Difference of two postive semi definite (psd) matrices

Question

Given two psd matrices $A$ and $B$, under what condition $A-B$ is also a psd matrix ? Both A and B are full-rank and of the form $T^HT$.

Answer 1

Ok so one quick condition i could think of is $\lambda_{\text{min}}(\pmb{A}) \geq \lambda_{\text{max}}(\pmb{B})$ because for any vector $\pmb{z}$, the quantity $\pmb{z}^T(\pmb{A} - \pmb{B})\pmb{z}$ should remain positive for any $\pmb{z}$. Keep in mind that $\pmb{z}^T\pmb{A}\pmb{z} \geq 0$ and $\pmb{z}^T\pmb{B}\pmb{z} \geq 0$ because both are psds \begin{equation} \begin{split} \pmb{z}^T(\pmb{A} - \pmb{B})\pmb{z} \geq 0 &\\ \Rightarrow \text{min}_{\pmb{z}} \pmb{z}^T(\pmb{A} - \pmb{B})\pmb{z} \geq 0 &\\ \Rightarrow \text{min}_{\pmb{z}} \pmb{z}^T\pmb{A}\pmb{z} - \text{max}_{\pmb{z}} \pmb{z}^T\pmb{B}\pmb{z}\geq 0 &\\ \Rightarrow \lambda_{\text{min}}(\pmb{A}) - \lambda_{\text{max}}(\pmb{B}) \geq 0 &\\ \Rightarrow \lambda_{\text{min}}(\pmb{A}) \geq \lambda_{\text{max}}(\pmb{B}) \end{split} \end{equation}