I just thought of a simple way to explain tensors. Imagine a linear function that accepts two numbers and returns a number, let’s call it f(x,y). Except there are two ways to imagine it:
Linear in both arguments combined: f(1,2)+f(1,3)=f(2,5). Every such function has the form f(x,y)=ax+by for some a and b, so the space of such functions is 2-dimensional. We say that the Cartesian product of R^1 and R^1 is R^2, because 1+1=2.
Linear in each argument when the other is fixed: f(1,2)+f(1,3)=f(1,5). Every such function has the form f(x,y)=axy for some a, so the space of such functions is 1-dimensional. We say that the tensor product of R^1 and R^1 is R^1, because 1*1=1.
In this case the tensor product is lower dimensional than the Cartesian product. But if we take say R^3 and R^3, then the Cartesian product will be R^6 and the tensor product will be R^9, because it will have separate coefficients a_(ij)*x_i*y_j.
I just thought of a simple way to explain tensors. Imagine a linear function that accepts two numbers and returns a number, let’s call it f(x,y). Except there are two ways to imagine it:
Linear in both arguments combined: f(1,2)+f(1,3)=f(2,5). Every such function has the form f(x,y)=ax+by for some a and b, so the space of such functions is 2-dimensional. We say that the Cartesian product of R^1 and R^1 is R^2, because 1+1=2.
Linear in each argument when the other is fixed: f(1,2)+f(1,3)=f(1,5). Every such function has the form f(x,y)=axy for some a, so the space of such functions is 1-dimensional. We say that the tensor product of R^1 and R^1 is R^1, because 1*1=1.
In this case the tensor product is lower dimensional than the Cartesian product. But if we take say R^3 and R^3, then the Cartesian product will be R^6 and the tensor product will be R^9, because it will have separate coefficients a_(ij)*x_i*y_j.