kemfys 0 Posted March 28, 2020 Share Posted March 28, 2020 Suppose I have a matrix P = \begin{bmatrix}x⊗x&x⊗y\\y⊗x&y⊗y\end{bmatrix} which is equal to another matrix Q = \begin{bmatrix}a&b\\c&d\end{bmatrix}. Is it possible to solve for x and y. Since xy and yx are equal to b and c and are not the same, the elements in the matrix P are outer product of two matrices x and y. Is it possible to solve for x and y such that xy and yx can be made predictable. Kindly let me know if I am not clear. Thank you for the kind help. Link to post Share on other sites

studiot 2240 Posted March 28, 2020 Share Posted March 28, 2020 31 minutes ago, kemfys said: Can matrix element be multiplication of two matrices We normally do this the other way round. It is called expansion of determinants. There are plenty of explanations to be had on google eg https://people.richland.edu/james/lecture/m116/matrices/determinant.html Link to post Share on other sites

taeto 93 Posted March 29, 2020 Share Posted March 29, 2020 (edited) 3 hours ago, kemfys said: Kindly let me know if I am not clear. It is given that \(b\) and \(c\) are not equal? From which ring(?) do you take your \(x\) and \(y,\) and how is \(x \otimes y\) defined? Do \(x\) and \(y\) have to be \(2 \times 2\) matrices? Edited March 29, 2020 by taeto Link to post Share on other sites

kemfys 0 Posted March 30, 2020 Author Share Posted March 30, 2020 Hi taeto, Thank you for kind response. I am not a mathematician, so I am not sure how to respond to the question based on ring. I am a physical chemist and I have some problem with me for which I am trying to develop a model using matrices. And my model requires that multiplication between two parameters be non-commutative and therefore I have chosen matrices. And to preserve the uniqueness in the result of matrix multiplication between tow matrices, I have chosen the multiplication to be outer product. Therefore, it is I who defined x⊗y = b and y⊗x = c. And the matrices x and y are 2×2 matrices. Actually, the intention is to extend it to any dimension. Please let me know if I am not clear. Thank you Link to post Share on other sites

taeto 93 Posted March 30, 2020 Share Posted March 30, 2020 (edited) Hi kemfys, actually that is exactly what a ring means. It is possible to add elements, and the addition is commutative. It is possible to multiply elements, but the product is not necessarily commutative. Matrices over fields like \(\mathbb{Q},\) \(\mathbb{R},\) and \(\mathbb{C},\) are examples of rings. To say that a matrix is "over a field" means that all the entries of the matrix belong to this field. In the case of a \(2\times 2\) matrix this would be all four entries. Matrices over a commutative ring like \(\mathbb{Z}\) still form a ring themselves. I think that I get that you want to consider matrices over rings that are not necessarily fields, or even commutative rings like \(\mathbb{Z}.\) Maybe you should simply work through some examples to experience how it works. It seems that if you have \(2\times 2\) matrices \(x\) and \(y\) as variables and you require your matrix \(P\) to equal the matrix \(Q,\) then it leaves you with four equations (expressions that should be equal to \(a,b,c,d\)) in eight unknowns (the entries of the matrices \(x\) and \(y\)). In the general case the answer should be yes, you can find a solution. Though usually the complete solution consists in a 4-dimensional solution space. Edited March 30, 2020 by taeto Link to post Share on other sites

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