Determinant of a Gram's $n \times n$ matrix

Question

If I have a $n \times n$ matrix, call it A, and I know that the determinant of the matrix $A_{n-1,n-1}>0$, how can I prove that det($A_{n,n})>0$, for the following special case? The idea is to prove, by induction, that $A$ is a positive-definite matrix.

I would also want to know which are the vectors $v_{i}$, $v_{j}$ that satisfy $(v_{i}|v_{j})=c^{|i-j|}$ i.e I would also want to find the set of independent (normalized) vectors that have $A$ for Gram's matrix.

Note: $0<c<1$.

Answer 1

Let $N$ be the upper triangular nilpotent Jordan block of size $n$. Then $$ (I-cN)\,A\,(I-cN)^T=\operatorname{diag}(1-c^2,\ldots,1-c^2,1)\succ0. $$ Therefore $A$ is positive definite, by Sylvester's law of inertia. Furthermore, we may rewrite the above as $A=V^TV$, where $$ V=\operatorname{diag}\left(\sqrt{1-c^2},\ldots,\sqrt{1-c^2},1\right)\left((I-cN)^{-1}\right)^T. $$ Hence the columns of $V$ satisfy $\langle v_i,v_j\rangle=a_{ij}=c^{|i-j|}$. More specifically, $V$ is the lower triangular matrix given by $$ v_{ij}=\begin{cases} 0 &\text{ if }j>i,\\ \sqrt{1-c^2}c^{i-j} &\text{ if }j\le i<n,\\ c^{n-j} &\text{ if }j\le i=n. \end{cases} $$

Answer 2

Substracting from rows $2$ to $n$ the previous row multiplied with $c$ leaves you with an upper triangular matrix, and the determinant is the product of the diagonal elements $$ 1 \cdot (1-c^2) \cdots (1-c^2) = (1-c^2)^{n-1} $$ which is positive for $0 < c < 1$.

Answer 3

By induction: the base case $n=1$ is easy. In general: write $$ A_n = \pmatrix{1 & x^T\\x&A_{n-1}} $$ Now, compute the Schur complement $$ A_n/A_{n-1} = 1 - x^TA_{n-1}^{-1}x $$ However, $x$ is simply the first column of $A_{n-1}$ multiplied by $c$. Thus, we have $$ A_n/A_{n-1} = 1 - x^T[A_{n-1}^{-1}x] = 1 - c\;x^Te_1 = 1 - c^2 $$ Because $A_{n-1}$ and $A_n/A_{n-1}$ are both positive (definite), we can conclude that $A_n$ is positive definite.