Does anyone know how to prove that the set of all $K\times K$ column finite matrices over a ring $R$ $[\mathrm{CFM}(R)]$ forms a ring? I am also confused about the definition of multiplication in $\mathrm{CFM}(R)$? Is it the standard matrix multiplication?
Infinite Matrices
449 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtThere are 2 best solutions below
On
Yes, multiplication is the standard matrix multiplication in the sense that if the matrices are indexed by $K \times K$, then $\sum_{j\in K} a_{ij}b_{jk}$ gives the $i,k$ entry of the product matrix.
You can either show this makes a multiplication directly, and that would not be substantially different from proving that sets of square matrices are rings under matrix multiplication. Or else you can work to show that it is just a matrix representation of $End(V_F)$ where $V$'s dimension is the cardinality of the set $K$ indexing the rows and columns of your matrices.
You would view the matrices as acting on the left of column vectors representing the coordinates for your vector space. The proof is not substantially different from the one you might use for the finite dimensional case.
Multiplication of $A=(A_{ij})$ and $B=(B_{ij})$ is defined by $$(AB)_{ij}=\sum_{k=1}^{\infty}{A_{ik}B_{kj}}$$ Since $B_{kj}=0$ for sufficiently large $k$, this is well defined. To see that it is column finite, note that with fixed $j$ the $A_{ik}$ for which $B_{kj}$ is not zero will vanish for sufficiently large $i$.
Associativity is a pain to prove directly, but straightforward. Addition is easily shown to turn the set into an abelian group, and what's left is to check if the distributive laws hold, which you should be able to do.