Kernel bookkeeping: why $\lambda_{\min}^+$ is subtle
A short conceptual note: “smallest positive” depends on the kernel, which can change as $T$ changes.
Two PSD operators, different kernels
Even if $A\preceq C$, it does not automatically follow that $\lambda_{\min}^+(C)\ge \lambda_{\min}^+(A)$, because $\lambda_{\min}^+$ is defined on $(\ker \cdot)^\perp$ and the kernel may shrink or grow.
Safe comparison condition
If $A\preceq C$ and $\ker(C)\subseteq \ker(A)$, then $(\ker A)^\perp \subseteq (\ker C)^\perp$ and $$\lambda_{\min}^+(C)\ge \lambda_{\min}^+(A).$$
In the triangle-selection setting
The upper term $L_1^{\uparrow}(T)$ is monotone in Loewner order as $T$ grows, but its kernel can shrink (new constraints kill harmonic directions). Thus $\lambda_{\min}^+(L_1^{\uparrow}(T))$ often increases — but the mechanism matters.
Moral: rank-one PSD updates give monotonicity of ordered eigenvalues, but understanding “the smallest positive” often requires tracking whether a previously-zero eigenvalue became positive (kernel rank change) versus the bottom positive eigenvalue moving upward.