Let's say $\alpha > 0$. I have a symmetric $n \times n$ matrix $A(\alpha)$ that is positive semidefinite, i.e.
$\forall x \in \mathbb R^{n},\; x^{t}A(\alpha)x \geq 0,$
and in addition I have that following dynamic that if I let $\alpha$ get larger, we have that the diagonal of $A(\alpha)$ tends to be a lot larger than its off-diagonal elements.
Question: In the case above, how can I justify that $A_{\text{per}}(\alpha)$ remains positive semidefinite, where $$\text{per}:\mathcal{M}(n\times n) \to \mathcal{M}(n\times n),\;\; \text{per}(A(\alpha)):=A_{\text{per}}(\alpha),$$
the function $\text{per}$ permutes some off-diagonal entries (in the same triangle and reflected across the diagonal to retain symmetry) and switches some off-diagonal signs (i.e. from $+$ to $-$ or from $-$ to $+$)?