Given a symmetric and positive-definite matrix $ C \in R^{M \times M} $ and another matrix $ B \in R^{N \times M} $: How can I select a matrix $ A \in R^{N \times N} $ such that $$ H = \begin{bmatrix} A & B \\ B^T & C \end{bmatrix}$$ becomes symmetric and positive-definite? I specifically need the solution for $N=1$.
Are there concrete constraints that $ A $ has to meet?
I tried to solve the problem (without success) using a rule to construct a positive definite matrix: $$ H = \begin{bmatrix} X & Y \\ Y^T & Z \end{bmatrix} \begin{bmatrix} X & Y \\ Y^T & Z \end{bmatrix}^T = \begin{bmatrix} X X^T + Y Y^T & X Y + Y Z^T \\ Y^T X^T + Z Y^T & Y^T Y + Z Z^T \end{bmatrix} $$ Hence, if we are able to apply the decompositions $B=X Y + Y Z^T$ and $C = Y^T Y + Z Z^T$, then we can easily construct $A=X X^T + Y Y^T$. It should be possible to express $Y$ and $Z$ with respect to $B$, $C$, and $X$ but I didn't make through, I struggled even in the simple case $N=M=1$ because of high-order polynomials... Maybe you have a completely different idea on how to approach this problem?
Thanks for your help!
Here's a short way for the $N=1$ case. Working with $H\in \mathbb R^{d\times d}:$
note that $H$ is graph isomorphic (permutation matrix similar) to $H'$
$H' = \begin{bmatrix} C & \mathbf b \\ \mathbf b^T & a \end{bmatrix}$
So $H\succ \mathbf 0$ iff $\mathbf H' \succ \mathbf 0$
now by Sylvester's Determinant Criterion $H'\succ \mathbf 0$ iff the determinant of all leading principal minors is $\gt 0$. We know this is true for the first $k\times k$ minors for $k\in\{1,2,...,k-1\}$ since $C \succ \mathbf 0$
Then by Sylvester's Criterion, we merely need to ensure
$\det\big(H'\Big) \gt 0$ and
and by block determinant forumla, we have
$\det\big(H'\Big)= \det\big(C\big)\cdot \det\big(a - \mathbf b^T C^{-1}\mathbf b\big)\gt 0$
since $\det\big(C\big)\gt 0$ this means you need to ensure that $a \gt \mathbf b^T C^{-1}\mathbf b$