A description/proof of that multiplication of matrix made of same size blocks by matrix made of multiplicable same size blocks is blocks-wise
Topics
About: matrix
The table of contents of this article
Starting Context
- The reader knows a definition of multiplication of matrices.
Target Context
- The reader will have a description and a proof of the proposition that the multiplication of any matrix made of any same size blocks by any matrix made of blocks of any multiplicable (with blocks of the former matrix) same size is blocks-wise.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any matrix, \(M_1\), made of the same \((r, s)\) size blocks and any matrix, \(M_2\), made of the same \((s, t)\) size blocks, the multiplication, \(M_1 M_2\), is blocks-wise, which means that as \(M_1\) is made of blocks, \(\{M_{1, i, j}\}\), and \(M_2\) is made of blocks, \(\{M_{2, j, k}\}\), \(M_1 M_2\) is made of blocks, \(\{\sum_j M_{1, i, j} M_{2, j, k}\}\).
2: Proof
Let us divide \(M_1 M_2\) in the \((r, t)\) size blocks and denote each block as \(M_{3, i, k}\). \(M_{3, i, k, l, m} = \sum_{n, o} M_{1, i, n, l, o} M_{2, n, k, o, m} = \sum_n (M_{1, i, n} M_{2, n, k})_{l, m} = (\sum_n (M_{1, i, n} M_{2, n, k}))_{l, m}\), which means that \(M_{3, i, k} = \sum_n (M_{1, i, n} M_{2, n, k})\).
