Let $A$ be an $n\times n$ dimensional Covariance matrix. Then $A$ can be any symmstric, positive-definite $n\times n$ matrix with elements between $0$ and $1$.
My understanding: The number $$ k=(1,1,\cdots,1)A^{-1}(1,1,\cdots,1)^{T} $$ represents the number of independent random variables in the system given by the $n$ random variables $x_1,\cdots,x_n$ modulo the covariance relations $\text{Cov}(x_i,x_j)=a_{i,j}$ in the Covariance matrix $A$.
Question: We can find some covariance matrix $A$ such that $k>n$. Why the number of independent random variables in the system given by the $n$ random variables $x_1,\cdots,x_n$ modulo the covariance relations $\text{Cov}(x_i,x_j)=a_{i,j}$ is larger than $n$? This contradicts the probability interpretation.