# Negating off-diagonal blocks retains positive-semidefiniteness?

I am trying to follow some notes that state $$M= \begin{bmatrix} A&B^T\\B&C \end{bmatrix} \succeq 0 \Longleftrightarrow M'= \begin{bmatrix} A&-B^T\\-B&C \end{bmatrix} \succeq 0$$ and I want to prove this to myself (just the $$\Rightarrow$$ direction because it's trivial to go the other way once one direction is proven).

Clearly, $$\begin{bmatrix} A&B^T\\B&C \end{bmatrix} \succeq 0\Longrightarrow A,C\succeq0$$ by computing $$x^TMx$$ (for $$x=\begin{bmatrix}v\\0\end{bmatrix}$$ and $$\begin{bmatrix}0\\v\end{bmatrix}$$) which is $$\geq0\ \ \forall\ v$$ by definition.

Then for a general $$x=\begin{bmatrix}x_1\\x_2\end{bmatrix}$$,

$$\begin{bmatrix} x_1^T&x_2^T \end{bmatrix} M \begin{bmatrix} x_1\\x_2 \end{bmatrix} = \underbrace{x_1^TAx_1}_{\geq0} + \underbrace{x_2^TCx_2}_{\geq0} +2x_2^TBx_1\geq2x_2^TBx_1$$

and

$$\begin{bmatrix} x_1^T&x_2^T \end{bmatrix} M' \begin{bmatrix} x_1\\x_2 \end{bmatrix} =x_1^TAx_1+x_2^TCx_2-2x_2^TBx_1\geq-2x_2^TBx_1.$$

Then I get stuck.

Just apply your assumption on $$M$$ to the vector $$[-x_1^T,x_2^T]$$.