$X:\Vert X\Vert_2<1 \iff\text{ matrix }\begin{bmatrix} I&X^*\\X&I\\\end{bmatrix} $ is positive Following question seems so simple, yet I could not come up with a solution. I started to think that there might be sth wrong with the question. Could you please take a look?
For a matrix $X:\Vert X\Vert_2<1 \iff \text{ matrix }\begin{bmatrix} I&X^*\\X&I\\\end{bmatrix} $ is positive
 A: Suppose $\|X\|_2 < 1$. Then, for any appropriately sized vectors $v,w$, we have 
$\begin{bmatrix}v^* & w^*\end{bmatrix}\begin{bmatrix}I&X^*\\X&I\end{bmatrix}\begin{bmatrix}v\\w\end{bmatrix}$ 
$= v^*Iv + v^*X^*w + w^*Xv + w^*Iw$ 
$= \|v\|_2^2+(Xv)^*w+w^*(Xv)+\|w\|_2^2$ 
$\ge \|v\|_2^2 - \|Xv\|_2 \cdot \|w\|_2 - \|w\|_2 \cdot \|Xv\|_2+\|w\|_2^2$ (by Cauchy Schwartz)
$\ge \|v\|_2^2 - \|v\|_2 \cdot \|w\|_2 - \|w\|_2 \cdot \|v\|_2+\|w\|_2^2$ (Since $\|X\|_2 < 1$)
$= \|v\|_2^2-2\|v\|_2 \cdot \|w\|_2 + \|w\|_2^2$
$= (\|v\|_2-\|w\|_2)^2$
$\ge 0$.
I'll leave it to you to verify that equality only holds if $v = 0$ and $w = 0$. This shows that $\begin{bmatrix}I&X^*\\X&I\end{bmatrix}$ is positive definite.
Now, suppose $\|X\|_2 \ge 1$. Then there exists a $v \neq 0$ such that $\|Xv\|_2 \ge \|v\|_2$. Let $w = -Xv$. 
Then, we have 
$\begin{bmatrix}v^* & w^*\end{bmatrix}\begin{bmatrix}I&X^*\\X&I\end{bmatrix}\begin{bmatrix}v\\w\end{bmatrix}$ 
$= v^*Iv + v^*X^*w + w^*Xv + w^*Iw$
$= v^*v - v^*X^*Xv - v^*X^*Xv + v^*X^*Xv$
$= v^*v - v^*X^*Xv$
$= \|v\|_2^2 - \|Xv\|_2^2$
$\le 0$.
This shows that $\begin{bmatrix}I&X^*\\X&I\end{bmatrix}$ is not positive definite.
