Matrix Inversion

I got the following error:

a = read.csv("mat.csv")
b = as.matrix(a)
tb = t(b)
bb = tb %*% b
dim(bb)
ibb = solve(bb)
bb %*% ibb
ibb = solve(bb)
Error in solve.default(bb) :
  system is computationally singular: reciprocal condition number =
1.77573e-19

Are there any ways to find more information about why it is singular?

Thanks.
Hello Wang

matrix bb is symmetric positive semidefinite, so
algebraically the eigenvalues are nonnegative.

I would use

bb <- crossprod(b)

to calculate bb (faster and possibly more accurate)

Look at eigen(bb,TRUE,TRUE)$values

(see ?eigen for the meaning of the arguments) to see how
many very small eigenvalues you have.  The number of zero
eigenvalues is equal to the number of linear relations
in the columns of b.

HTH

rksh

I got the following error:

a = read.csv("mat.csv")
b = as.matrix(a)
tb = t(b)
bb = tb %*% b
dim(bb)
ibb = solve(bb)
bb %*% ibb

ibb = solve(bb)
Error in solve.default(bb) :
  system is computationally singular: reciprocal condition number =
1.77573e-19

Are there any ways to find more information about why it is singular?

Thanks.
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I got the following error:

a = read.csv("mat.csv")
b = as.matrix(a)
tb = t(b)
bb = tb %*% b
dim(bb)
ibb = solve(bb)
bb %*% ibb

ibb = solve(bb)

Error in solve.default(bb) :
  system is computationally singular: reciprocal condition number =
1.77573e-19

Are there any ways to find more information about why it is singular?

Thanks.

Yes. Since the matrix is positive semidefinite by construction, I'd
probably go for chol(bb, pivot=TRUE), then the first "rank" elements of
"pivot" gives you a maximal subset of linearly independent columns, and
you can proceed by something like lm(b[,-subset]~b[,subset]) to see what
the linear dependencies are.

(Other approaches could be eigen() and svd().)
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