What is the condition for n by n symmetric matrix with 1 on diagonal and x everywhere else to be positive definite

Question

What is the condition on x for n by n symmetric matrix with 1 on diagonal and x everywhere else to be positive definite

Let the matrix be

\begin{bmatrix} 1 & x & x & ... &x\\ x & 1 & x & ... &x\\ x & x & 1 & ... &x\\ .. & .. & ..& ...& ..\\ x & x & x & ... &1 \end{bmatrix}

Answer 1

The eigenvalues of this matrix are $1-x$ and $1+(n-1)x$. For both to be positive, we must have

$$-\frac1{n-1}\lt x\lt 1\;.$$

Answer 2

My long answer would be based on the fact that all the pivots of Positive-Definite matrix are always greater than 0

Now when $n=1$, matrix is $ 1 X 1 $, and pivot is $1$

When $n=2$, row reduced echelon form is \begin{bmatrix} 1 & x\\ 0 & 1-x^2 \end{bmatrix}

and hence its pivots are $1, (1-x^2)$

When $n=3$, row reduced echelon form is, \begin{bmatrix} 1 & x & x \\ 0 & 1-x^2 & x(1-x) \\ 0 & 0 & \frac{(1-x)(1+2x)}{(1+x)} \end{bmatrix} and hence its pivots are $1, (1-x^2), \frac{(1-x)(1+2x)}{(1+x)}$

When $n=4$, row reduced echelon form is, \begin{bmatrix} 1 & x & x & x\\ 0 & 1-x^2 & x(1-x) & x(1-x)\\ 0 & 0 & \frac{(1-x)(1+2x)}{(1+x)} & \frac{(1-x)x}{(1+x)}\\ 0 & 0 & 0 & \frac{(1-x)(1+3x)}{(1+2x)} \end{bmatrix} and hence its pivots are $1, (1-x^2),\frac{(1-x)(1+2x)}{(1+x)}, \frac{(1-x)(1+3x)}{(1+2x)}$

Generalizing the results for $n$, pivots are $1, \frac{(1-x)(1+x)}{(1+0x)},\frac{(1-x)(1+2x)}{(1+x)}, \frac{(1-x)(1+3x)}{(1+2x)}, ... ,\frac{(1-x)(1+(n-1)x)}{(1+(n-2)x)}$

For matrix to be positive-definite, all the pivots should be greater than $0$

Clearly first pivot, $(1)$ is $>0$

for second pivot, $(1-X^2) > 0 \implies -1<x<1$

for third pivot, $\frac{(1-x)(1+2x)}{(1+x)} > 0 \implies \frac{-1}{2}<x<1$

for fourth pivot, $\frac{(1-x)(1+3x)}{(1+2x)} > 0 \implies \frac{-1}{3}<x<1$

Generalizing this way, for n pivots, $\frac{-1}{(n-1)}<x<1$