Vtyjkyuk

Suppose $a_0, a_1, ... , a_n-1$ are real numbers from $(0; 1)$, such that $sum_k=0^n-1 a_k=1$. Suppose $A = (c_ij)$ is a $n times n$ matrix with entries $c_ij = a_(i-j)%n$, where $%$ is modulo operation. Is it always true that $lim_m to infty A^m = frac1n beginpmatrix 1 & 1 & cdots & 1 \1 & 1 & cdots & 1 \ vdots & vdots & ddots & vdots \ 1 & 1 & cdots & 1 endpmatrix$?

This statement is true for $n = 2$:
Suppose $A = beginpmatrix a_0 & 1 - a_0 \ 1 - a_0 & a_0 endpmatrix = beginpmatrix 1 & -1 \ 1 & 1endpmatrix^-1beginpmatrix 1 & 0 \ 0 & 1-2a_0 endpmatrixbeginpmatrix 1 & -1 \ 1 & 1 endpmatrix$. Because $a_0$ is in $(0; 1)$, $1-2a_0$ is in $(-1;1)$. Thus $$lim_m to infty A^m = beginpmatrix 1 & -1 \ 1 & 1endpmatrix^-1beginpmatrix 1 & 0 \ 0 & 0endpmatrixbeginpmatrix 1 & -1 \ 1 & 1 endpmatrix = frac12beginpmatrix 1 & 1 \ 1 & 1 endpmatrix$$

However, I do not know how to prove this statement for arbitrary $n$.

Any help will be appreciated.

This follows directly from the Perron-Frobenius theorem.

Your matrix $A$ is a circulant matrix:

$$A = beginbmatrix
a_0 & a_1 & cdots & a_n-1\
a_n-1 & a_0 & cdots & a_n-2\
vdots & vdots & cdots & vdots \
a_1 & a_2 & cdots & a_0\
endbmatrix$$

$A$ is known to have eigenvalues equal to $lambda_j = sum_k=0^n-1 a_komega_j^k$ with eigenvectors $beginbmatrix 1 \ omega_j \ omega_j^2 \ vdots \ omega_j^n-1endbmatrix$, where $omega_j = e^frac2pi ijn$ for $j = 0, ldots n-1$.

Therefore we can diagonalize $A$ as follows:

$$A = P^-1DP = beginbmatrix
1 & 1 & cdots & 1\
omega_0 & omega_1 & cdots & omega_n-1\
vdots & vdots & cdots & vdots \
omega_0^n-1 & omega_1^n-1 & cdots & omega_n-1^n-1
endbmatrix^-1
beginbmatrix
lambda_0 & 0 & cdots & 0 \
0 & lambda_1 & cdots & 0\
vdots & vdots & cdots & vdots \
0 & 0 & cdots & lambda_n-1\
endbmatrix
beginbmatrix
1 & 1 & cdots & 1\
omega_0 & omega_1 & cdots & omega_n-1\
vdots & vdots & cdots & vdots \
omega_0^n-1 & omega_1^n-1 & cdots & omega_n-1^n-1
endbmatrix$$

The triangle inequality for the eigenvalues gives

$$|lambda_j| = left|sum_k=0^n-1 a_komega_j^kright| le sum_k=0^n-1 a_k |omega_j|^k = sum_k=0^n-1 a_k = 1$$

with equality holding if and only if $a_0, a_1omega_j, ldots, a_n-1omega_j^n-1$ lie on the same side of a single line through the origin. Clearly this is true iff $j = 0$ so the eigenvalues satisfy $lambda_0 = 1$ and $|lambda_j| < 1$ for $j = 1, ldots, n-1$.

Hence letting $mtoinfty$ in $A^m = P^-1D^mP$ gives

beginalign
lim_mtoinfty A^m &= beginbmatrix
1 & 1 & cdots & 1\
omega_0 & omega_1 & cdots & omega_n-1\
vdots & vdots & cdots & vdots \
omega_0^n-1 & omega_1^n-1 & cdots & omega_n-1^n-1
endbmatrix^-1
beginbmatrix
1 & 0 & cdots & 0 \
0 & 0 & cdots & 0\
vdots & vdots & cdots & vdots \
0 & 0 & cdots & 0\
endbmatrix
beginbmatrix
1 & 1 & cdots & 1\
omega_0 & omega_1 & cdots & omega_n-1\
vdots & vdots & cdots & vdots \
omega_0^n-1 & omega_1^n-1 & cdots & omega_n-1^n-1
endbmatrix\
&= beginbmatrix
1 & 1 & cdots & 1\
omega_0 & omega_1 & cdots & omega_n-1\
vdots & vdots & cdots & vdots \
omega_0^n-1 & omega_1^n-1 & cdots & omega_n-1^n-1
endbmatrix^-1 beginbmatrix 1 \ 1 \ vdots \ 1endbmatrix
endalign

You can calculate this by hand, or notice that the columns of $lim_mtoinfty A^m$ satisfy the system $Px = beginbmatrix 1 \ 1 \ vdots \ 1endbmatrix$, which you can solve.

The sum $sum_k=0^n-1 omega_k^j$ is equal to $0$ here, which would suggest that the limit doesn't have to be of the form you specified.