Finding valid covariance matrices Question 1:
Let $A$ be a $m\times m$ matrix, $m\geq2$, such that the diagonal is only ones and off diagonals are only one fixed constant $a$ in $\mathbb{R}$.
For each $m$, for what $a$ is $A$ a covariance matrix?
Question 2:
Suppose a random vector $\vec{X}$ has a multivariate normal distribution with mean $\vec{0}$ (the zero vector) and a covariance matrix such as $A$ above. Then $\vec{X}$ is echangeable.
Show that $\vec{X}$ can be constructed such that $\sum_{i}^{m}X_{i}=0$. What is $a$ then?
 A: Given
\begin{align}
A=
\begin{pmatrix}
1 & a & \ldots & a\\
a & 1 & \ldots & a\\
\vdots& \vdots & \ddots &\vdots\\
a & a & \ldots  & 1
\end{pmatrix}
\end{align}
then it follows
\begin{align}
x^TAx  = \sum^n_{j=1}x_j^2+ 2a\sum_{\substack{i< j}} x_ix_j = \sum^n_{j=1}(1-a)x_j^2 + a\sum_{i, j} x_ix_j \geq 0
\end{align}
for all $x\in \mathbb{R}^n$ if $A$ is a covariance matrix. Next, observe we could further rewrite the expression to get
\begin{align}
\sum^n_{j=1}(1-a)x_j^2 + a\sum_{i, j} x_ix_j
=&\ (1-a)\sum^n_{j=1}x_j^2+a\left[\sum^n_{j=1}x_j\right]^2 \geq0.
\end{align}
If $1-a< 0$, then we immediately see that 
\begin{align}
\frac{a}{a-1}\left[\sum^n_{j=1}x_j\right]^2 \geq \sum^n_{j=1}x_j^2
\end{align}
which is false if $\sum^n_{j=1} x_j = 0$ and $x_j \neq 0$ for some $j$. Thus, we have that $1-a \geq 0$. 
Next, consider the case $a< -\frac{1}{n-1}$, then we immediately see
\begin{align}
\left(1-\frac{1}{a} \right)\sum^n_{j=1} x_j^2=\frac{a-1}{a} \sum^n_{j=1} x_j^2 \geq \left[\sum^n_{j=1}x_j\right]^2 \ \ (\ast)
\end{align}
but we have
\begin{align}
1-\frac{1}{a}< n
\end{align}
which means
\begin{align}
n\sum^n_{j=1} x_j^2> \left[\sum^n_{j=1}x_j\right]^2. \ \ (\ast \ast)
\end{align}
However, this is false if we consider 
\begin{align}
x_j = \frac{1}{\sqrt{n}}
\end{align} 
since the right-hand side of $(\ast\ast)$ has to be $n$. 
Lastly, for the range $ -\frac{1}{n-1}\leq a \leq 1$, we see that $(a-1)/a\geq n$. Then by Cauchy-Schwarz, it follows $(\ast)$ will always hold. 
Hence when $ -\frac{1}{n-1}\leq a \leq 1$, we have that $A$ is a positive semi-definite matrix, i.e. a covariance matrix. 
A: Your second question is answered by citing the fact that a multivariate normal distribution is determined by its expected value and its variance (or its matrix of covariances, it you like to call it that).
That every non-negative-definite square matrix is a matrix of covariances can be proved by using the "spectral theorem", which says that every real symmetric matrix can be diagonalized by an orthogonal matrix.  Take the square root of each of the diagonal entries (which must be non-negative, since otherwise the matrix is not non-negative-definite), and you get a non-negative-definite symmetric square root $A^{1/2}$ of your non-negative-definite symmetric matrix $A$ with real entries. Let $Z$ be an $n\times 1$ matrix of real random variables with expected value $0$ and each entry having variance $1$ and the covariances being $0$.  Then $A^{1/2}Z$ has covariance matrix $A$.
So the question reduces to this: For which values of $a$ is the $m\times m$ matrix non-negative definite?
$\ldots\,$and now I'm gonna crash.  Tomorrow I'll post something to the effect that the off-diagonals must be $\ge -1/(m-1)$, and that that is enough.
A: It has to satisfy the Cauchy-Schwarts inequality.
That is $\sigma_X\sigma_Y \ge |\operatorname{cov}(X,Y)|$
If the main daigonal is all 1's then $\sigma_X^2 = 1$ for all of your variables.
What does that mean for $a.$?
Actually, merely satisfying Cauchy-Shwartz is not enough (in general)
Really, the matrix must be positive semi-definite.
