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\title{Block Multiplication of Matrices}
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This note describes multiplication of block (partitioned matrices).  A
special case gives a representation of a matrix as a sum of rank one
matrices. Suppose \( [n] = (1,2,\dots, n)\) is the (ordered) sequence of
integers from $1$ to $n$. An \emph{ordered partition} (my term) is a
set of ordered subsets \( J = (J_1, J_2,\dots, J_p)\)  which come from
putting marks at $p-1$ places in the ordered list  \( (1,2,\dots, n)\)
at arbitrary
locations, for example 
\[
J_1 = (1,2,3), J_2=(4), J_3=(5,6)
\]
is an ordered partition of \([6]\). The \emph{size} of $J$ is $n$ and
the \emph{size} of $J_k$ is the cardinality of f $J_k$.  These are
denoted by $|J|, |J_k|$. 

\begin{definition}  Let $A$ be a matrix. Let  \(A(J_r;K_s)\),
denote the submatrix of $A$ with entries from
row $J_r$ and columns $K_s$. This defines a paritioning of $A$ and we
call $A$ a partitioned matrix.
\end{definition}

\begin{theorem}
Let \(J, K, L\) be ordered partitions of size $p,q,r$
respectively. Let $A,B$ and be partioned matrices with the partitions of
$A$ defined by $J,K$ and the partitions of $B$ defined by $K,L$.  Let
$C=AB$. Let the number of partitions in $K$ be $m$. Then
\[
C(J_r;L_t) = \sum _{s=1}^{s=m}A(J_r;K_s)B(K_s;L_t).
\]
\end{theorem}

\begin{proof}
The proof will be illustrated in a particular case.  The argument is
general and can be used to prove the theorem.  Let take as an example
a partition that includes the index $2\in J_1$ and the
index $4 \in L_2$.  We want to identify the term $\ds
c_{24}$ in the product matrix $C$.  This entry is found by summing
terms found by multiply entries in row $2$ of $A$ times corresponding
entries in column $4$ of $B$.  This sum can be computed by adding the
products of the entries in row $2$ in blocks $A(J_1;K_s)$ times the
entries in column $4$ in the blocks $B(K_s;L_2)$.  With an obvious
simplified notation the theorem can be written
\[
C_{rt} = \sum_s A_{rs}B_{st},
\]
and the formula reads formally as if the entries $\ds A_{rs}, B_{st}$
are scalars.

\end{proof}

\begin{corollary}
Let $A=[A_1,A_2, \dots, A_n], B=[B_1,B_2,\dots, B_n]^T$ be matrices
with $n$ column vectors  and $n$ row vectors respectively. Then 
\[
AB= A_1B_1^T+A_2B_2^T + \cdots + A_nB_n^T.
\]
\end{corollary}


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