4 ms·
The better line of thought here might be "A*B is not multiplication, it's function composition". It literally looks nothing like multiplication when carried ou
by prionassembly 6y ago
The better line of thought here might be "A*B is not multiplication, it's function composition".
It literally looks nothing like multiplication when carried out manually.
- eigenket 6y agoIt looks exactly like scalar multiplication when your matrices are diagonal.
- prionassembly 6y agoThen: all matrices look like the identity matrix if applied to a suitably rescaled eigenvector.
- eigenket 6y agoWe're talking about matrix/matrix multiplication here, not matrix/vector, right?
- prionassembly 6y agoI was talking about matrix/matrix composition (rather than multiplication) at first. Then I talked about matrix application. A matrix is a function. Not all functions can be represented by matrices (although all smooth functions can be represented by a Taylor series that sums over matrices (and "the sum C=A+B" really means "the function C such that Cx = Ax+Bx for all x in the range of both A and B"))
- eigenket 6y agoI rather dislike this way of saying things. A matrix is a matrix, its a table of numbers, and one can usefully define operations like addition and multiplication on them. A linear map is a linear map, it maps between vector spaces and it obeys some nice axioms. You can define addition and composition as operations on them. It is a quite interesting and non-trivial theorem that if you fix a particular choice of basis then you get "for free" a bijection between linear maps and matrices. The bijection between matrices and linear maps is completely dependent on the basis you choose, however, and there certainly isn't a canonical way to choose the basis. Often it is natural to change basis to make it easier solve some particular problem, and then the matrix that represents a particular linear map will change, but the properties of the linear map won't change (for example its rank, kernel, eigenvectors/values etc).
- nimish 6y agoNot the zero matrix, or any non-invertible matrix over a field with 0 characteristic. The Jordan-Chevalley decomposition makes the difference precise -- the nilpotent part (off diagonal 1's in the jordan normal form)
- mantap 6y agoThe mathematical term is "product". It's a product.