Finding $\text{diag} \lbrace \textbf{s} \rbrace$ that makes block matrix positive definite

Question

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.

Answer 1

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 $0$.

1. 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.

2. 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)...

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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.