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 $A_1, A_2, \cdots A_{n-1}$ ?

  • 1
    $\begingroup$ 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? $\endgroup$
    – fedja
    Nov 30, 2021 at 1:57
  • $\begingroup$ Thanks for your comment, actually i miss the assumption that A is strictly diagonally dominant, otherwise it fails to be definite positive $\endgroup$ Nov 30, 2021 at 15:03

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:

Picture for the graphs of \det A_j for N=4

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 $n=11$.

Picture for the graphs of \det A_j for 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 $(\dagger)$.


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

  • $\begingroup$ 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. $\endgroup$ Nov 30, 2021 at 22:04
  • $\begingroup$ 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. $\endgroup$
    – dan_fulea
    Dec 1, 2021 at 11:40

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