(Remark: this is a "scholastic" question - I'm reviewing my implementation of factor analysis procedures; I'm not looking for good approximations for an actual survey/actual data or the like.)
There are different methods of estimating individual variances for the items in a covariance matrix $C$ with $m$ rows and columns; I know that method of using $D$, the reciprocal of the diagonal of the inverse of $C$. Well, it shall result in a Heywood-case/in negative definiteness of the remaining matrix if I simply remove that variance from the diagonal of the covariance matrix ($C - D$ is surely negative definite); but I can iteratively determine the greatest possible part in $D \cdot 1/r$ to be removed which still keeps the covariance $C - D/r$ positive semidefinite. This gives then a certain sum of that itemspecific variances ($s_1=sum(D/r)$).
Another method is to get the least principal axis $A_m$ , take the eigenvalue $\lambda _m $ , then norm the other axes to the same length $B_k = A_k \cdot \lambda_m / \lambda_k $ and in the diagonal of $ B \cdot B^T$ we get (equal) itemspecific variances. Note that again $ C - B \cdot B^T $ has reduced rank, and thus "all individual variance" is removed - however, the sum of all that itemspecific variances $s_2$ is usually much smaller than than $s_1$.
After that two different solutions, already leading to different amounts of overall itemspecific variance removed, I experimented with further different methods and one gives $s_3$ which is even greater than $s_1$.
And having now a handful of further methods with different values $s_j$, the question naturally occurs:
Q: is there a specific method, which allows to extract the maximally possible sum of individual variances of a covariance matrix, and if there is a special method, how is it defined?
To see, that the differences between the methods are not simply neglectable I add an example with some test-covariance matrix.
Overview, comparision of 4 methods:
Detail 2: I'm surprised that the shape of the approaching of the maximum has such a spike - I'd expect some smooth "top of a normal-curve" here: