Positive defnite matrices Hei guys,
I would like to check with you a prove I have
Let the matrix $X_1 = \begin{bmatrix}
    A_1 & B_1\\
    B_1 ^ T & A_1\\
  \end{bmatrix}$ be positive definite. Based on this I would like to show that the matrix $X_2 = \begin{bmatrix}
    X_1 & B_2\\
    B_2 ^ T & A_1\\
  \end{bmatrix}$, where $B_2 = \begin{bmatrix}
    0\\
    B_1\\
  \end{bmatrix}$ is also positive definite.
The prove I have looks like this:


*

*Based on the assumptions on $X_1$ and the Schur Lemma we get that $A_1, A_1 - B_1 ^ T \times A_1 ^ {-1} \times B_1$ and $A_1 - B_1 \times A_1 ^ {-1} \times B_1 ^ T$ are positive definite. 

*In order for $X_2$ to be positive definite one has to show (based on the same lemma) that $X_1 - B_2 \times A_1 ^ {-1} \times B_2 ^ T$ is positive definite.
This is equivalent with showing that the matrix:


$\begin{bmatrix}
    A_1 & B_1\\
    B_1 ^ T & A_1 - B_1 \times A_1 ^ {-1} \times B_1 ^ T\\
  \end{bmatrix}$ is positive definite.
For me this is true on if the matrix $- B_1 \times A_1 ^ {-1} \times B_1 ^ T$ is positive definite.
I would like to ask you if is possible $X_2$ to be positive definite with the $- B_1 \times A_1 ^ {-1} \times B_1 ^ T$ being positive definite?
Thanks,
Bogdan.
 A: The result seems to follow from the Sylvester criterion for positive definiteness and determinant identity for Schur complement. 
 You need $A_1-B_1 A^{-1} B^T_1$ to be positive definite (which is positive definite, because $X_1$ is, as you say), but $-B_1 A^{-1}_1B^T_1$ is not positive definite if $X_1$ is positive definite. In other words, $-B_1 A^{-1}_1B^T_1$ is never positive definite if $A_1$ is. 
A: I have to say I got lost somewhere on the way, but was thinking about the following alternative proof:


*

*Assuming that $X_1$ is positive definite we would get that $\begin{bmatrix}
    x & y
  \end{bmatrix} \times \begin{bmatrix}
    A & B\\
    B^T & A\\
  \end{bmatrix} \times \begin{bmatrix}
    x ^ T\\ y ^ T
  \end{bmatrix} \geq 0$. This is equivalent with: $x A x^T + 2x B y^T + y A y^T \geq 0, \forall x, y$ (vectors with proper sizes). 

*For $X_2$ to be positive definite (based on the same definition) would imply that: $u X_1 u^T + 2u B_2 v^T + v A v^T \geq 0, \forall u, v$. Choosing $u = (x, y), v = z$ we get that $x A x^T + 2x B y^T + y A y^T + 2y B z^T + z A z^T \geq 0$. 
I fail to see why the second inequality is true $\forall x, y, z$ assuming that the first one is true $\forall x, y$.
This makes me to think that $X_2$ in general is not positive definite, but rather for some specific B with some specific properties.
