Eigenvalues of a matrix with repeating pattern of entries

Question

I observed that if

$$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$$

with non-zero eigenvalues $\alpha$ and $\beta$, then

$$\begin{bmatrix} A & A\\ A & A \end{bmatrix}$$

has eigenvalues $2 \alpha$, $2 \beta$, and $0$. Also,

$$\begin{bmatrix} A & A & A \\ A & A & A \\ A & A & A \end{bmatrix}$$

has eigenvalues $3 \alpha$, $3 \beta$, $0$. Therefore, my conjecture is that for some $r$, $A^{[r]}$ has eigenvalues $(r+1) \alpha$, $(r+1) \beta$, $0$.

Is it correct? Is there some theorems related to this? How about their eigenvectors? Can you please send me links that can help me with this kind of problem?

PS. This is my first time asking here. I am an undergrad math student. Please help me. Thank u so much.

Answer 1

Let

$$\begin{bmatrix} \mathrm A & \mathrm A & \dots & \mathrm A\\ \mathrm A & \mathrm A & \dots & \mathrm A\\ \vdots & \vdots & \ddots & \vdots\\ \mathrm A & \mathrm A & \dots & \mathrm A\end{bmatrix} = 1_k 1_k^\top \otimes \mathrm A$$

be a $k \times k$ block matrix, where $\otimes$ denotes the Kronecker product. Let the spectrum of $\rm A$ be $\{\alpha,\beta\}$.

Since the eigenvalues of rank-$1$ matrix $1_k 1_k^\top$ are $0$ (with multiplicity $k-1$) and $k$ (with multiplicity $1$), then the eigenvalues of block matrix $1_k 1_k^\top \otimes \mathrm A$ are

• $0$ with multiplicity $2 k - 2$.
• $k \alpha$ with multiplicity $1$.
• $k \beta$ with multiplicity $1$.

Answer 2

Yes, your conjecture is correct. Suppose that $A$ is an $n\times n$ matrix and that you consider the $(nm)\times(nm)$ matrix $$ B=\begin{bmatrix}A&A&\dots&A\\ A&A&\dots&A\\ \vdots&\vdots&\ddots&\vdots\\ A&A&\dots&A \end{bmatrix} $$ that is a block matrix consisting of $m\times m$ copies of $A$.

We can determine the eigenvalues of this matrix by inspection. We first note that $\operatorname{rank}(A)=\operatorname{rank}(B)$. This follows from the fact $B$ has lots of copies of the same rows. In fact, if $i$ and $j$ have $i\equiv j\pmod n$, then the rows $i$ and $j$ are copies of each other. Putting this together, with simple row operations, we can row reduce $B$ to $$ \begin{bmatrix}A&A&\dots&A\\ 0&0&\dots&0\\ \vdots&\vdots&\ddots&\vdots\\ 0&0&\dots&0 \end{bmatrix} $$ Then, we observe that the ranks of $A$ and $B$ are the same. Moreover, this row reduction exposes that at least $n(m-1)$ eigenvalues of $B$ are $0$. (Normally, row operations mess up eigenvalues, but this doesn't happen with zero eigenvalues)

Now, let's deal with the eigenvalues from $A$. Suppose that $\lambda$ is an eigenvalue from $A$ with eigenvector $v$. Now, consider the vector $$ w=\begin{bmatrix} v\\v\\\vdots\\v \end{bmatrix} $$ consisting of $m$ copies of $v$. Then $$ Bw=\begin{bmatrix}A&A&\dots&A\\ A&A&\dots&A\\ \vdots&\vdots&\ddots&\vdots\\ A&A&\dots&A \end{bmatrix}\begin{bmatrix} v\\v\\\vdots\\v \end{bmatrix} =\begin{bmatrix} Av+Av+\dots+Av\\ Av+Av+\dots+Av\\ \vdots\qquad\vdots\\ Av+Av+\dots+Av \end{bmatrix}, $$ where each block row has $m$ copies of $Av$. In other words, this simplifies to $$ \begin{bmatrix} mAv\\ mAv\\ \vdots\\ mAv \end{bmatrix}. $$ Since $v$ is an eigenvector for $A$, we can further simplify to $$ \begin{bmatrix} m\lambda v\\ m\lambda v\\ \vdots\\ m\lambda v \end{bmatrix}=m\lambda w. $$ Therefore, $m\lambda$ is an eigenvalue for $A$.

Now, you have to convince yourself that we haven't double counted, but I'll leave that for you. If you want to, try to find the eigenvectors for the zeros we found above and observe that the eigenvectors are in distinct spaces. Hint: if $i\equiv j\pmod n$, use the vector which is $1$ in row $i$, $-1$ in row $j$, and zero elsewhere.