Product of binary matrices with binary eigenvalues

Question

Consider two binary matrices with obvious patterns: $$ C= \begin{bmatrix} 1 &0 &0 &0 &0 &0 &0\\ 1 &0 &0 &0 &0 &0 &0\\ 0 &1 &0 &0 &0 &0 &0\\ 0 &1 &0 &0 &0 &0 &0\\ 0 &0 &1 &0 &0 &0 &0\\ 0 &0 &1 &0 &0 &0 &0\\ 0 &0 &0 &1 &0 &0 &0 \end{bmatrix} $$ and $$ T= \begin{bmatrix} 1 &1 &0 &0 &0 &0 &0\\ 0 &1 &1 &0 &0 &0 &0\\ 0 &0 &1 &1 &0 &0 &0\\ 0 &0 &0 &1 &1 &0 &0\\ 0 &0 &0 &0 &1 &1 &0\\ 0 &0 &0 &0 &0 &1 &1\\ 0 &0 &0 &0 &0 &0 &1 \end{bmatrix} $$

The eigenvalues of the matrices $T^n C,n=0,1,2,3$ are zeros and consecutive powers of $2$ equal to $0,1,2,4$. I'd like to have a proof of the generalization of this fact for matrices of larger size with the same patterns.

Note, the entries of $T^n C$ in the left upper corner are zeros and binomial coefficients for the power $n+1$.

A motivation for this question is in Binary eigenvalues matrices and continued fractions

Answer 1

The left upper block of the matrix $T^n C$ can be conjugated to upper triangular one with the eigenvalues on diagonal by Pascal triangle matrix.

@Suvrit https://mathoverflow.net/questions/258284/is-the-matrix-left2m-choose-2j-i-right-i-j-12m-1-nonsingular

Answer 2

Some thoughts:

Note that $T = I + N$, where $I$ is the identity matrix and $$ N = \pmatrix{0&1\\&0&1\\&&0&1\\&&&0&1\\&&&&0&1\\&&&&&0&1\\&&&&&&0} $$ Notably, $N^7 = 0$. Because $NI = IN$, we can compute $T^n = (I + N)^n$ by binomial expansion. That is, we have $$ T^n = \binom n0 I + \binom n1 N + \cdots + \binom n6 N^6 $$ We can verify that $T^n$ is therefore the upper triangular Toeplitz matrix for which, in the notation of the linked wiki page, we have $a_{-k} = \binom nk$ whenever $0 \leq k \leq n$ and all other entries are $0$.

With that, we may compute $$ T^n C = \pmatrix{\binom n0 + \binom n1 & \binom n2 + \binom n3 & \binom n4 + \binom n5 & \binom n6 &0&0&0\\ \binom n0 & \binom n1 + \binom n2 & \binom n3 + \binom n4 & \binom n5 & 0&0&0\\ 0 & \binom n0 + \binom n1 & \binom n2 + \binom n3 & \binom n4 & 0&0&0\\ 0 & \binom n0 & \binom n1 + \binom n2 & \binom n3 & 0&0&0\\ 0 & 0 & \binom n0 + \binom n1 & \binom n2 & 0&0&0\\ 0 & 0 & \binom n0 & \binom n1 & 0&0&0\\ 0 & 0 & 0 & \binom n0 & 0&0&0\\} $$