Suppose A is a covariance matrix whose diagonal elements are same, i.e. $A\_{1,1}=A\_{2,2}=\cdots=A\_{N,N}$, can we conclude that A is full rank?
Suppose the absolute values of the off-diagonal elements in A are all smaller than the diagonal elements in A.
Thanks!



Rank of covariance matrix whose diagonal elements are same
No. Let $a,b,c,d$ be complex numbers of modulus $1$, chosen so that $a\neq c$, $b\neq d$, and $ab^\*\neq cd^\*$. Form the column vectors $v=(1,a,b)^T$ and $w=(1,c,d)^T$. Then the matrix
$$
vv^\* +ww^\*=\begin{pmatrix} 2 & a^\*+c^\* & b^\*+d^\* \\ a+c & 2 & ab^\*+cd^\* \\ b+d & a^\*b+c^\*d & 2 \end{pmatrix}
$$
has rank two, but all the off-diagonal entries have modulus strictly less than $2$. 


*EDIT*: A simple example would be
$$
\begin{pmatrix} 1 & \frac35 & \frac45 \\ \frac35 & 1 & 0 \\ \frac45 & 0 & 1\end{pmatrix}
$$
where one can replace $(\frac35, \frac45)$ with any pair $(a,b)$ such that $|a|^2+|b|^2=1$.


