Correlation matrix with same pairwise correlation coefficient Question
Correlation matrix. Consider  random variables with the same pairwise correlation
coefficient $\rho_n$. Find the highest possible value of $\rho_n$ for
a) n=3
b) n=4
c) general, n $\geq$ 2
HINT: Correlation matrix must be positive semi-definite. 
My Workings
This is what I infer from "same pairwise coefficients":
$$
    \begin{pmatrix}
    1 & \rho_n & \rho_n \\
    \rho_n & 1 & \rho_n \\
    \rho_n & \rho_n & 1 \\
    \end{pmatrix}
$$
Because a correlation matrix is positive semi-definite, all principal minors have to be positive
For n=3, the principal minors calculations yield 
$∣H_1∣$=1
$∣H_2∣$=1-$\rho_n^2$ $\geq 0$ $\Rightarrow $ $\rho_n$ $\leq 1$
$∣H_3∣=\rho_n^2$ - $\rho_n$ $\geq 0$ $\Rightarrow $ $\rho_n$ $\geq 1$
$∣H_4∣=(1^3$+2$\rho_n^3$)-(3$\rho_n^2$)$\geq 0$ $ . $ I solved for the roots and found 1
Conclusion: For n=3, max $\rho_n$=1
I did the same method for n=4 and found max $\rho_n$=1 again
My Problem
Result looks too simple and false. Method is tedious
I have no idea how to do the general case. (by induction ?)
Thank you for your help
$\\$
INDUCTION ATTEMPT
$H_0$: n=2
$$
    \begin{pmatrix}
    1 & \rho_2 \\
    \rho_2 & 1 \\
    \end{pmatrix}
$$
1-$\rho_2^2 \geq 0 $ $\Rightarrow - \frac{1}{2-1} \leq \rho_2 \leq 1$
$H_n$: Pn: Suppose that for a nxn correlation matrix $ A_n$ with same pairwise coefficients $ \rho_n$, $ -\frac{1}{n-1} \leq \rho_n \leq 1 $ holds
$H_{n+1}$:
$ -\frac{1}{n-1} \leq \rho_n \leq 1 $
$ \Leftarrow \Rightarrow$ $- \frac{1}{(n+1)-1} \leq \rho_{n+1} \leq 1 $
$ \Leftarrow \Rightarrow$ $- \frac{1}{n} \leq \rho_{n+1} \leq 1 $
And because for all n>1, $- \frac{1}{n} \geq -\frac{1}{n-1}$
$ -\frac{1}{n-1} \leq \rho_{n+1} \leq 1 $
 A: A symmetric matrix is positive semidefinite if and only if all eigenvalues are nonnegative.
This type of correlation matrix has eigenvalues $\lambda_1 =1 +(n-1) \rho_n$ of multiplicity $1$ and $\lambda_2 =1 - \rho_n$ of multiplicity $n-1$.  By inspection it is easy to see that the vector $e_1 + e_2 + \ldots + e_n = (1,1,\ldots,1)^T$ is an eigenvector corresponding to $\lambda_1$.  Since the matrix is real symmetric the remaining eigenvectors are orthogonal.  Again by inspection, we see that $e_1-e_2, e_1 - e_3, \ldots e_1 - e_n$ are the remaining eigenvectors all corresponding to $\lambda_2$.
The conditions for the matrix to be positive semidefinite are 
$$1 - \rho_n \geqslant 0, \,\, 1 +(n-1)\rho_n  \geqslant 0,$$
which implies for all $n \geqslant 2$,
$$-\frac{1}{n-1} \leqslant \rho_n \leqslant 1$$
Using Principal Minors
Principal minors are determinants of submatrices obtained by deleteing rows and columns with the same indexes.
For $n=3$: 
The three first-order principal minors are obtained by deleting (1) the first and second rows and columns,  (2) the first and third rows and columns, and (3) the second and third rows and columns.  All have value $1$.
The three second-order principal minors are obtained by deleting (1) the first row and column,  (2) the second row and column, and (3) the third row and column.  All have value
$$\begin{vmatrix}
1 & \rho_3\\
\rho_3 & 1
\end{vmatrix} = 1 - \rho_3^2$$
The third-order principal minor is the full determinant
$$D_3 = \begin{vmatrix}
1 & \rho_3 & \rho_3\\
\rho_3 & 1 & \rho_3 \\ \rho_3 &  \rho_3 &1
\end{vmatrix} = 1 - 3\rho_3^2 + 3 \rho_3^3 = (1 +2 \rho_3)(1- \rho_3)^2$$
To ensure that all principal minors are nonnegative we must have 
$$1 - \rho_3^2 \geqslant 0 \implies \rho_3 \leqslant 1, \\ (1 +2 \rho_3)(1- \rho_3)^2 \implies \rho_3 \geqslant - 1/2,$$
and the largest  possible value is $\rho_3 = 1$ if the matrix is positive semi-definite.
For n = 4:
Because all off-diagonal entries are identical the principal minors of first-, second- and third-order are identical (replacing $\rho_3$ with $\rho_4$) to those obtained for $n =3$, imposing the requirement $-1/2 \leqslant \rho_4 \leqslant 1$.
The fourth-order principal minor is the determinant
$$D_4 = \begin{vmatrix}
1 & \rho_4 & \rho_4 &\rho_4\\
\rho_4 & 1 & \rho_4 & \rho_4\\ \rho_4 &  \rho_4 &1 &\rho_4 \\ \rho_4 &  \rho_4 &\rho_4 &1 
\end{vmatrix} =  \begin{vmatrix}
1 - \rho_4 & \rho_4 - 1 & 0 & 0\\
\rho_4 & 1 & \rho_4 & \rho_4\\ \rho_4 &  \rho_4 &1 &\rho_4 \\ \rho_4 &  \rho_4 &\rho_4 &1 
\end{vmatrix} $$
where the RHS is obtained by subtracting the second row from the first row in the LHS.
Expanding we get 
$$D_4 = (1 - \rho_4) \begin{vmatrix}
1 & \rho_4 & \rho_4\\  \rho_4 &1 &\rho_4 \\   \rho_4 &\rho_4 &1 
\end{vmatrix}- (\rho_4 -1) \begin{vmatrix}
\rho_4 & \rho_4 & \rho_4\\  \rho_4 &1 &\rho_4 \\   \rho_4 &\rho_4 &1 
\end{vmatrix}$$
Using the result for $D_3$ for the first determinant on the RHS and subtracting the third column from the first column in the second determinant we get
$$D_4 = (1-\rho_4)(1 + 2\rho_4)(1- \rho_4)^2 - (\rho_4-1)(\rho_4 - 1)\rho_4(\rho_4- 1) \\ = (1 + 3\rho_4)(1- \rho_4)^3$$
Given that we must have $\rho_4 \leqslant 1$, the only new constraint obtained by imposing $D_4 \geqslant 0$ is $\rho_4 \geqslant -1/3$.
For $n >4$:
Proceeding inductively we find the same conditions obtained for $n -1$ and the additional condition obtained by examining the $n$th-order principal minor. Using the same manipulations as above this can be expanded in two terms involving lower-order determinants leading to 
$$D_n = (1 + (n-1) \rho_n) (1 - \rho_n)^{n-1}$$ 
For all $n$ the largest possible value ensuring that the matrix is positive semi-definite is $\rho_n = 1$.
