If we had two square matrices and we have taken the say (A \odot B) that is hadarmad product of A and B there are some 0 in the product now how to find (A \odot B)^(-1) is theit any and technique we can apply to find the inverse
https://math.stackexchange.com/questions/1321879/is-there-a-formula-for-the-inverse-of-hadamard-product
The inverse of AoB is A^{-1}oB^{-1}. Please take a look in the attachment.
Dear Anton Vrdoljak
The link doesn't give the right answer. It shows some discussions.
Regards
Dear Wiwat Wanicharpichat
The attachment includes mistakes!
The matrix J, according to your definition, is not invertible.
It has determinant zero.
The included formula is not correct.
What if one of the two matrices is not invertible, but the Hadamard product is invertible!!
Counterexample:
A= [ 1 , 1
2 , 2]
B= [ 1, 0
0 , 1]
A is not invertible.
B is the identity which is invertible.
A⊗B = [ 1 0
0 2 ]
A⊗B is invertible.
Best regards
Dear Sriram Vaidyanathan
Your request has no general answer. But you may consider special cases.
For example:
A = [3 , 4
5 , 6 ]
B = [ 5 6
3 4]
Both A and B are invertible but the Hadamard product is not invertible.
A⊗B = [ 15 24
15 24 ]
Special cases:
(1) If A and B are invertible diagonal matrices, then A⊗B and B⊗A are invertible, and (A⊗B)-1 = A-1 ⊗ B-1 = B-1 ⊗ A-1 = (B⊗A)-1
(2) If A is diagonal matrix without zero entries.
B is any matrix such that its diagonal doesn't include zeros,
then (A⊗B)-1 = A-1 ⊗ (diagB) -1 , where diag(B ) is the diagonal of B.
And more.
Best Regards
Dear Issam Kaddoura
This problem is not Cayley product, the binary operation is Hadamard product. Therefore the matrix J is invertible under this product.
In your counterexample:
A= [ 1 , 1
2 , 2]
is invertible under Hadamard product. The inverse under Hadamard product is
A^{-1} = [ 1/1 , 1/1
1/2 , 1/2],
But it is noninvertible under Cayley product.
Now
B= [ 1, 0
0 , 1]
has some entry is zero B is noninvertible under Hadamard product because the matrix
C = [ 1/1, 1/0
1/0, 1/1]
is undefine, since 1/0 is divided by zeor.
But det B = 1
under Cayley product B is invertible and B{-1} is itself.
Best regards
Dear Sriram Vaidyanathan
The Hadamard product is sometime called the entrywise product. If there are some 0 in the product we cannot find (A \odot B)^(-1) under Hadamard product.
Best Regards
Dear Wiwat Wanicharpichat
I think you are confused about the definition of the Hadamard product.
Could you please see
https://en.wikipedia.org/wiki/Hadamard_product_(matrices)
Next, we continue our discussion.
Best wishes
Dear Issam Kaddoura
The Hadamard product is sometime called the entrywise product
Best Regards
Dear Wiwat Wanicharpichat
It seems you are talking about the inverse of the matrix related to the operation ⊗, and I am talking about the usual inverse.
I assume that this is the request of Sriram Vaidyanathan.
Otherwise, the problem is trivial.
In other words, the question is to find (A⊗B)-1 (if exists), where the inverse is the usual inverse.
Best regards
.
The question is to find (A⊗B)-1 (if exists), where the inverse is the usual INVERSE.
KIND HELP
operation ⊗ inside is the hadarmad product kind help if any general result
Here A and B are square matrices of same dimension
So, I got you right.
I hope that you find my notes are useful. Read them carefully.
Regards
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