Find matrix $A\in \mathcal{M}_n (\mathbb{N})$ such that $A^k =\left( \sum_{i=1}^{k}10^{i-1} \right)A$.

There is a partial answer, which corresponds to the value $k=2.$

Firstly, since $$\det A^k= {\det}^k A = \underbrace{111\dots11}_k\det A,$$ then the easiest case of the solution is $$\det A=0,\tag1$$ as in the given example.

Let us consider the possible dimensions $n$ of the matrix $A.$

$\color{brown}{\textbf{Case n=1.}}$

The case is trivial, it does not correspond with the task statement.

Also, the equation $a^k = \underbrace{111\cdot11}_k$ has not solutions.

This fact excludes solutions in the form $A=aE,$ where $\;E\;$ is an arbitrary unit matrix (or transformed unit matrix).

$\color{brown}{\textbf{Case n=2.}}$

The equation $$\begin{pmatrix} a & b \\ c & d\end{pmatrix}^2 = 11\begin{pmatrix} a & b \\ c & d\end{pmatrix},$$ or $$\begin{pmatrix} a^2+bc & b(a+d) \\ c(a+d) & bc+d^2\end{pmatrix}^2 = \begin{pmatrix} 11a & 11b \\ 11c & 11d\end{pmatrix},$$ \begin{cases} a+d=11\\ bc = ad, \end{cases}

leads to the solutions in the matrix forms of $$\begin{cases} \begin{pmatrix} 2 & 9 \\ 2 & 9\end{pmatrix}, \begin{pmatrix} 2 & 2 \\ 9 & 9\end{pmatrix}, \begin{pmatrix} 2 & 3 \\ 6 & 9\end{pmatrix}, \begin{pmatrix} 2 & 6 \\ 3 & 9\end{pmatrix}; \\[4pt] \begin{pmatrix} 9 & 9 \\ 2 & 2\end{pmatrix}, \begin{pmatrix} 9 & 2 \\ 9 & 2\end{pmatrix}, \begin{pmatrix} 9 & 3 \\ 6 & 2\end{pmatrix}, \begin{pmatrix} 9 & 6 \\ 3 & 2\end{pmatrix}; \\[4pt] \begin{pmatrix} 3 & 8 \\ 3 & 8\end{pmatrix}, \begin{pmatrix} 3 & 3 \\ 8 & 8\end{pmatrix}, \begin{pmatrix} 3 & 4 \\ 6 & 8\end{pmatrix}, \begin{pmatrix} 3 & 6 \\ 4 & 8\end{pmatrix}; \\[4pt] \begin{pmatrix} 8 & 8 \\ 3 & 3\end{pmatrix}, \begin{pmatrix} 8 & 3 \\ 8 & 3\end{pmatrix}, \begin{pmatrix} 8 & 4 \\ 6 & 3\end{pmatrix}, \begin{pmatrix} 8 & 6 \\ 4 & 3\end{pmatrix}; \\[4pt] \begin{pmatrix} 4 & 7 \\ 4 & 7\end{pmatrix}, \begin{pmatrix} 4 & 4 \\ 7 & 7\end{pmatrix}, \begin{pmatrix} 7 & 7 \\ 4 & 4\end{pmatrix}, \begin{pmatrix} 7 & 4 \\ 7 & 4\end{pmatrix}; \\[4pt] \begin{pmatrix} 5 & 6 \\ 5 & 6\end{pmatrix}, \begin{pmatrix} 5 & 5 \\ 6 & 6\end{pmatrix}, \begin{pmatrix} 6 & 6 \\ 5 & 5\end{pmatrix}, \begin{pmatrix} 6 & 5 \\ 6 & 5\end{pmatrix}. \end{cases}\tag2$$

For example, $$\begin{pmatrix} 2 & 6 \\ 3 & 9\end{pmatrix}^2 = \begin{pmatrix} 22 & 66 \\ 33 & 99\end{pmatrix},$$

All the solutions satisfies to the next conditions:

• the sum of rows(columns) divides to 11;
• the rows(columns) are collinear.

$\color{brown}{\textbf{Case n=3.}}$

Let us search non-trivial solutions in the form of $$A = \begin{pmatrix} k & a & b \\ ky & ay & by \\ kz & az & bz \tag3\end{pmatrix},$$ then WLOG \begin{cases} bz = 11-k-ay\\[4pt] a \le y, \quad b\le z, \end{cases} and this lead to the basic equalities in the forms of \begin{align} &\begin{pmatrix} 1 & 1 & 1 \\ y & y & y \\ 10-y & 10-y & 10-y\end{pmatrix}^2 = 11\,\begin{pmatrix} 1 & 1 & 1 \\ y & y & y \\ 10-y & 10-y & 10-y\end{pmatrix}, \qquad (y=1,2,\dots,9);\\[4pt] &\begin{pmatrix} 1 & 1 & 2 \\ 10-2z & 10-2z & 20-4z \\ z & z & 2z \end{pmatrix}^2 = 11\begin{pmatrix} 1 & 1 & 2 \\ 10-2z & 10-2z & 20-4z \\ z & z & 2z \end{pmatrix},\qquad (z=2,3,4);\\[4pt] &\begin{pmatrix} 1 & 1 & 3 \\ 1 & 1 & 3 \\ 3 & 3 & 9 \end{pmatrix}^2 = \begin{pmatrix} 11 & 11 & 33 \\ 11 & 11 & 33 \\ 33 & 33 & 99 \end{pmatrix};\\[4pt] &\begin{pmatrix} 1 & 2 & 1 \\ y & 2y & y \\ 10-2y & 20-4y & 10-2y\end{pmatrix} = 11\begin{pmatrix} 1 & 2 & 1 \\ y & 2y & y \\ 10-2y & 20-4y & 10-2y\end{pmatrix},\qquad (y=2,3,4);\\[4pt] &\begin{pmatrix} 1 & 2 & 2 \\ 2 & 4 & 4 \\ 3 & 6 & 6\end{pmatrix}^2 = \begin{pmatrix} 11 & 22 & 33 \\ 22 & 44 & 44 \\ 33 & 66 & 66\end{pmatrix};\\[4pt] &\begin{pmatrix} 1 & 2 & 2 \\ 3 & 6 & 6 \\ 2 & 4 & 4 \end{pmatrix}^2 = \begin{pmatrix} 11 & 22 & 22 \\ 33 & 66 & 66 \\ 22 & 44 & 44\end{pmatrix};\\[4pt] &\color{brown}{\mathbf{\begin{pmatrix} 1 & 3 & 1 \\ 3 & 9 & 3 \\ 1 & 3 & 1\end{pmatrix}^2 = \begin{pmatrix} 11 & 33 & 11 \\ 33 & 99 & 33 \\ 11 & 33 & 11\end{pmatrix};}}\\[4pt] &\begin{pmatrix} 2 & a & 9-a \\ 2 & a & 9-a \\ 2 & a & 9-a\end{pmatrix}^2 = 11 \begin{pmatrix} 2 & a & 9-a \\ 2 & a & 9-a \\ 2 & a & 9-a\end{pmatrix}, \qquad (a=1,2,\dots,8);\\[4pt] &\begin{pmatrix} 3 & a & 8-a \\ 3 & a & 8-a \\ 3 & a & 8-a\end{pmatrix}^2 = 11 \begin{pmatrix} 3 & a & 8-a \\ 3 & a & 8-a \\ 3 & a & 8-a\end{pmatrix}, \qquad (a=1,2,3,4);\\[4pt] &\begin{pmatrix} 1 & 2 & 8 \\ 1 & 2 & 8\\ 1 & 2 & 8\end{pmatrix}^2 = \begin{pmatrix} 11 & 22 & 88 \\ 11 & 22 & 88 \\ 11 & 22 & 88\end{pmatrix};\\[4pt] &\begin{pmatrix} 1 & 3 & 7 \\ 1 & 3 & 9 \\ 1 & 1 & 9\end{pmatrix}^2 = \begin{pmatrix} 11 & 33 & 77 \\ 11 & 33 & 77 \\ 11 & 33 & 77\end{pmatrix};\\[4pt] \end{align} etc.

Besides, $$\begin{pmatrix} 0 & 0 & 0 \\ 3 & 3 & 6 \\ 4 & 4 & 8\end{pmatrix}^2 = \begin{pmatrix} 0 & 0 & 0 \\ 33 & 33 & 66 \\ 44 & 44 & 88\end{pmatrix},$$ $$\color{brown}{{ \begin{pmatrix} 4 & 3 & 2 \\ 4 & 3 & 2 \\ 8 & 6 & 4 \end{pmatrix}^2 = \begin{pmatrix} 44 & 33 & 22 \\ 44 & 33 & 22 \\ 88 & 66 & 44 \end{pmatrix}. }}\tag4$$ At the same time, $$\color{brown}{{ \begin{pmatrix} 2 & 3 & 4 \\ 2 & 3 & 4 \\ 4 & 6 & 8 \end{pmatrix}^2 = 13_{\text{dec}} \begin{pmatrix} 2 & 3 & 4 \\ 2 & 3 & 4 \\ 4 & 6 & 8 \end{pmatrix} = \begin{pmatrix} 22 & 33 & 44 \\ 22 & 33 & 44 \\ 44 & 66 & 88 \end{pmatrix}_{12} }}\tag5$$ in twelve-digit number system.

Besides, this kind of matrices can be obtained, using transformations of the solutions.

$\color{brown}{\mathbf{Case\ n\ge 4.}}$

Solutions in the form of \begin{pmatrix} k & a & b & c & \dots \\ kz & az & bz & cz & \dots \\ ky & ay & by & cy & \dots \\ kx & ax & bx & cx & \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix} can be obtained from the solutions of the equation $$k + az + by + cx + \dots = 11.$$

So there are a lot of solutions with a strictly positive elements. For example, $$\color{brown}{\mathbf{ \begin{pmatrix} 1&2&1&2&1 \\ 2&4&2&4&2 \\ 1&2&1&2&1 \\ 2&4&2&4&2 \\ 1&2&1&2&1 \end{pmatrix}^2 =\begin{pmatrix} 11&22&11&22&11 \\ 22&44&22&44&22 \\ 11&22&11&22&11 \\ 22&44&22&44&22 \\ 11&22&11&22&11 \end{pmatrix}.}}\tag6 $$ Looks perfect the solution for $n=11:$ $$\color{brown}{\mathbf{ \begin{pmatrix} 1&1&1&1&1& 1 &1&1&1&1&1 \\ 1&1&1&1&1& 1 &1&1&1&1&1 \\ 1&1&1&1&1& 1 &1&1&1&1&1 \\ 1&1&1&1&1& 1 &1&1&1&1&1 \\ 1&1&1&1&1& 1 &1&1&1&1&1 \\ 1&1&1&1&1& 1 &1&1&1&1&1 \\ 1&1&1&1&1& 1 &1&1&1&1&1 \\ 1&1&1&1&1& 1 &1&1&1&1&1 \\ 1&1&1&1&1& 1 &1&1&1&1&1 \\ 1&1&1&1&1& 1 &1&1&1&1&1 \\ 1&1&1&1&1& 1 &1&1&1&1&1 \end{pmatrix}.}}\tag7$$

If $n>11,$ then solutions should contain zeros.

$\color{brown}{\textbf{Allowed transformations of matrices.}}$

Allowed transformations of matrices are transposition and sparsing.

There are two kinds of the allowed sparsing:

• Inserting of the zero row and zero before or after diagonal element of matrix;
• Substitution of each matrices element $a$ to the $2\times2$ matrix in the form of $$\begin{pmatrix} a & 0 \\ 0& a \end{pmatrix}\tag8.$$

In particular, the matrices in the forms of $$\begin{pmatrix} 0 & 0 & 0 \\ 0 & a & b \\ 0 & c & d \end{pmatrix}, \begin{pmatrix} a & 0 & b \\ 0 & 0 & 0 \\ c & 0 & d \end{pmatrix}, \begin{pmatrix} a & b & 0 \\ c & d & 0 \\ 0 & 0 & 0 \end{pmatrix}, \tag9$$ where $a,b,c,d$ correspond to the $2\times2$ solutions $(2),$ are the solutions in the $3\times3$ case.

Here's a solution for all $k$: take a $(k-1) \times (k-1)$ matrix $A$ with $A_{k-1,1} = \underbrace{11\dots1}_k$, $A_{i,i+1} = 1$ for $i=1,\dots,k-2$, and all other entries $0$. For example, for $k=6$, take the following $5 \times 5$ matrix: $$ \begin{bmatrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 111111 & 0 & 0 & 0 & 0 \end{bmatrix} $$ This works because $A$ satisfies $A \vec{e}_i = \vec e_{i-1}$ for $i=2,\dots,k-1$, and $A \vec e_1 = \underbrace{11\dots1}_k \vec e_{k-1}$. Therefore, for any $i$, $A^{k-1} \vec e_i = \underbrace{11\dots1}_k \vec e_i$, so $A^{k-1} = \underbrace{11\dots1}_kI$, and $A^k = \underbrace{11\dots1}_kA$.