matrix inequality proof [completion of squares]

Question

Can someone help me to prove this?

$\begin{bmatrix} 0 & B^\top W^\top \\ WB & 0 \end{bmatrix} \leq \begin{bmatrix} B^\top Q B & 0 \\0 & W^\top Q^{-1}W \end{bmatrix}$

with $Q$ positive definite matrix of suitable dimensions. I guess it comes from the completion of squares. Thanks! How can I found an upper bound forthe following matrix, based on the former result?

$\delta \begin{bmatrix} B+ B^\top & W^\top B^\top \\ WB & 0 \end{bmatrix}$, with $\delta \in \mathbb{R} $.

Answer 1

Do a difference and use Gaussian Elimination Method for blocked Matrices.

Answer 2

The matrix $\begin{bmatrix} B^\top Q B & -B^\top W^\top \\WB & W^\top Q^{-1} W\end{bmatrix} = \begin{bmatrix} -B^\top & 0 \\ 0 & W^\top \end{bmatrix}\begin{bmatrix} Q & I \\ I & Q^{-1} \end{bmatrix}\begin{bmatrix} -B & 0 \\ 0 & W\end{bmatrix}$ is positive semidefinite if Q is positive definite (Q can not be positive semidefinite because Q is invertible).