Hello, Does someone know what exactly happens in background when we do [1, 2, 3, 4] / [1, 2, 3, 4]. For [1, 2, 3, 4] ./ [1, 2, 3, 4] is clear that it’s doing element-wise division as shown below.
But what about [1, 2, 3, 4] / [1, 2, 3, 4] ?
It’s doing a matrix division:
help?> /
[...]
──────────────────────────────────────────────────────
A / B
Matrix right-division: A / B is equivalent to (B' \ A')' where \ is the
left-division operator. For square matrices, the result X is such
that A == X*B.
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What I know about matrix division is that its basically multiplication with inverse of other Matrix so like X = A * inv(B) is same as X = A/B and I guess it is specific for square matrices.
Now I played around with \ (left division operator) and it seems its works in same way just changes dividend and divisor. So, A/B == B\A.
If thats the case, I will expect
julia> [1 2 3 4]\ [1 2 3 4]
4×4 Matrix{Float64}:
0.0333333 0.0666667 0.1 0.133333
0.0666667 0.133333 0.2 0.266667
0.1 0.2 0.3 0.4
0.133333 0.266667 0.4 0.533333
and
julia> [1 2 3 4]/ [1 2 3 4]
1×1 Matrix{Float64}:
0.9999999999999999
to be same which is true in case of sqaure matrix. So, what I’m wondering is how exactly that division happens when I do [1 2 3 4] / [1 2 3 4] ?
Essentially if A and B are vectors of the same length there are two ways of multiplying them, either as
- (1 by n matrix) * (n by 1 matrix) = (1 x 1 matrix), or
- (n by 1 matrix) * (1 by n matrix) = (n by n matrix).
The two possibilities are what you’re seeing here.
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In my case [1 2 3 4] \ [1 2 3 4] both are 1*4 Matrix. And In case of vector I think it always taken as n*1 matrix if we try to multiply it to a matrix example below.
I understood what you were trying to say about matrix multiplication but I still dont see it answering my question about how that division take place. 
.
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For rectangular matrices this division corresponds to multiplication by the Moore-Penrose pseudoinverse. None of these results are going to make much sense unless you know that. You can see the pseudoinverse of a matrix (or vector) with the function LinearAlgebra.pinv.
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The exact way can be found with @edit [1, 2, 3, 4] / [1, 2, 3, 4], which shows me this on my machine:
function (/)(A::AbstractVecOrMat, B::AbstractVecOrMat)
size(A,2) != size(B,2) && throw(DimensionMismatch("Both inputs should have the same number of columns"))
return copy(adjoint(adjoint(B) \ adjoint(A)))
end
As the docstring says, it takes the adjoint and then does left division, taking the adjoint of the result again.
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Just to flesh this out a touch: the pseudoinverse of a vector is the transpose divided by it’s squared magnitude (see here). ie if v is a column vector then pinv(v) is a row vector, and vice-versa.
The left (\) and right (/) division is just whether you invert the first or second argument, so you get the two situations I first described: multiplication of (1 x n) with (n x 1) or the other way around.
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Thanks I reproduced result with concept of pseudo-inverse.
I will put results here. So, it can be helpful for others 
Most important think is this :
where x* is conjugate transpose.
julia> [1, 2, 3, 4] / [1, 2, 3, 4]
4×4 Matrix{Float64}:
0.0333333 0.0666667 0.1 0.133333
0.0666667 0.133333 0.2 0.266667
0.1 0.2 0.3 0.4
0.133333 0.266667 0.4 0.533333
Now issue was how that division occurs. In normal scenarios we find A/B by A * B-1 .
So, similarly we can do here by finding pseudo-inverse for [1, 2, 3, 4]
julia> INV = [1 2 3 4] ./ ([1 2 3 4] * [1 ,2 ,3, 4])
1×4 Matrix{Float64}:
0.0333333 0.0666667 0.1 0.133333
julia> [1, 2, 3, 4] * INV
4×4 Matrix{Float64}:
0.0333333 0.0666667 0.1 0.133333
0.0666667 0.133333 0.2 0.266667
0.1 0.2 0.3 0.4
0.133333 0.266667 0.4 0.533333
Let me know if you guys don’t see any issue.
Thanks everyone for your inputs 
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Yes. You picked that up quickly. More generally for a wide matrix A with full rank the pseudoinverse is A^\dagger = A^T (AA^T)^{-1} and A^\dagger b gives the unique solution of minimum norm to an underdetermined system Ax=b. For a tall matrix A with full rank it is A^\dagger = (A^T A)^{-1} A^T and A^\dagger b gives a unique least squares solution to the overdetermined system Ax=b. Even more generally, for any size matrix that is not necessarily of full rank, multiplying by the pseudoinverse gives the least squares solution of minimum norm, which is unique.
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