How to prove the positive-definiteness of this matrix? I have this matrix
$$ C=
\left(
\begin{array}{c|c}
A+\alpha I_m & -A-\alpha I_m \\
\hline
-A-\alpha I_m & A+\alpha I_m
\end{array}
\right)\\
A=
\left(
\begin{array}{cccc}
(y_1,y_1) &(y_2,y_1)& \cdots&(y_m,y_1)   \\
(y_1,y_2) &(y_2,y_2)&\cdots&(y_m,y_2) \\
\vdots  & \vdots  & \ddots & \vdots  \\
(y_1,y_m) &(y_2,y_m)&\cdots&(y_m,y_m) \\
\end{array}
\right)
$$
  where $y_i\in L^2(\Omega)$ for all $i=1,\dots,m$ ,  $\alpha\geq 0$ and  $(\cdot,\cdot)$ denote the inner product in  $L^2(\Omega)$ with $\Omega\subset\mathbb{R}^2$.
I would like to prove this matrix is Positive-definite, any suggestions? 
 A: Since $A$ is a Gram matrix, $A$ is positive semidefinite, so $A+\alpha I$ is positive semidefinite for $\alpha \ge 0$.
Then, for any vector $w = \begin{bmatrix}x\\y\end{bmatrix} \in \mathbb{R}^{2m}$ (where $x,y \in \mathbb{R}^m$) we have 
$w^TCw$ $= \begin{bmatrix}x^T & y^T\end{bmatrix}\begin{bmatrix}A+\alpha I & -A-\alpha I \\ -A-\alpha I & A+\alpha I\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}$
$= x^T(A+\alpha I)x - x^T(A+\alpha I)y - y^T(A+\alpha I)x + y^T(A+\alpha I)y$
$= x^T(A+\alpha I)(x-y) - y^T(A+\alpha I)(x-y)$ $= (x-y)^T(A+\alpha I)(x-y) \ge 0$,
where we have used the fact that $A+\alpha I$ is positive semidefinite. 
Since $w^TCw \ge 0$ for all $w \in \mathbb{R}^{2m}$ and $C$ is symmetric, we have that $C$ is positive semidefinite.

Note that for any $x \in \mathbb{R}^m$, the vector $w = \begin{bmatrix}x\\x\end{bmatrix} \in \mathbb{R}^{2m}$ satisfies $w^TCw = 0$. 
Thus, $m$ eigenvalues of $C$ are zero. So, $C$ is not positive definite.
A: One approach is to note that 
$$
C= \pmatrix{1&-1\\-1&1} \otimes
(A+\alpha I)
$$
where $\otimes$ denotes the Kronecker product. From there, it suffices to note that the Kronecker product of positive semidefinite matrices is positive semidefinite.
