Proving that $M_p(M_q (K)) \cong M_{pq} (K)$.

Question

My textbook finishes the proof of one of the theorems with the following fact: $$ M_p(M_q (K)) \cong M_{pq} (K), $$ where $K$ is a field, and it says that it is true by "block multiplication isomorphism". However, I was not able to find the information about this. Could you please help me to prove this isomorphism?

Answer 1

Quite sketchy: identify each matrix $M\in M_q(K)$, $M=(M_{ij})_{i,j=1,\dots,q}$ with a vector

$$M\leftrightarrow v(M):=(M_{11},\dots,M_{1q},M_{21},\dots,M_{2q},\dots\dots,M_{qq})$$

in $K^{q^2}$. By definition, $M_{ij}\in K$ for all $i,j$.

Then repeat the same trick with $M_p(M_q(K))$: each $N\in M_p(M_q(K))$ is a matrix $N=(N_{ij})_{i,j=1,\dots,p}$ now with

$$N_{ij}\in M_q(K), $$

i.e.

$$N_{ij}\leftrightarrow v(N_{ij})=(N_{ij, 11},\dots,N_{ij, 1q},N_{ij,21},\dots,N_{ij,2q},\dots\dots,N_{ij,qq}),$$

for all $i,j=1,\dots, p$.

In summary, each $N\in M_p(M_q(K))$ is identified with the vector

$$N\leftrightarrow v(N):=(v(N_{11}),v(N_{12}),\dots,v(N_{pp}))=\\(\underbrace{N_{11, 11},\dots,N_{11, 1q},N_{11,21},\dots,N_{11,2q},\dots\dots,N_{11,qq}}_{=v(N_{11})}, \underbrace{N_{12, 11},\dots,N_{12, 1q},N_{12,21},\dots,N_{12,2q},\dots\dots,N_{12,qq}}_{=v(N_{12})},\dots\dots, \underbrace{N_{qq, 11},\dots,N_{qq, 1q},N_{qq,21},\dots,N_{qq,2q},\dots\dots,N_{qq,qq}}_{=v(N_{pp})}), $$

which is a vector in $K^{pq}$. All identifications are isomorphisms, and we are done.