$\mathbb{Q}\in\mathbb{S}^{Nm\times Nm}$ ($\mathbb{S}$ denotes the symmtric matrix), $Q\in\mathbb{S}^{m\times m}$, $Q\geq 0$. Can we find some large $M$ such that for $N>M$ we always have $\mathbb{Q}\geq 0$? If no, what about $Q> 0$? \begin{equation*} \begin{aligned} \mathbb{Q}\triangleq&\begin{smallmatrix}1\\\vdots\\\vdots\\N\end{smallmatrix} \left(\begin{smallmatrix} Q & 0 & \cdots & 0 \\ 0 & Q & \cdots & 0 \\ \vdots & \vdots & \ddots &\vdots\\ 0 & 0 & \cdots & Q \\ \end{smallmatrix}\right) - \frac{1}{N}\left(\begin{smallmatrix} Q & Q & \cdots & Q \\ Q & Q & \cdots & Q \\ \vdots & \vdots & \ddots &\vdots\\ Q & Q & \cdots & Q \\ \end{smallmatrix}\right) \end{aligned} \end{equation*} It is related to my research. If this result holds true, then it will help me save a lot of works. Otherwise, I should prove my result via another tidious way. Thus, I really appreciate your concerning. Thank you very much!
If $N = 2^n$, I have proved that $\mathbb{Q}\geq 0$ if $Q\geq 0$ by using the following result to computate its eigenvalue. \begin{equation} \det\left(\begin{smallmatrix}A&B\\B&A\end{smallmatrix}\right) = \det(A-B)\det(A+B) \end{equation}
However, I cannot get a further result.
Edited
Consider a arbitrary vector $x = (x^T_1,\cdots,x_N^T)^T$. Then \begin{equation*} \begin{aligned} x^T\mathbb{Q}x=&x^T\begin{smallmatrix}1\\\vdots\\\vdots\\N\end{smallmatrix} \left(\begin{smallmatrix} Q & 0 & \cdots & 0 \\ 0 & Q & \cdots & 0 \\ \vdots & \vdots & \ddots &\vdots\\ 0 & 0 & \cdots & Q \\ \end{smallmatrix}\right)x - \frac{1}{N}x^T\left(\begin{smallmatrix} Q & Q & \cdots & Q \\ Q & Q & \cdots & Q \\ \vdots & \vdots & \ddots &\vdots\\ Q & Q & \cdots & Q \\ \end{smallmatrix}\right)x\\ =&(x^T_1Qx_1+\cdots+x_N^TQx_N) - \frac{1}{N}(x_1+\cdots+x_N)^TQ(x_1+\cdots+x_N)\\ \color{red}\geq&(x^T_1Qx_1+\cdots+x_N^TQx_N) - \frac{1}{N}\left(N(x^T_1Qx_1+\cdots+x_N^TQx_N)\right)\geq 0 \end{aligned} \end{equation*} The red $\color{red}\geq$ holds if $Q\geq 0$.
I'm having trouble with the notation. By $Q>0$, do you mean all entries are positive? If so, how do you expect to find a number, for which $0-Q_{ij}\geq 0$? Also, what is the vector written to the left of the diagonal matrix? Also, the last identity with the determinant is false. The correct result would be $det(A-BA^{-1}B)\cdot det(A)$