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What is the isomorphism function in $M_m(M_n(\mathbb R))\cong M_{mn}(\mathbb R)$. I tried this $[[a_{ij}]_{kl}]\mapsto[a_{ijkl}]$ , but I couldn't prove all steps.

Ian Coley
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Bbbh
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  • Are you allowed to assume that "block multiplication" of matrices works, or are you certain you have to prove that? – rschwieb Apr 24 '13 at 17:31
  • What are you writing $a_{ijkl}$ to mean? The entries of an $mn\times mn$ matrix will only have two indices... – rschwieb Apr 24 '13 at 17:33
  • @rschwieb To clarify, the original text was $[aijkl]$, so I attempted that notation as to mirror the author's intent (perhaps). – Ian Coley Apr 24 '13 at 17:36
  • @FrankMcGovern Ah, thanks for letting me know. I didn't notice that. Still, the original text has the same problem I'm asking about :) – rschwieb Apr 24 '13 at 17:37

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Hint: Starting with an element $[a_{ij}]_{kl}$ in $M_m(M_n(\Bbb R))$, the "obvious" homomorphism is to just "erase all the squares around $n\times n$ submatrices" and let all the contents form an $mn\times mn$ matrix $b$.

For example, the elements of $[a_{i,j}]_{1,1}$ would wind up being $b_{i,j}$ in the "big" matrix $b$ with components $b_{\cdot,\cdot}$. Then $[a_{i,j}]_{1,2}$ would wind up being $b_{i,n+j}$.

Then also $[a_{i,j}]_{2,1}$ would be $b_{n+i,j}$.

And $[a_{i,j}]_{4,5}$ would be $b_{3n+i,4n+j}$.

Can you see the pattern?

Once you grasp the indexing, you can take a look at why the multiplication works. Here is the relevant stuff (with no explanation, because I want you to think about it.) Below, $c$ is another $mn\times mn$ matrix like $b$ is for a matrix with entries $[d_{i,j}]_{k,l}$.

$$([b][c])_{pn+i,qn+j}:=\sum_{k=1}^{mn}b_{pn+i,k}c_{k,qn+j}=\sum_{r=0}^{m-1}\sum_{s=1}^{n}b_{pn+i,rn+s}c_{rn+s,qn+j}$$

Can you see $([a]_{i,s}[d]_{s,j})_{p+1,q+1}$ in that expression? This is basically expressing "block multiplication."

Alternatively, if you are comfortable with the fact that matrix rings can be interpreted as linear transformations on vector spaces, you could restate the entire problem in terms of linear transformations of an $n$ dimensional vector space $V$, and the tranformations of $V^m$. This would ameliorate the need for matrix multiplication, but you would have to be comfortable with the translation.

rschwieb
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  • And what about multiplicaiton? This is the most interesting part – Norbert Apr 24 '13 at 17:44
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    @Norbert I wouldn't exactly classify writing the details about how the matrix multiplication falls out from this indexing as "interesting" but it is certainly "the challenging part." Still, you've made me realize that I could probably post a large piece of that verification that would still require attention to understand. So I'll do that! – rschwieb Apr 24 '13 at 17:54
  • I might be wrong, but I think it should say in the first line, $[a_{ij}]_{kl}$ in $A$ in $M_m(M_n(\Bbb R))$ – Sally G Jun 07 '19 at 15:35