Monotonicity of the determinant of symmetric Toeplitz Matrices

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 be a Kac-Murdock-Szegő (KMS) matrix, e.g., for n=4

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

• Something is strange: if $A_1$ (or $A_k$) can cross into non-positive domain, then at the moment of crossing the determinant is zero, so the inequality you propose fails. So either they are all positive definite for $0<a<1$ (and the corresponding requirement is automatic), or the conjectured inequality fails. However if $A=\begin{bmatrix}1&0&1&1\\0&1&0&1\\1&0&1&0\\1&1&0&1\end{bmatrix}$ and $x=[1,1,-1,-1]$, then $(Ax,x)=-2<0$, so we are already way out of positive definite domain. What am I missing? Nov 30, 2021 at 1:57
• Thanks for your comment, actually i miss the assumption that A is strictly diagonally dominant, otherwise it fails to be definite positive Nov 30, 2021 at 15:03

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)

• Thanks a lot for the exhaustive answer! So if i get this right, the proposed inequality works as long as "a" is close enough to zero. I miss some details about how obtaining the term $a^{2j}$, but probably it's my fault. Nov 30, 2021 at 22:04
• Yes, near zero the inequality is valid. For the computation of the determinants, this should be simple. For a matrix $X$ of shape $n\times n$ we have $$\det X=\sum_{\sigma\in S(n)}\pm x_{1\sigma(1)}x_{2\sigma(2)}\dots x_{n\sigma(n)}\ ,$$the sign of the term involving $\sigma$ being the sign of the permutation $\sigma$. For the matrix $A_{j-a}(a)$ products obtained from this expansion have some diagonal-$1$-factors (if any) and else only factors $a^M$ with $M\ge j$ that must come from different rows and columns. $1=1^n$ comes from diagonal. $a^{2j}$ from two diagonally reflected $a^j$'s. Dec 1, 2021 at 11:40