question about SVD
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Jim Newton
Apr 4, 2021, 11:58:54 AM4/4/21
to Scala Breeze
I notice that the `svd` function gives me a decomposition, (u,s,v), where `s` seems to be the vector consisting of eigenvalues, but u*v is not always the inverse. Sometimes it seems to be the negative of the inverse. Is that intended. As I recall in the singular value decomposition u and v are inverses of each other.
Am I confused?
Alec Zorab
Apr 6, 2021, 4:52:27 AM4/6/21
to Scala Breeze
from memory, I think the breeze SVD gives you (u, s, vt)
On Sun, 4 Apr 2021 at 16:58, Jim Newton <jimka...@gmail.com> wrote:
I notice that the `svd` function gives me a decomposition, (u,s,v), where `s` seems to be the vector consisting of eigenvalues, but u*v is not always the inverse. Sometimes it seems to be the negative of the inverse. Is that intended. As I recall in the singular value decomposition u and v are inverses of each other.Am I confused?
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Jim Newton
Apr 7, 2021, 3:52:13 AM4/7/21
to Scala Breeze
Some experimentation seems to show that it is v, not vt.
The following test passes. So I think svt returns (u,s,v) such that u*diag(s)*v equal the original matrix.
test("breeze"){
import breeze.linalg._
import scala.util.Random
def maxabsnorm(dim:Int, m:DenseMatrix[Double]):Double = {
(0 until dim).foldLeft(m(0,0)){(acc:Double,row:Int) =>
val x = (0 until dim).foldLeft(m(0,0)){
(acc:Double,col:Int) => math.max(acc,math.abs(m(row,col)))}
math.max(acc,x)
}
}
for{dim <- 1 to 4} {
val dm1: DenseMatrix[Double] = DenseMatrix.tabulate(dim, dim)((_, _)=>Random.between(-10.0, 10.0))
val svd.SVD(u, s, v) = svd(dm1)
val dm2 = u * diag(s) * v
val dist:Double = maxabsnorm(dim,dm2 - dm1)
assert(dist < 0.001, s"dist=$dist dm1=[$dm1] dm2=[$dm2] diff=[${dm2-dm1}]")
}
}
import breeze.linalg._
import scala.util.Random
def maxabsnorm(dim:Int, m:DenseMatrix[Double]):Double = {
(0 until dim).foldLeft(m(0,0)){(acc:Double,row:Int) =>
val x = (0 until dim).foldLeft(m(0,0)){
(acc:Double,col:Int) => math.max(acc,math.abs(m(row,col)))}
math.max(acc,x)
}
}
for{dim <- 1 to 4} {
val dm1: DenseMatrix[Double] = DenseMatrix.tabulate(dim, dim)((_, _)=>Random.between(-10.0, 10.0))
val svd.SVD(u, s, v) = svd(dm1)
val dm2 = u * diag(s) * v
val dist:Double = maxabsnorm(dim,dm2 - dm1)
assert(dist < 0.001, s"dist=$dist dm1=[$dm1] dm2=[$dm2] diff=[${dm2-dm1}]")
}
}
Alec Zorab
Apr 7, 2021, 7:35:28 AM4/7/21
to Scala Breeze
That's interesting. The naming in the source implies it's vt. https://github.com/scalanlp/breeze/blob/master/math/src/main/scala/breeze/linalg/functions/svd.scala#L27
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Jim Newton
Apr 7, 2021, 8:17:08 AM4/7/21
to Scala Breeze
is my `maxabsnorm` function buggy? I could not figure out how to use the built-in norm function to measure the norm of dm2-dm1.
Darren Wilkinson
Apr 8, 2021, 11:19:36 AM4/8/21
to Scala Breeze
It is vt - the function returns (u, s, vt). But it is correct that u * diag(s) * vt is the original matrix. I think that this is all fine.
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