Given random Hermitian matrices $A_1,A_2,A_3,A_4$，they satisfy：
$$
\text{tr}(A_i)\in[0,1],\quad i=1,2,3,4.\quad \text{tr}(A_1+A_2)=\text{tr}(A_3+A_4)=1
$$
Given Hermitian variables $X_1,X_2,X_3,X_4$, consider the following positive semi-definite programming problem:
$$
\begin{aligned}
\min_{X_i}\quad&\text{tr}\sum_{i=1}^4A_iX_i \\
\mathrm{s.t.}\quad &X_1+X_3\succeq0,X_1+X_4\succeq0,X_2+X_3\succeq0,X_2+X_4\succeq0\\
&\text{tr}\sum_{i=1}^4 X_i=k
\end{aligned}
$$
How to prove that when I arbitrarily change the value of $k$ in the interval $(0,\infty)$, there always exists $A_1,A_2,A_3,A_4$, such that the sign of the optimal value of the objective function will not change.
Note: All $A_i$ and $X_i$ are square matrices.
Edit: From the first and last inequality constraints we get:
$$
\sum_i X_i\succeq 0
$$
This shows that $\sum_iX_i$ is a positive semi-definite matrix. Because we are discussing the sign of the objective function, all $X_i$ here are not zero matrices, otherwise the objective function value will be $0$. So there is $\text{tr}\sum_iX_i>0$, denoted as $k$.
We have previously shown that for any equality constraint $\text{tr}\sum_iX_i=k(k>0)$, the sign of the optimal value of the objective function does not change. And from the inequality constraints we can also deduce this equality. So the equality constraint can be omitted.
Best Answer
Here's a simple method of seeing it. Do a change of variables so that $Y_i = X_i / k$. The problem becomes $$ \begin{aligned} \min_{Y_i}\quad& k \,\text{tr}\sum_{i=1}^4A_iY_i \\ \mathrm{s.t.}\quad &Y_1+Y_3\succeq0,Y_1+Y_4\succeq0,Y_2+Y_3\succeq0,Y_2+Y_4\succeq0\\ &\text{tr}\sum_{i=1}^4 Y_i=1 \end{aligned} $$
You can prove that this new problem is equivalent to the original problem, in particular they will have the same optimal value. However, in this new form it is clear that $k$ just rescales the objective value by some fixed positive constant and can therefore not affect the sign of the result.