Linear Algebra – Monotonicity of Determinant of Symmetric Toeplitz Matrices

Question

For simplicity, i focus on a particular Toeplitz symmetric matrix, so let $A_{ij} = a^{|i-j|}$ for $i,j=1,\dots,n$ and $0<a<1$ be a Kac-Murdock-Szegő (KMS) matrix, e.g., for n=4

\begin{equation}
A = \left[
\begin{matrix}
1 & a & a^2 & a^3 \\
a & 1 & a & a^2 \\
a^2 & a & 1 & a \\
a^3 & a^2 & a & 1
\end{matrix}
\right]
\end{equation}
Consider the following matrix $A_1$ obtained from $A$ setting to zero both the subdiagonal and the superdiagonal as
\begin{equation}
A_1 = \left[
\begin{matrix}
1 & 0 & a^2 & a^3 \\
0 & 1 & 0 & a^2 \\
a^2 & 0 & 1 & 0 \\
a^3 & a^2 & 0 & 1
\end{matrix}
\right]
\end{equation}
when $a$ is chosen so that $A_1$ is definite positive, we have $\det(A) < \det(A_1)$.
Continuing (under the def.positive assumption) nulling the other additional off diagonals elements we obtain
\begin{equation}
A_2 = \left[
\begin{matrix}
1 & 0 & 0 & a^3 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
a^3 & 0 & 0 & 1
\end{matrix}
\right]
\end{equation}
Hence, we have $(1-a^2)^3 =\det(A) < \det(A_1)<\det(A_2) < 1$, intuitively increasing the sparsity of the matrix its determinant goes to 1.

My question is: how this fact can be proved, or alternatively to establish that $(1-a^2)^{(n-1)}$ is the minimum value attained by the determinant of the above defined matrices $A_1, A_2, \cdots A_{n-1}$ ?

Answer 1

Let us compute and plot the functions of the parameter $a$ defined as the determinants of the matrices $A_0$, $A_1$, $A_2$, $A_3$: $$ \begin{aligned} \det A_0 &= \begin{vmatrix}1&a&a^2&a^3\\a&1&a&a^2\\a^2&a&1&a\\a^3&a^2&a&1\end{vmatrix} = 1-3a^2+3a^4-a^6=(1-a)^3(1+a)^3 \ ,\\[2mm] \det A_1 &= \begin{vmatrix}1&0&a^2&a^3\\0&1&0&a^2\\a^2&0&1&0\\a^3&a^2&0&1\end{vmatrix} = 1-2a^4+a^8-a^6 =(1-a^3-a^4)(1+a^3-a^4) \ ,\\[2mm] \det A_2 &= \begin{vmatrix}1&0&0&a^3\\0&1&0&0\\0&0&1&0\\a^3&0&0&1\end{vmatrix} = 1-a^6=(1-a)(1+a)(1-a+a^2)(1+a+a^2) \ ,\\[2mm] \det A_3 &= \begin{vmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{vmatrix} =1 \ . \end{aligned} $$ A plot of these functions of $a\in(0,1)$ is:

So the claimed inequality $(1-a^2)^3=\det A_0(a)\le \det A_1(a)\le\det A_2(a)\le \det A_3(a)=1$ is valid for $n=4$ only in some neighborhood of zero.

Alternatively, let us consider the case $n=11$.

---

So let us consider only the functions in a neighborhood of zero. In $a=0$ all determinants of the $n\times n$-matrices $A_j(a)$ are one. Then it is enough to figure out the first non-trivial term in $a,a^2,a^3,\dots$ in the Taylor expansion of $\det A_j(a)$, $j$ running from $0$ to $n-1$. It turns out that we have: $$\tag{$\dagger$}$$ $$ \begin{aligned} \det A_0(a) &= 1 -(n-1)a^2 + O(a^4)\ ,\\ \det A_1(a) &= 1 -(n-2)a^4 + O(a^6)\ ,\\ \det A_2(a) &= 1 -(n-3)a^6 + O(a^8)\ ,\\ &\text{ and so on }\\ \det A_{j-1}(a) &= 1 -(n-j)a^{2j} + O(a^{2(j+1)})\ ,\\ &\text{ till finally we have}\\ \det A_{n-2}(a) &= 1-a^{2(n-1)}\ ,\\ \det A_{n-1}(a) &= 1\ . \end{aligned} $$ To see this, consider the matrix $A_{j-1}(a)$, $$ A_{j-1}(a)= \begin{bmatrix} 1 &0&0&\dots&0&a^j&a^{j+1}&\dots\\ 0&1 &0&0&\dots&0&a^j&\ddots\\ 0&0&1 &0&0&\dots&0&\ddots\\ \vdots&0&0&1&0&0&0&\ddots\\ 0&\vdots&0&0&1&0&0&\ddots\\ a^j&0&\vdots&0&0&1&0&\ddots\\ a^{j+1}&a^j&0&\vdots&0&0&1&\ddots\\ \vdots&\ddots&\ddots&\ddots&\ddots&\ddots&\ddots&\ddots\\ \end{bmatrix} $$ Its determinant is a sum of terms corresponding to permutations in the symmetric group $S(n)$ with $n!$ elements. Excepting $1$ (corresponding to the trivial permutation), there is no term in degree $<2j$. To obtain terms in degree $a^{2j}$ we need a permutation that takes from each row (column) a non-zero entry, and two of these entries are $a^j$, all other must be diagonal ones. So it is a transposition, one among $(1,j+1)$, $(2,j+2)$, ... $(n-j,n)$. And there is no chance to produce $(\pm 1)^{n-2}\cdot a^j\cdot a^{j+1}$ using other permutations. This shows $(\dagger)$.

$\square$

---

Sage code used to produce the graphs:

def A(n, k, a):
    return matrix(n, n, [ 0 if j != jj and abs(j-jj) <= k 
                          else a^abs(j-jj) 
                          for j in range(n) for jj in range(n)])

def mygraphs(N):
    P = plot([])
    for j in range(N):
        P += plot( lambda s: A(N, j, s).det() , (0, 1)
                   , color=((j+1)^2/N^2, (j+1)/N, 1-j/N)
                   , legend_label=f'{j}' )
    return P

mygraphs(4)
mygraphs(11)