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I think the author's argument becomes more clear when you consider multiplication in rings other than the integers, for example (square) matrices. The product o
by y7 6y ago
I think the author's argument becomes more clear when you consider multiplication in rings other than the integers, for example (square) matrices. The product of two matrices A*B does not correspond to repeated addition, and it is not commutative (A*B does not equal B*A in general).
I can see that having an engrained belief that multiplication is defined via addition becomes problematic at some point when learning about mathematics. However, this is true for a lot of basic properties that hold over the integers and not in other domains, so I'm not really convinced that it's actually wrong to teach kids about multiplication this way.
- prionassembly 6y agoThe 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.
- jmull 6y agoIt's just that matrix multiplication is not the same thing as scaler multiplication. I think you need to understand that no matter how you define or think of scaler multiplication.
- anticensor 6y agoHadamard multiplication of matrices is a scalar multiplication. https://en.wikipedia.org/wiki/Hadamard_product_(matrices) https://en.wikipedia.org/wiki/Hadamard_product_(matrices)