lu! is not In-place ?

[Disatiler]

Hello everyone,

Recently I need to do the LU factorization "in-place" for a matrix.
I know function with ! will represent the "in-place" case but the result (in type of LU) looks to not be store in the original matrix.

Here a MWE :

using LinearAlgebra

function main()
    A = rand(10, 10)
    println(A)
    lu!(A, NoPivot())
    println("LU factorization of A:")
    println(A)
end

main()

The matrix A is the same at the end.

Best regard,

[araujoms]

The matrix A is not the same at the end, you should look more closely. Or create a copy with B = copy(A), and look at A-B.

In any case, !, does not mean that the result will be stored there, only that the argument A will be used up to save memory. In the case of lu! it is not even possible to store the result in A, there's not enough space. You still need to do F = lu!(A), the result will be stored in F.

In the future, please ask this kind of question on Slack or Discourse.

[dkarrasch]

A few seconds late, but here is an explicit example:

julia> A = randn(3, 3)
3×3 Matrix{Float64}:
  1.39501    0.99615   -0.14752
  1.25005   -0.670471   0.580276
 -0.172159   0.556769   0.202036

julia> lu!(A, NoPivot())
LU{Float64, Matrix{Float64}, Vector{Int64}}
L factor:
3×3 Matrix{Float64}:
  1.0        0.0       0.0
  0.896084   1.0       0.0
 -0.12341   -0.434842  1.0
U factor:
3×3 Matrix{Float64}:
 1.39501   0.99615  -0.14752
 0.0      -1.56311   0.712466
 0.0       0.0       0.493641

julia> A
3×3 Matrix{Float64}:
  1.39501    0.99615   -0.14752
  0.896084  -1.56311    0.712466
 -0.12341   -0.434842   0.493641

It's just the first row that is untouched.