23
$\begingroup$

If $A,B$ are $2 \times 2$ matrices of real or complex numbers, then

$$AB = \left[ \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]\cdot \left[ \begin{array}{cc} b_{11} & b_{12} \\ b_{21} & b_{22} \end{array} \right] = \left[ \begin{array}{cc} a_{11}b_{11}+a_{12}b_{21} & a_{11}b_{12}+a_{12}b_{22} \\ a_{21}b_{11}+a_{22}b_{21} & a_{22}b_{12}+a_{22}b_{22} \end{array} \right] $$

What if the entries $a_{ij}, b_{ij}$ are themselves $2 \times 2$ matrices? Does matrix multiplication hold in some sort of "block" form ?

$$AB = \left[ \begin{array}{c|c} A_{11} & A_{12} \\\hline A_{21} & A_{22} \end{array} \right]\cdot \left[ \begin{array}{c|c} B_{11} & B_{12} \\\hline B_{21} & B_{22} \end{array} \right] = \left[ \begin{array}{c|c} A_{11}B_{11}+A_{12}B_{21} & A_{11}B_{12}+A_{12}B_{22} \\\hline A_{21}B_{11}+A_{22}B_{21} & A_{22}B_{12}+A_{22}B_{22} \end{array} \right] $$ This identity would be very useful in my research.

$\endgroup$
  • 3
    $\begingroup$ Yes it does if the "blocking" is "conforming". $\endgroup$ – Algebraic Pavel May 9 '14 at 14:12
  • 1
    $\begingroup$ Yes. (The blocking is "confirming" in the situation you have given.) Discussed in detail in §6.12 of cip.ifi.lmu.de/~grinberg/primes2015/sols.pdf (specifically Exercise 38 and Remark 6.73; search for "block-matrix notation" if these numbers change). $\endgroup$ – darij grinberg Dec 4 '15 at 10:16
10
$\begingroup$

It depends on how you partition it, not all partitions work. For example, if you partition these two matrices

$$\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}, \begin{bmatrix} a' & b' & c' \\ d' & e' & f' \\ g' & h' & i' \end{bmatrix} $$

in this way

$$ \left[\begin{array}{c|cc}a&b&c\\ d&e&f\\ \hline g&h&i \end{array}\right], \left[\begin{array}{c|cc}a'&b'&c'\\ d'&e'&f'\\ \hline g'&h'&i' \end{array}\right] $$

and then multiply them, it won't work. But this would

$$\left[\begin{array}{c|cc}a&b&c\\ \hline d&e&f\\ g&h&i \end{array}\right] ,\left[\begin{array}{c|cc}a'&b'&c'\\ \hline d'&e'&f'\\ g'&h'&i' \end{array}\right] $$

What's the difference? Well, in the first case, all submatrix products are not defined, like $\begin{bmatrix} a \\ d \\ \end{bmatrix}$ cannot be multiplied with $\begin{bmatrix} a' \\ d' \\ \end{bmatrix}$

So, what is the general rule? (Taken entirely from the Wiki page on Block matrix)

Given, an $(m \times p)$ matrix $\mathbf{A}$ with $q$ row partitions and $s$ column partitions $$\begin{bmatrix} \mathbf{A}_{11} & \mathbf{A}_{12} & \cdots &\mathbf{A}_{1s}\\ \mathbf{A}_{21} & \mathbf{A}_{22} & \cdots &\mathbf{A}_{2s}\\ \vdots & \vdots & \ddots &\vdots \\ \mathbf{A}_{q1} & \mathbf{A}_{q2} & \cdots &\mathbf{A}_{qs}\end{bmatrix}$$

and a $(p \times n)$ matrix $\mathbf{B}$ with $s$ row partitions and $r$ column parttions

$$\begin{bmatrix} \mathbf{B}_{11} & \mathbf{B}_{12} & \cdots &\mathbf{B}_{1r}\\ \mathbf{B}_{21} & \mathbf{B}_{22} & \cdots &\mathbf{B}_{2r}\\ \vdots & \vdots & \ddots &\vdots \\ \mathbf{B}_{s1} & \mathbf{B}_{s2} & \cdots &\mathbf{B}_{sr}\end{bmatrix}$$

that are compatible with the partitions of $\mathbf{A}$, the matrix product

$ \mathbf{C}=\mathbf{A}\mathbf{B} $

can be formed blockwise, yielding $\mathbf{C}$ as an $(m\times n)$ matrix with $q$ row partitions and $r$ column partitions.

$\endgroup$
  • $\begingroup$ Use \pmatrix{a&b&c\\d&e&f\\g&h&i} $\endgroup$ – Berci May 9 '14 at 14:37
  • $\begingroup$ @Berci I know that. But getting those horizontal and vertical lines is the difficult part. $\endgroup$ – The very fluffy Panda May 9 '14 at 14:38
  • $\begingroup$ @PandaBear You can use $\color{blue}{\text{colors}}$ :) $\endgroup$ – Algebraic Pavel May 9 '14 at 14:45
  • $\begingroup$ @PavelJiranek Your comment is funny looking. $\endgroup$ – The very fluffy Panda May 9 '14 at 14:46
  • 3
    $\begingroup$ I think this is wrong. You can't partition both of them same way. If you partition after x rows in first matrix , you've to partition after x columns (not rows ) in the second matrix. Otherwise while multiplying you'll have to multiply mn block with another mn block which is not possible. (you need np block) Try it with your example. $\endgroup$ – A Googler Oct 1 '15 at 18:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.