vector transpose equality

[uniment]

julia> @show [1;2;3]  == [1 2 3]'
       @show [1;2;3]' == [1 2 3]''
       @show [1 2 3]  == [1;2;3]'
       @show [1 2 3]' == [1;2;3]''
[1; 2; 3] == ([1 2 3])' = false
([1; 2; 3])' == (([1 2 3])')' = true
[1 2 3] == ([1; 2; 3])' = true
([1 2 3])' == (([1; 2; 3])')' = false

[jakobnissen]

That seems right to me. What's the issue?

[barucden]

I think this boils down to the fact that a single-column matrix is never equal to a (column) vector:

julia> X = ones(Int, 3, 1)
3×1 Matrix{Int64}:
 1
 1
 1

julia> X == X[:, 1]
false

[uniment]

@jakobnissen maybe i'm rusty but i'm unaware of any mathematical objects x and y for which x = y' but x' ≠ y''

[adienes]

julia> [1,2,3] == [1,2,3]''
true

julia> [1 2 3] == [1 2 3]''
true

[adienes]

sorry, I misread your comment. I do actually see the core issue, which is that

julia> [1 2 3] == [1,2,3]'
true

basically since LinearAlgebra.Adjoint <: AbstractArray it uses the AbstractArray fallbacks for == (and hash of course) which check axes and elements. arguably, Adjoints should not equal non-adjoint arrays of the same shape, although I would bet that ends up being pretty breaking.

either way though, this would be an issue for LinearAlgebra not Base

[uniment]

arguably, Adjoints should not equal non-adjoint arrays of the same shape

no this behavior is desired. Adjoint is just a lazy view representing the conjugate transposed matrix, and it's desirable that a lazy view should equal a realized matrix whose dimensions and values are the same:

julia> [1 2 3 4; 5 6 7 8]' == [1 5; 2 6; 3 7; 4 8]
true

as @barucden pointed out, the underlying issue is that a Vector is not equal to any Matrix despite being transposable to an object that equals a Matrix:

julia> [1;2;3] == [1;2;3;;]
false

julia> [1;2;3]' == [1;2;3;;]' == [1 2 3]
true

since transposing is understood to create a matrix from another matrix, the fact that a Vector can be transposed to an object that equals a Matrix implies that it is considered to "be" a matrix, or at least to be promotable to one, but equality has a different opinion of the matter.

if a vector could equal a matrix, then for consistency across tensor ranks we would want these also to evaluate to true:

julia> [1;2;3] == [1;2;3;;]
false

julia> [1;2;3;;] == [1;2;3;;;]
false

julia> [1;2;3;;;] == [1;2;3;;;;]
false

the options seem to be:

1. allow different-rank tensors to be equal as suggested above, as long as their entries are in the same dimensions and their values are equal (would it break anything?)
2. make adjoint for Vector undefined (would definitely break things)
3. leave unchanged, understanding the special behavior of Vector to effectively promote to Matrix for transpose but not for equality, and live with the fact that there exist objects x,y in the language for which x == y and x' != y' (and x'' == y'', x''' != y''', x'''' == y'''', etc.)