Finding $\text{diag} \lbrace \textbf{s} \rbrace$ that makes block matrix positive definite Let
$$ X =\begin{pmatrix} \Sigma & \Sigma - \text{diag} \lbrace \textbf{s} \rbrace \\ \Sigma - \text{diag} \lbrace \textbf{s} \rbrace & \Sigma \end{pmatrix} $$
where $\textbf{s}$ is a $p$ dimensional vector, and $\Sigma$ is a positive semi-definite or positive definite matrix (symmetric) covariance matrix.
I need to find $\textbf{s}$ such that $X$ is a positive definite matrix (in either cases of $\Sigma$).
In the positive definite case of $\Sigma$, we have the Schur's complement:
$$
2\text{diag} \lbrace \textbf{s} \rbrace-\text{diag} \lbrace \textbf{s} \rbrace \Sigma^{-1} \text{diag} \lbrace \textbf{s} \rbrace
$$
Choosing $\textbf{s}$ to make the Schur's complement positive definite will then make $X$ positive definite. However, I'm not sure how I can exactly find such a $\textbf{s}$ for the above complement...
In the positive semi-definite case of $\Sigma$, we can't apply Schur's complement and I'm not actually sure if it at all would be possible to choose $\textbf{s}$ such that $X$ would be positive definite.
So I'm just wondering if anyone has any ideas about how I can address the cases above.
 A: Edit: One can obtain a simple expression (see (1)) for the associated quadratic form and a necessary condition for definite positivity. Here is how:
Let $D=diag(S)$
Let us consider quadratic form :
$$W^TXW=\begin{pmatrix}U&V\end{pmatrix}\begin{pmatrix}\Sigma&\Sigma - D\\ \Sigma - D&\Sigma \end{pmatrix}\begin{pmatrix}U\\V\end{pmatrix}$$
$$W^TXW=U^T\Sigma U+2U^T(\Sigma - D)V+V^T\Sigma V $$
$$W^TXW=\underbrace{(U+V)^T \Sigma (U+V)}_{\text{always} \ge 0} - 2 U^T D V\tag{1}$$
We want expression (1) to be $>0$ whatever $U$, $V$ non simultaneously equal to $0$.

*

*A first result is that necessarily, all (diagonal) entries of $S$ must be $>0$.

Otherwise, let $i$ be an index such that  $s_{ii} \le 0$; by defining $U$ as the vector with a $1$ in the $i$th position, and zeros elswehere, then setting $V:=-U$, the first term in (1) would be $0$ and the second term would be equal to $2s_{ii} \le 0$ contradicting the fact that $X$ is (or has to be) positive definite.


*A second, negative, result: this necessary condition is certainly not sufficient: everything can happen! Let us consider the case where $\Sigma$ has all its entries with small size, and $S$ has large entries; let us take $U=V$ equal to the vector with all its entries equal to $1$ ; the second term "dwarves" the first one in relationship (1)...


First version, based on an entrywise approach:
Consider the case of a $2 \times 2$ blocks.
Matrix
$$X=\left(\begin{array}{cc|cc}a&    b& (a - e)&     b\\
     b&     d&     b& (d - f)\\
\hline
 (a - e)&     b&     a&     b\\
     b& (d - f)&     b&     d \end{array}\right)$$
has characteristic polynomial :
$$(x-e)(x-f)(x^2+(e + f- 2a - 2d)x+ (4ad - 4b^2 - 2af - 2de + ef))$$
As $X$ is positive definite iff all its eigenvalues are $>0$, we must have :

*

*both $e>0$ and $f>0$.


*the roots of the quadratic factor in (1) are real and $>0$.
This is equivalent to the fact that the
sum $S=-(e + f- 2a - 2d)$ and the product $P=(4ad - 4b^2 - 2af - 2de + ef)$ of its roots are $>0$ (we haven't to check that the discriminant is $>0$ because a symmetrical matrix has only real roots), i.e., the conditions are
$$\begin{cases}
e+f&<&2\underbrace{(a+d)}_{\text{trace}(\Sigma)}\\4\underbrace{(ad-b^2)}_{\det(\Sigma)}&>& 2af + 2de - ef
\end{cases}$$
Remark: the eigenvectors associated with eigenvalue $e$, resp. $f$ is $\begin{pmatrix}1\\0\\-1\\0\end{pmatrix}$, resp. $\begin{pmatrix}0\\1\\0\\-1\end{pmatrix}.$
Important remark: the fact that the diagonal values of $D$ are eigenvalues of matrix $X$ is not restricted to the particular case of $2 \times 2$ blocks: it is true for any size $n \times n$ of blocks. The proof is straightforward by exhibiting  associated eigenvectors of the same structure as before: entries $1$ in $k$th position and $-1$ in $(n+k)$th position and $0$ elsewhere.
