rdiv!(A', svd(B))

Specific Domains Numerics
,
1

Is it possible to rdiv! by an SVD factorization? As in

using LinearAlgebra
A = ones(3)
B = randn(3, 3)
Blu = lu!(copy(B))
rdiv!(A', Blu) # works
Bsvd = svd!(copy(B))
rdiv!(A', Bsvd) # fails, no such method

In other words, I am asking if there is a technical reason preventing this, or it simply wasn’t added (yet, as of 1.13.3).

2

So basically you want: X \leftarrow X F^{+}, where F = U \Sigma V^\*. Then you can make use of:

\left( X F^{+} \right)^{\top} = \left( F^{+} \right)^{\top} X^{\top} = \left( F^{\top} \right)^{+} X^{\top}

So this should work:

LinearAlgebra.rdiv!(X::AbstractMatrix, F::SVD) = (ldiv!(F', X'); X)

With your example I get

julia> rdiv!(A', Bsvd)
1×3 adjoint(::Vector{Float64}) with eltype Float64:
 1.2252  -2.33595  0.699014

p.s. this works because ldiv!(F::SVD, B::AbstractVecOrMat) does exist: