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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?

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    $\begingroup$ That matrix is actually just positive semidefinite. The matrix $C$ has $m$ eigenvalues that are $0$ and $m$ eigenvalues of the form $\lambda+\alpha$ where $\lambda$ is an eigenvalue of $A$. $\endgroup$
    – JimmyK4542
    Commented Jan 9, 2017 at 3:16
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    $\begingroup$ ^Correction: There are $m$ eigenvalues in the form $2(\lambda+\alpha)$ (not $\lambda+\alpha$) for each eigenvalue $\lambda$ of $A$. $\endgroup$
    – JimmyK4542
    Commented Jan 9, 2017 at 17:06

2 Answers 2

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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.

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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.

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