We know that the tensor product of $P_n~~and~~P_m$ may be identified with the space $P_{n,m}$ of polynomials in 2 variables. Prove that if A and B are differentiation on $P_n~~and~~P_m$ respectively, and if C = A$\bigotimes$ B, then C is mixed partial diffentiation, that is , if z is in $P_{n,m}, then \(Cz=\dfrac{\partial^2z}{\partial s\partial t}$
Please, help

Question

We know that the tensor product of $P_n~~and~~P_m$ may be identified with the space $P_{n,m}$ of polynomials in 2 variables. Prove that if A and B  are differentiation on $P_n~~and~~P_m$ respectively, and if C = A$\bigotimes$ B, then C is mixed partial diffentiation, that is , if z is in $P_{n,m}, then \(Cz=\dfrac{\partial^2z}{\partial s\partial t}$
Please, help

loser66 · Answer

@primeralph

loser66 · Answer

so far, I got :
let z in $P_{n,m}$ there exists x  $\in P_n$ and y  $\in P_m$ such that z = x(s) $\bigotimes$y(t)

loser66 · Answer

C(z) = A $\bigotimes$ B (x(s)$\bigotimes$ y(t)) = A x(s) $\bigotimes$ B y(t)

loser66 · Answer

and stuck.:)

primeralph · Answer

@Loser66  I can't really help you with this. I don't do much of this.

loser66 · Answer

It's ok, friend. Thanks for reply. :)