
> Q = H1 %*% H2
> Q
[,1] [,2] [,3]
[1,] -0.8571429 0.3942857 0.33142857
[2,] -0.4285714 -0.9028571 -0.03428571
[3,] 0.2857143 -0.1714286 0.94285714
Now let's verify the Q computed above is correct:
> qr.Q(Aqr)
[,1] [,2] [,3]
[1,] -0.8571429 0.3942857 0.33142857
[2,] -0.4285714 -0.9028571 -0.03428571
[3,] 0.2857143 -0.1714286 0.94285714
Looks good! We can also verify QR is equal to A.
> R = qr.R(Aqr) # extract R from Aqr$qr
> Q %*% R
[,1] [,2] [,3]
[1,] 12 -51 4
[2,] 6 167 -68
[3,] -4 24 -41
A Householder reflection is computed as Hi = I - piviviT, where I is the identity matrix, pi is the corresponding entry in $qraux
, and vi is as follows:
qr
)There is another twist when using the LAPACK routine in R: column pivoting is used, so the decomposition is solving a different, related problem: AP = QR, where P is a permutation matrix.
This section does the same example as before.
> A <- matrix(c(12, 6, -4, -51, 167, 24, 4, -68, -41), nrow=3)
> Bqr = qr(A, LAPACK=TRUE)
> Bqr
$qr
[,1] [,2] [,3]
[1,] 176.2554964 -71.1694118 1.668033
[2,] -0.7348557 35.4388886 -2.180855
[3,] -0.1056080 0.6859203 -13.728129
[snip...]
$qraux
[1] 1.289353 1.360094 0.000000
$pivot
[1] 2 3 1
attr(,"useLAPACK")
[1] TRUE
[snip...]
Notice the $pivot
field; we will come back to that. Now we generate Q from the information the Aqr
.
> p = Bqr$qraux # for convenience
> v1 = matrix(c(1, Bqr$qr[2:3,1]))
> v1
[,1]
[1,] 1.0000000
[2,] -0.7348557
[3,] -0.1056080
> v2 = matrix(c(0, 1, Bqr$qr[3,2]))
> v2
[,1]
[1,] 0.0000000
[2,] 1.0000000
[3,] 0.6859203
> H1 = I - p[1]*v1 %*% t(v1) # I - p[1]*v1*v1^T
> H2 = I - p[2]*v2 %*% t(v2) # I - p[2]*v2*v2^T
> Q = H1 %*% H2
[,1] [,2] [,3]
[1,] -0.2893527 -0.46821615 -0.8348944
[2,] 0.9474882 -0.01602261 -0.3193891
[3,] 0.1361660 -0.88346868 0.4482655
Once again, the Q computed above agrees with the R-provided Q.
> qr.Q(Bqr)
[,1] [,2] [,3]
[1,] -0.2893527 -0.46821615 -0.8348944
[2,] 0.9474882 -0.01602261 -0.3193891
[3,] 0.1361660 -0.88346868 0.4482655
Finally, let's compute QR.
> R = qr.R(Bqr)
> Q %*% R
[,1] [,2] [,3]
[1,] -51 4 12
[2,] 167 -68 6
[3,] 24 -41 -4
Notice the difference? QR is A with its columns permuted given the order in Bqr$pivot
above.
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