Vampire Matrices

On Halloween in 2020, Matt Parker of the YouTube Channel Standup Maths posted a video about Vampire Matrices. In this early definition of the term, Parker defined a (single) Vampire Matrix to be a square matrix consisting of positive integers such that the square of the matrix is equal to its element-wise concatenation. As a $2\times 2$ example, \[ \mat{cc} 3 & 4 \\ 6 & 8 \rix^2 = \mat{cc} 33 & 44 \\ 66 & 88 \rix \] and a $3\times 3$ example, \[ \mat{ccc} 1 & 3 & 1 \\ 3 & 9 & 3 \\ 1 & 3 & 1 \rix^2 = \mat{ccc} 11 & 33 & 11 \\
33 & 99 & 33 \\
11 & 33 & 11 \rix. \] As Parker illustrates in his video, himself and other mathematicians on YouTube and Twitter noticed that these matrices satisfy the equation \begin{equation}\label{eq1} A^2 = 11 A .\end{equation} If we allow for two-digit integer entries (by this I mean strictly two-digit numbers to rule out things like $04$ which is really just $4$, a one-digit number), then the matrix equation of interest becomes \begin{equation}\label{eq2} A^2 = 101 A .\end{equation} One such matrix is \[ \mat{cc} 60 & 30 \\ 82 & 41 \rix^2 = \mat{cc} 6060 & 3030 \\ 8282 & 4141 \rix. \]

Can we have a pair of matrices for which their product is equal to their element-wise concatenation? Yes! For example, \[ \mat{cc} 4 & 8 \\ 2 & 3 \rix \mat{cc} 8 & 8 \\ 2 & 7 \rix = \mat{cc} 48 & 88 \\ 22 & 37 \rix. \] Here entries are one-digit integers and so the corresponding matrix equation is \begin{equation}\label{eq3} AB = 10A + B. \end{equation}

Now, we start to see the pattern. Equation \eqref{eq3} reduces to \eqref{eq1} if $A=B$. And in the last example if we had strictly two-digit integers in the matrices on the left-hand side, then \eqref{eq3} would be $AB = 100A + B$ and then becomes \eqref{eq2} in the special case when $A=B$.

To clarify some things and make this development more rigorous, let’s define some terms.

• Let $N_d = \{n \in \mathbb{N} \mid n \text{ has exactly } d \text{ digits}\}$.
• For example, $N_1 = \{ 1, \dots, 9\}$ and $N_2 = \{10, \dots, 99\}$.
• Let $N_d^{n \times n}$ be the set of $n\times n$ matrices with entries in $N_d$.
• A set of matrices $\{A_1, \dots, A_k\}$ in $N_d^{n \times n}$ is said to satisfy the multiplication concatenation property (MCP) if $\prod_{i=1}^k A_i \; = \; \sum_{i=1}^k 10^{d(k-i)} A_i$.

We have proven some elementary properties so far about matrices $A \in N_d^{n\times n}$ which satisfy the MCP by themselves. That is, the set of matrices is just $A_1=A_2=A$ so that $A^2 = (10^d+1)A$.

1. Eigenvalues of $A$ are $0$ and $10^d+1$, with multiplicities $n-1$ and $1$, respectively. This means the trace of $A$ is $10^d+1$.
2. Therefore, $A$ is singular. In fact, for these matrices to satisfy the MCP it is sufficient and necessary that $A$ is singular and has trace $10^d+1$. In the case of two or more matrices satisfying the MCP, this biconditional statement is no longer true.

This project is a work in progress, and I welcome anyone interested in “joining it” to reach out to me! Here are some ideas I have on what to pursue:

1. Is there a size limit on the matrices for which $A^2 = (10^d+1)A$? We have found examples of size $n=2, 3, 4$.
2. Can we think about solving the matrix equation in the MCP? That is, can we generate them in some manner?
3. Does there exist some matrix $B$ so that $\{ A_1, \dots, A_k\}$ satisfies the MCP but $\{ A_1, \dots, A_k,B\}$ does not? Here, we might allow for $B$ to depend on the $A$ matrices.