Let us have a PSD matrix $A\in\mathbb{R}^{n\times n}$ and let us define the square root of this matrix as
$$A^{1/2} = V\times D^{1/2} \times V^{-1},$$
where $A=V\times D\times V^{-1}$ is the eigendecomposition of $A$ and $D^{1/2}$ is element-wise square root of matrix $D$ having eigenvalues of $A$ on its diagonal.
I would be interested, if there is something to say about the relation of $\textsf{nne}(A)$ and $\textsf{nne}(A^{1/2})$ where $\textsf{nne}(\cdot)$ stands for the number of negative elements of the matrix. Furthermore, is there something to say about the relation of maximum elements of matrices $A$ and $A^{1/2}$? I known that for any PSD matrix it holds that $|a_{ij}| \leq \max\{a_{ii},a_{jj}\}$, so it is sufficient to focus on diagonal elements only.
So far, my computational results for some randomly generated matrices suggest that $\max_{ij} A \geq \max_{ij} A^{1/2}$ and $\textsf{nne}(A) \geq \textsf{nne}(A^{1/2})$, at least when all eigenvalues of $A$ are $\lambda_i\geq1$, but even with $1\geq \lambda_i\geq0$ I was not able to produce counter example so far.
So is this in general true or is there some counter example to this?
Update #1:
As pointed out by Alex Ravsky in the comments, the inequality about maximum element fails for a diagonal matrix with all eigenvalues $<1$. So now the question is, whether this fact somehow transfers to the question about $\textsf{nne}(\cdot)$ or not. If yes, does the above hold at least for matrices with $\lambda_{min}\geq1$?
Best Answer
The question is equivalent to asking if for PSD matrices $A$, we have $nne(A^2)\ge nne(A)$. The answer is "certainly not always". Indeed, suppose it is the case. Then raising $A$ to high powers should result in more and more negative entries. However, those entries are just $(A^n e_i,e_j)=\sum_k\lambda_k^n\langle e_i,v_k\rangle\langle e_j,v_k\rangle$ ($n$ is a large power of $2$). Now choose the eigenvectors $v_k$ of $A$ so that $\lambda_1$ is the largest eigenvalue and $\langle v_1,e_j\rangle>0$ for all $j$. That still allows for negative entries if the dimension is not too small. However, in this case a sufficiently high power of $A$ will have all entries positive. For a particular example in $\mathbb R^3$, take $v_1=(1,1,1)/\sqrt 3$, $v_2=(1,-1,0)/\sqrt 2$ and $v_3=(1,1,-2)/\sqrt 6$ with $\lambda_1=1.1, \lambda_2=1$ and $\lambda_3=0.01$, say.