Assume having a symmetric real matrix $A$ and a skew-symmetric matrix $\Delta = [0 1; -1, 0 ]$, such that the following generalized matrix inequality holds in the PSD sense:
$$\pm \frac{i}{2} \Delta\preceq A.$$
Given the fact that for $ \frac{i}{2} \Delta$ the eigenvalues are $\lambda_{1,2} = \pm 1/2$ and for $ -\frac{i}{2} \Delta$ the eigenvalues are $\lambda_{1,2} = \mp 1/2$, does this inequality imply: $$\frac{1}{2}I \preceq A?$$ If yes (or no), then can you explain what is the difference between them? Thanks.
Consider the following family of matrices $A_t:=\begin{pmatrix} 1 & t \\ t & 1 \end{pmatrix}$ for $t\in\mathbb{R}$. Obviously this is Hermitian (even symmetric Hermitian if that is what you really meant).
It is clear that the characteristic poly is $P_t(\lambda)=\begin{vmatrix} 1-\lambda & t \\ t & 1-\lambda\end{vmatrix} = (1-\lambda)^2-t^2=\big((1-t)-\lambda\big)\big((1+t)-\lambda\big)$, so the eigenvalues are $\lambda=1\pm t$.
Now consider the characteristic polys for $A_t\pm\tfrac{i}{2}\Delta$, which we denote by $Q_t$. We get $Q_t(\lambda)=\begin{vmatrix}1-\lambda & t\pm\tfrac{i}{2} \\ t\mp\tfrac{i}{2} & 1-\lambda\end{vmatrix}=(1-\lambda)^2-(t-\tfrac{i}{2})(t+\tfrac{i}{2})=\lambda^2-2\lambda+\tfrac{3}{4}-t^2$. The discriminant for this is $\delta=1+4t^2$, so the eigenvalues will be $1\pm \sqrt{t^2+\tfrac{1}{4}}$.
We want $A_t\pm\tfrac{i}{2}\Delta$ to be psd, so we need that $t^2\leq\tfrac{3}{4}$, i.e., that $t\in[-\tfrac{\sqrt{3}}{2},\tfrac{\sqrt{3}}{2}]$. Note however that if we pick $t\in(\tfrac{1}{2},\tfrac{\sqrt{3}}{2}]$, then $A_t$ will have an eigenvalue smaller than $\tfrac{1}{2}$, meaning that $A-\tfrac{1}{2}I$ is not psd.