Bounds on off-diagonal entries of a correlation matrix Assume that all the entries of an $n \times n$ correlation matrix which are not on the main
diagonal are equal to $q$. Find upper and lower bounds on the possible values of $q$.
I know that the matrix should be positive semidefinite but how to proceed to get the upper and lower bounds?
Thanks!
 A: Consider $n$ unit-variance random variables $X_1, X_2, \ldots X_n$ with the property that $\operatorname{cov}(X_i,X_j) = q$ for all $i \neq j$. Then,
the covariance matrix of these random variables is the same as the
correlation matrix.  Now
$$\begin{align*}
\operatorname{var}(X_1+X_2+\cdots+X_n)
&= \sum_{i=1}^n \operatorname{var}(X_i) 
+ 2\sum_{i=1}^n\sum_{j=i+1}^n\operatorname{cov}(X_i,X_j)\tag{1}\\
&= n + n(n-1)q\\
&\geq 0
\end{align*}$$
and so it must be that
$$q \geq -\frac{1}{n-1}$$
as Michael Hardy noted in a succinct comment on the question. The upper bound
is, of course, $q \leq 1$. Both bounds are achievable. Obviously, if
all the $X_i$ are the same random variable $X$, then $q = 1$.
For the lower bound, suppose that the $X_i$ are independent
unit-variance random variables so that they enjoy the desired constant
correlation with $q=0$. For each $i$, set $Y_i = X_i-\bar{X}$ where
$$\bar{X} = \frac{1}{n}\sum_{i=1}^n X_i.$$
Then, 
$$\operatorname{var}(Y_i) = \left(\frac{n-1}{n}\right)^2
+ (n-1)\left(\frac{1}{n}\right)^2 = \frac{n-1}{n}$$
while for $i \neq j$,
$$\begin{align}
\operatorname{cov}(Y_i,Y_j) &= \operatorname{cov}(X_i - \bar{X}, Y_j- \bar{X})\\
&= \operatorname{cov}(X_i,X_j) - \operatorname{cov}(X_i,\bar{X})
- \operatorname{cov}(X_j,\bar{X})+ \operatorname{var}(\bar{X})\\
&= 0 - \frac{1}{n} - \frac{1}{n} + \frac{1}{n}\\
&= -\frac{1}{n}
\end{align}$$
showing that all the correlation coefficients do indeed have
the minimum value
$$ \frac{-1/n}{\sqrt{(n-1)/n}\sqrt{(n-1)/n}} = -\frac{1}{n-1}.$$

Returning to $(1)$, note that if the correlation coefficients are
not required to all have the same value, then from $(1)$, we get
that the sum of the $n(n-1)$ correlations must be at least $-n$.
Thus, the average of the $n(n-1)$ correlations is at least
$-1/(n-1)$ and since at least one correlation must be as large
as the average, we can assert that

In any collection of $n$ random variables $X_1, X_2, \ldots, X_n$
  with finite variance, there must be at least
  one pair of random variables $(X_i,X_j)$ (with $i\neq j$) for
  which
  $$\operatorname{cov}(X_i,X_j) \geq -\frac{1}{n-1}$$

A: Since it's a correlation matrix, the diagonal entries are equal to 1 and the off-diagonal entries are in $[-1,1]$.  Now write the matrix as $aP+bQ$ where $P$ is the $n\times n$ matrix in which every entry is $1/n$, so it's the matrix of the orthogonal projection onto the line where all components of the vector are equal, and $Q = I - P$.  Then you can exploit the fact that $P$ and $Q$ are complementary orthogonal projections onto spaces of dimensions $1$ and $n-1$.  From that it follows that the matrix $aP+bQ$ can be diagonalized as
$$
\begin{bmatrix}
a \\ & b \\ & & b \\ & & & b \\ & & & & \ddots
\end{bmatrix}
$$
This should be a covariance matrix.  To see that, recall that (1) a correlation matrix is a covariance matrix in which the diagonal entries are all 1, and (2) if $A$ is the matrix of covariances of a random vector $X$, the $MAM^\top$ is the matrix of covariances of $MX$ ($M$ need not generally be a square matrix, but in this case it is).
Since the diagonal matrix above is a covariance matrix, $a$ and $b$ cannot be negative.  So what must $q$ be in order that $a$ and $b$ be nonnegative?
A: A general scheme for the answer is immediately obvious by generalization of the following example. Assume the correlation-matrix $R$ of size nxn where in the example n=5 and $R=L \cdot L^T$ . Then define L with a unknown value $a$
$$ L=\begin{bmatrix} 
   a&a&a&a&.&.&.&.&.&.    \\
  -a&.&.&.&a&a&a&.&.&.     \\
   .&-a&.&.&-a&.&.&a&a&.   \\ 
   .&.&-a&.&.&-a&.&-a&.&a   \\ 
   .&.&.&-a&.&.&-a&.&-a&-a   \\ 
\end{bmatrix}
$$
Then all offdiagonal entries in $R=L \cdot L^T$ are $r_{k,j}=-a^2$ and the diagonal entries are $r_{k,k}=4 a^2$. To have $r_{k,k}=1$ we must have $a=\sqrt{1 \over 4} $ and thus $q = r_{k,j}=-{1 \over 4}$.
It is immediately obvious how this is generalized, so for some $n$ we have $q=-{1 \over n-1}$     
Unfortunately, this is only an illustrative example so far. It would be nice to show, that this defines indeed also the highest possible value for $-q$, but I do not see it at the moment how this could be done in a similarly obvious manner ...
