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Show that the matrix, $$ A= \begin{bmatrix} 1 & \rho & \dots & \rho \\ \rho & 1 & \dots & \rho \\ \vdots & \rho & \ddots & \vdots \\ \rho & \dots & \rho & 1 \end{bmatrix} $$ is positive-definite if and only if $\frac{-1}{n-1}<\rho<1$.

I know that one way would be to show that all eigenvalues are positive but I couldn't produce anything usefuol yet. I would be glad for some ideas or rough proof sketch.

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  • $\begingroup$ For a proof that the common correlation $\rho$ must have value at least as large as $frac{1}{n-1}$, see this answer of mine on stats.SE. $\endgroup$ – Dilip Sarwate Apr 14 '18 at 22:22
  • $\begingroup$ Strictly speaking this is only true for $n > 1$. $\endgroup$ – Thoth Apr 15 '18 at 3:50
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Setting aside the degenerate case $n=1$ for which $A$ is positive-definite for all $\rho$, here's a sketch of a purely mathematical solution,

Consider the decomposition of your matrix,

$$A(\rho)=\rho{\bf 11}^T + (1-\rho)I.$$

Let $A_k(\rho)$ denote the upper-left $k\times k$ submatrix of $A(\rho)$.

Then $A(\rho)$ being positive-definite is equivalent to all of its leading principal minors $\det A_k(\rho)$ for $1\leq k\leq n$ being positive.

Therefore we must show that for all $1\leq k\leq n$,

\begin{align} \det A_k(\rho)&=\det (\rho{\bf 11}_k^T + (1-\rho)I_k)\\ &=(1-\rho)^k+k\rho(1-\rho)^{k-1} \end{align}

is positive on the open interval $(-\frac{1}{n-1},1)$, and that for any point outside this interval there exists a $k$ such that it's non-positive, where the second equality follows by the matrix determinant lemma.

Hint: take the derivative with respect to $\rho$.

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  • $\begingroup$ Very nice proof. Thank you! $\endgroup$ – drogan Apr 17 '18 at 12:19

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