condest {Matrix}  R Documentation 
“Estimate”, i.e. compute approximately the CONDition number of
a (potentially large, often sparse) matrix A
.
It works by apply a fast randomized approximation of the 1norm,
norm(A,"1")
, through onenormest(.)
.
condest(A, t = min(n, 5), normA = norm(A, "1"),
silent = FALSE, quiet = TRUE)
onenormest(A, t = min(n, 5), A.x, At.x, n,
silent = FALSE, quiet = silent,
iter.max = 10, eps = 4 * .Machine$double.eps)
A 
a square matrix, optional for 
t 
number of columns to use in the iterations. 
normA 
number; (an estimate of) the 1norm of 
silent 
logical indicating if warning and (by default) convergence messages should be displayed. 
quiet 
logical indicating if convergence messages should be displayed. 
A.x , At.x 
when 
n 

iter.max 
maximal number of iterations for the 1norm estimator. 
eps 
the relative change that is deemed irrelevant. 
condest()
calls lu(A)
, and subsequently
onenormest(A.x = , At.x = )
to compute an approximate norm of
the inverse of A
, A^{1}
, in a way which
keeps using sparse matrices efficiently when A
is sparse.
Note that onenormest()
uses random vectors and hence
both functions' results are random, i.e., depend on the random
seed, see, e.g., set.seed()
.
Both functions return a list
;
condest()
with components,
est 
a number 
v 
the maximal 
The function onenormest()
returns a list with components,
est 
a number 
v 
01 integer vector length 
w 
numeric vector, the largest 
iter 
the number of iterations used. 
This is based on octave's condest()
and
onenormest()
implementations with original author
Jason Riedy, U Berkeley; translation to R and
adaption by Martin Maechler.
Nicholas J. Higham and Françoise Tisseur (2000). A Block Algorithm for Matrix 1Norm Estimation, with an Application to 1Norm Pseudospectra. SIAM J. Matrix Anal. Appl. 21, 4, 1185–1201.
William W. Hager (1984). Condition Estimates. SIAM J. Sci. Stat. Comput. 5, 311–316.
data(KNex, package = "Matrix")
mtm < with(KNex, crossprod(mm))
system.time(ce < condest(mtm))
sum(abs(ce$v)) ##  v _1 == 1
## Prove that  A v  =  A  / est (as v = 1):
stopifnot(all.equal(norm(mtm %*% ce$v),
norm(mtm) / ce$est))
## reciprocal
1 / ce$est
system.time(rc < rcond(mtm)) # takes ca 3 x longer
rc
all.equal(rc, 1/ce$est) # TRUE  the approximation was good
one < onenormest(mtm)
str(one) ## est = 12.3
## the maximal column:
which(one$v == 1) # mostly 4, rarely 1, depending on random seed