# Product of two matrices in block form: step in proof

Let $$R$$ be a commutative ring with $$1$$ and $$M_{k,l}(R)$$ denote set of matrices of size $$k\times l$$ over $$R$$.

$$A\in M_{m,n}(R)$$ and $$B\in M_{n,p}(R)$$. Partition $$m$$, $$n$$ and $$p$$ as $$m=m_1+\cdots + m_s, \,\,\,\, n=n_1+\cdots + n_t, \,\,\,\, p=p_1+\cdots + p_u.$$ Let $$A=[A_{ij}]$$ and $$B=[B_{ij}]$$ be partition of $$A$$ and $$B$$ so that $$A_{ij}\in M_{m_i,n_j}(R)$$ and $$B_{ij}\in M_{n_i,p_j}(R)$$.

Claim: $$C=AB$$ has partition $$[C_{ij}]$$ with $$C_{ij}=A_{i1}B_{1j} + \cdots + A_{in}B_{nj}.$$

Proof: (1) Consider $$(\alpha,\beta)$$-th entry of $$C$$:

$$ent_{\alpha\beta}(C)=a_{\alpha 1}b_{a\beta} + \cdots + a_{\alpha n} b_{n\beta}.$$

(2) $$C$$ is $$m\times p$$ matrix, and it has a partition given by $$m=m_1+\cdots + m_s$$ and $$p=p_1+\cdots + p_t$$.

(3) In this partition of $$C$$, the $$ent_{\alpha\beta}(C)$$ will lie in block (submatrrix) $$C_{ij}$$, of size $$m_i\times p_j$$ so $$ent_{\alpha\beta}(C)=ent_{\omega\tau}(C_{ij}).$$ (4) Thus we have a partition $$row_{\alpha}(A)$$ and $$col_{\beta}(B)$$ as $$row_{\alpha}(A)=[row_{\omega}(A_{i1}) \,\, \cdots \,\, row_{\omega}(A_{it})$$ and $$col_{\beta}(B)=\begin{bmatrix} col_{\tau}(B_{1j})\\ \vdots \\ col_{\tau}(B_{tj})\end{bmatrix}.$$

Q. I do not understand (4). Can one explain why the $$\alpha$$-th row of $$A$$ and $$\beta$$-th column of $$B$$ get partitioned in this way?

Ref: Algebra by Adkins and Weintraub (p. 191)

The matrix $$A$$ looks like this in block form:

$$\begin{array}{cc} 1&\\2&\\ \vdots&\\ m_1& \\ m_1 + 1& \\ \vdots& \\[1ex] 1 + \sum_{h=1}^{i-1}m_h&1 \\ \vdots&\vdots \\ \alpha&\omega \\ \vdots&\vdots \\ \sum_{h=1}^i m_h&m_i\\[1ex] \vdots& \\ m - m_s \\ m - m_s + 1\\ \vdots& \\m-1&\\ m \end{array} \begin{array}{|ccc|c|ccc|c|ccc|}\hline a_{1,1}&\cdots&a_{1,n_1}&\cdots&&\cdots&&\cdots&a_{1,n-n_t+1}&\cdots&a_{1,n}\\ a_{2,1}&&&&&&&&\\ \vdots&&&&&&&&&\\ a_{m_1,1}&&&&&&&&&\\ \hline a_{m_1+1,1}&&&&&&&&&\\ \vdots&&&&&&&&&\\[0.5ex] \hline (A_{1,j})_{1,1}&&&&&&&&&\\ \vdots&&&&&&&&&\\[0.5ex] (A_{i,1})_{\omega,1}&\cdots&(A_{i,1})_{\omega,n_1}&\cdots& (A_{i,j})_{\omega,1}&\cdots&(A_{i,j})_{\omega,n_i}&\cdots& (A_{i,t})_{\omega,1}&\cdots&(A_{i,t})_{\omega,n_t}\\ \vdots&&&&&&&&\\[0.5ex] (A_{1,j})_{m_i,1}&&&&&&&\\ \hline \vdots&&&&&&&&\\ a_{m-m_s,1}&&&&&&&&\\ \hline a_{m-m_s+1,1}&&&&&&&&\\ \vdots&&&&&&&&\\ a_{m-1,1}&&&&&&&&\\ a_{m,1}&\cdots&a_{m,n_1}&\cdots&&\cdots&&\cdots&a_{m,n-n_t+1}&\cdots&a_{m,n}\\ \hline \end{array}$$

The numbers down the left side are row index numbers, rows of the entire matrix on the far left and rows of the block $$A_{i,1}$$ alongside the block $$A_{i,1}.$$

The row of elements $$a_{\alpha,1},$$ $$a_{\alpha,2},\ldots, a_{\alpha,n}$$ (row $$\alpha$$ of the matrix $$A$$) is found in one of the rows of block matrices. Suppose this is the $$i$$th row of block matrices, consisting of the blocks $$A_{i,1},\ldots,A_{i,t},$$ which are shown from left to right across the middle of the block format above.

Row $$\alpha$$ of the entire matrix $$A$$ contains entries from some row of $$A_{i,1},$$ which we decide to call row $$\omega.$$ The same row of the large matrix also contains row $$\omega$$ of the blocks $$A_{i,2},\ldots,A_{i,t}.$$

If we take row $$\omega$$ of each of the blocks $$A_{i,1},A_{i,2},\ldots,A_{i,t}$$ and line them up in a single long row from left to right, we get back row $$\alpha$$ of $$A,$$ as illustrated by the row of elements $$(A_{i,1})_{\omega,1}, \ldots, (A_{i,1})_{\omega,n_1}, \ldots, (A_{i,j})_{\omega,1}, \ldots, (A_{i,j})_{\omega,n_i}, \ldots, (A_{i,t})_{\omega,1}, \ldots, (A_{i,t})_{\omega,n_t}$$ across the middle of the block-formatted matrix above. That is,

$$row_\alpha(A)=[row_\omega(A_{i,1}) \ \cdots \ row_\omega(A_{i,t})].$$

Similarly, column $$\beta$$ of matrix $$B$$ span columns of block matrices within it, taking elements from the same column of each of those blocks, which you can then stack up in one big column to recover column $$\beta$$ of matrix $$B.$$