The matrix multiplication of square matrices of different order is often claimed to be impossible. Yet, if the order of one matrix is divisible by the order of the other, a natural multiplication rule is visible. The bigger matrix simply should be considered a "matrix of matrices" or, alternatively, in small matrix all elements should be replaced with equivalent diagonal $m\times m$ (in this case, $2\times2$) square matrices:

$\left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right).\left( \begin{array}{cc} \left( \begin{array}{cc} {a_1} & {b_1} \\ {c_1} & {d_1} \\ \end{array} \right) & \left( \begin{array}{cc} {a_2} & {b_2} \\ {c_2} & {d_2} \\ \end{array} \right) \\ \left( \begin{array}{cc} {a_3} & {b_3} \\ {c_3} & {d_3} \\ \end{array} \right) & \left( \begin{array}{cc} {a_4} & {b_4} \\ {c_4} & {d_4} \\ \end{array} \right) \\ \end{array} \right)=\left( \begin{array}{cccc} a & 0 & b & 0 \\ 0 & a & 0 & b \\ c & 0 & d & 0 \\ 0 & c & 0 & d \\ \end{array} \right).\left( \begin{array}{cccc} {a_1} & {b_1} & {a_2} & {b_2} \\ {c_1} & {d_1} & {c_2} & {d_2} \\ {a_3} & {b_3} & {a_4} & {b_4} \\ {c_3} & {d_3} & {c_4} & {d_4} \\ \end{array} \right)$ $=\left( \begin{array}{cc} \left( \begin{array}{cc} a {a_1}+{a_3} b & a {b_1}+b {b_3} \\ a {c_1}+b {c_3} & a {d_1}+b {d_3} \\ \end{array} \right) & \left( \begin{array}{cc} a {a_2}+{a_4} b & a {b_2}+b {b_4} \\ a {c_2}+b {c_4} & a {d_2}+b {d_4} \\ \end{array} \right) \\ \left( \begin{array}{cc} {a_1} c+{a_3} d & {b_1} c+{b_3} d \\ c {c_1}+{c_3} d & c {d_1}+d {d_3} \\ \end{array} \right) & \left( \begin{array}{cc} {a_2} c+{a_4} d & {b_2} c+{b_4} d \\ c {c_2}+{c_4} d & c {d_2}+d {d_4} \\ \end{array} \right) \\ \end{array} \right)$ $=\left( \begin{array}{cccc} a \text{a1}+\text{a3} b & a \text{b1}+b \text{b3} & a \text{a2}+\text{a4} b & a \text{b2}+b \text{b4} \\ a \text{c1}+b \text{c3} & a \text{d1}+b \text{d3} & a \text{c2}+b \text{c4} & a \text{d2}+b \text{d4} \\ \text{a1} c+\text{a3} d & \text{b1} c+\text{b3} d & \text{a2} c+\text{a4} d & \text{b2} c+\text{b4} d \\ c \text{c1}+\text{c3} d & c \text{d1}+d \text{d3} & c \text{c2}+\text{c4} d & c \text{d2}+d \text{d4} \\ \end{array} \right)$

So, in certain circumstances this is possible. Thus, my question is: can this somehow be generalized to the matrices of orders that are not divisors of each other? Possibly, the resulting matrix would be of the order which is the least common multiple of the orders of the terms?

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    $\begingroup$ We can define all sorts of things... To my mind, the question is what properties we expect/require a thing to have... not just a formulaic definition. Can you clarify? $\endgroup$ Commented Jun 29, 2022 at 22:19
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    $\begingroup$ In case it helps, what you describe above is basically tensor multiplication. $\endgroup$ Commented Jun 29, 2022 at 22:21
  • $\begingroup$ @paulgarrett my primary aim was to see how this definition of matrix multiplication (of different orders) can be generalized to empty matrix (which is unclear whether can be done at all). $\endgroup$
    – Anixx
    Commented Jun 29, 2022 at 22:21
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    $\begingroup$ Multiplication on a set $X$ is a map $X\times X\to X$. What you have defined is different, it is a map of the form $X\times Y\to Y.$ What you have defined is some sort of action of 2 by 2 matrices on 4 by 4 matrices. Although I am not sure whether it has any nice properties, and your choice of the order in some of the product puzzles me. For example, I guess you don't have $(A\cdot B)\cdot C=A\cdot(B\cdot C).$ $\endgroup$ Commented Jun 29, 2022 at 22:32
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    $\begingroup$ @Anixx It is very, very far from being the same. $\endgroup$ Commented Jun 29, 2022 at 22:35

1 Answer 1


I went ahead and made a Mathematica code for this kind of multiplication. The code below can multiply square matrices of ANY order (except $0$).

DotG[A_, B_] := 
   IdentityMatrix[LCM[Length[A], Length[B]]/Length[A]]] . 
   IdentityMatrix[LCM[Length[A], Length[B]]/Length[B]]]

This multiplication is associative and generalizes the dot product of matrices of the same order.


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