341: Multiplication of Matrix Made of Same Size Blocks by Matrix Made of Multiplicable Same Size Blocks Is Blocks-Wise

<The previous article in this series | The table of contents of this series | The next article in this series>

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
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

---

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 $M_{3,i,k}=\sum _n(M_{1,i,n}{M}_{2,n,k})$.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>