When will matrix multiplication become just “concatenation”?

Saw below entertaining matrix multiplication examples.

Obviously they are just coincidences. But I am curious when does below hold?

$$$$\begin{pmatrix} a_1 & b_1 \\ c_1 & d_1 \end{pmatrix} \times \begin{pmatrix} a_2 & b_2 \\ c_2 & d_2 \end{pmatrix} = \begin{pmatrix} \overline{a_1 a_2} & \overline{b_1 b_2} \\ \overline{c_1 c_2} & \overline{d_1 d_2} \end{pmatrix}$$$$

here $$\overline{a_1 a_2}$$ means gluing the integers together, not multiplications.. assuming we are working on integer matrices.

• One instance of this is a "vampire matrix" where the matrices being multiplied are the same. For example, $$\pmatrix{3 & 4\\6 & 8} \cdot \pmatrix{3 & 4\\6 & 8} = \pmatrix{33 & 44\\66 & 88}.$$ Matt Parker has a nice youtube video about these. – Ben Grossmann Jan 18 at 15:23
• Don't know if it will help, but "concatenate" is a technical term we use in programming to refer to "gluing" the integers together. I believe it is used in the mathematics of languages as well. – Cort Ammon Jan 19 at 0:12
• Agree that searchability would be improved if we substitute "concatenation" for "glues" in the title and post. – Daniel R. Collins Jan 19 at 7:07

Let $$A$$ denote the first matrix in the product, and let $$B$$ denote the second. I will consider only the case in which the entries of $$B$$ have one digit.

In this case, the "gluing" property of this matrix multiplication can be written as $$AB = 10A + B.$$ Note that this equation can be rearranged into $$AB - 10 A - B + 10I = 10I \implies\\ (A - I)(B - 10 I)= 10 I,$$ where $$I$$ denotes the identity matrix. Now, suppose that we select a matrix $$B$$ and we want a corresponding matrix $$A$$. We have $$(A - I)(B - 10 I) = 10 I \implies\\ A = 10(B - 10 I)^{-1} + I.$$ Note that this equation only has a solution if $$\det(B - 10 I) \neq 0$$. Now, a question remains: how do we ensure that $$10 (B - 10 I)^{-1}$$ is an integer matrix? As it turns out, this will hold for a given integer matrix $$B$$ if and only if the determinant of $$B - 10 I$$ divides $$10$$.

In fact, we can generate pairs of matrices with non-negative integer entries that have the gluing property via the following steps:

• Find a matrix $$C$$ whose diagonal entries satisfy $$-10 \leq c_{ii} \leq -1$$ and whose off-diagonal entries are a single digit positive number such that the determinant of $$C$$ is either $$-1$$, $$-2$$, $$-5$$, or $$-10$$.
• Take $$B = 10 I + C$$ and $$A = 10C^{-1} + I$$.

For example, the matrix $$C = \pmatrix{-4 & 3\\7 &-4}$$ has determinant $$16 - 21 = -5$$, which divides $$10$$. The corresponding matrix $$B$$ is $$B = 10 I + \pmatrix{-4 & 3\\7 &-4} = \pmatrix{6&3\\7&6}.$$ The associated matrix $$A$$ is $$10C^{-1} + I = \frac{10}{-5} \cdot \pmatrix{-4 & -3\\-7 & -4} + \pmatrix{1&0\\0&1} = \pmatrix{9 & 6\\14 & 9}.$$ If we compute the product, we indeed find that $$\pmatrix{\color{blue}{9} & \color{blue}{6}\\ \color{blue}{14} & \color{blue}{9}} \cdot \pmatrix{\color{green}{6} & \color{green}{3}\\ \color{green}{7} & \color{green}{6}} = \pmatrix{\color{blue}{9}\color{green}{6} & \color{blue}{6}\color{green}{3}\\ \color{blue}{14}\color{green}{7} & \color{blue}{9}\color{green}{6}}.$$

An interesting phenomenon: if $$B$$ is a single digit matrix for which there exists an $$A,B$$ pair with this "gluing" property, we will have $$A = B$$ if and only if $$B$$ is a "vampire matrix" (cf. my comment on the question), which holds if and only if $$B$$ has eigenvalues $$0,11$$, which holds if and only if $$C = B - 10 I$$ has eigenvalues $$-10,1$$, which holds if and only if $$C$$ has determinant $$-10$$ and trace $$-9$$.

• This is really a great fun (matrix-algebra has been many years a favourite for me). (+1) and I would like to give more... – Gottfried Helms Jan 18 at 21:25
• @GottfriedHelms Thank you, I am glad you enjoyed it! – Ben Grossmann Jan 18 at 22:45

Call the two matrices (in blue and in green respectively) $$A$$ and $$B$$. Assuming for simplicity that we want $$B$$ to consist of single digits only, we can write the property we're looking for as a matrix equation: $$AB = 10A + B$$.

Moving everything to one side, we can add a fourth term to make things easier to factor: $$AB - 10A - B + 10I = 10I$$. This can be written as $$(A - I)(B - 10I) = 10 I$$. So, in terms of $$B$$, we have $$A = 10(B - 10 I)^{-1} + I.$$ This gives us a formula for $$A$$ no matter which $$B$$ we pick, though out of $$10^4$$ possibilities for $$B$$, my computer only found $$429$$ for which $$A$$ consists of nonnegative integers.

So, let's calculate: first, how can we write $$\overline{a_1a_2}$$? In fact, $$a_1$$ and $$a_2$$ are simply, as you said, "glued" together. So, we simply have $$\overline{a_1a_2} = 10a_1+a_2$$.
$$\begin{pmatrix} a_1 & b_1 \\ c_1 & d_1 \end{pmatrix} \times \begin{pmatrix} a_2 & b_2 \\ c_2 & d_2 \end{pmatrix} = \begin{pmatrix} \overline{a_1 a_2} & \overline{b_1 b_2} \\ \overline{c_1 c_2} & \overline{d_1 d_2} \end{pmatrix} = \begin{pmatrix} 10a_1+a_2 & 10b_1+b_2 \\ 10c_1+c_2 & 10d_1+d_2 \end{pmatrix}$$
Now, let's name this "equation" $$(E)$$ with every number being an integer. Then, by calculating the product, we have $$(E) \iff \begin{pmatrix} a_1a_2 + b_1c_2 & a_1b_2 + b_1d_2\\ c_1a_2 + d_1c_2 & c_1b_2 + d_1d_2 \end{pmatrix} = \begin{pmatrix} 10a_1+a_2 & 10b_1+b_2 \\ 10c_1+c_2 & 10d_1+d_2 \end{pmatrix}$$
Finally, we are able to identify the coefficients and get a system of equations: $$\boxed{(E) \iff \begin{cases} a_1a_2+b_1c_2 = 10a_1+a_2\\ c_1a_2+d_1c_2 = 10c_1+c_2 \\ a_1b_2+b_1d_2 = 10b_1+b_2\\ c_1b_2 + d_1d_2 = 10d_1+d_2 \end{cases}}$$