# How to find $\min\limits_{{X:\,det(I+X)\geq a}} tr(X\Sigma)$ given diagonal $\Sigma$ and real $a$?

Given $$\Sigma=$$diag$$\{\sigma_1,\sigma_2,\cdots,\sigma_n\}$$ and $$a\in\mathbb{R}$$. Find $$\begin{array}{ll}\min\limits_{{X:\,det(I+X)\geq a}} tr(X\Sigma).\end{array}$$

My attempt: If $$X=$$diag$$\{x_1,x_2,\cdots,x_n\},$$ then using AM-GM inequality $$\begin{equation}tr(X\Sigma)=\sum_{i=1}^nx_i\sigma_i=\sum_{i=1}^n(1+x_i)\sigma_i-\sum_{i=1}^n\sigma_i\geq\frac{1}{n}\sqrt[n]{\prod_{i=1}^n(1+x_i)\sigma_i}-\sum_{i=1}^n\sigma_i\geq\frac{a}{n}\sqrt[n]{\prod_{i=1}^n\sigma_i}-\sum_{i=1}^n\sigma_i.\end{equation}$$

Your computation contains $$2$$ mistakes; moreover, your problem is ill posed.

i) We assume that, for every $$i$$, $$\sigma_i\geq 0$$ and $$\sigma_1>0$$.

Note that $$tr(X\Sigma)=\sum_ix_{i,i}\sigma_i$$ where $$X=[x_{i,j}]$$. With the sole condition on $$X$$, $$\det(I+X)\geq a$$, the lower bound of $$tr(X\Sigma)$$ is $$-\infty$$; indeed, choose $$X_t=diag(-tI_{2k},0_{n-2k})$$ where $$t\rightarrow +\infty$$. Thus, we must add a condition on $$X$$; for example, its eigenvalues are real and $$\geq -1$$.

ii) We assume also that $$a\geq 0$$.

Let $$f(X)=tr(X\Sigma)$$ and $$\phi(X)=\log(\det(I+X))-\log(a)$$. We calculate the critical poins of the problem: "find $$\min(f(X))$$ under the condition $$\phi(X)=0$$".

$$Df_X-\lambda D\phi_X:H\rightarrow tr(H\Sigma)-\lambda tr(H(I+X)^{-1})$$. The critical points $$X$$ satisfy $$\Sigma=\lambda(I+X)^{-1}$$, that is, $$I+X=\lambda\Sigma^{-1}$$.

Thus $$\det(I+X)=\lambda^n/\Pi_i \sigma_i=a$$ and $$\lambda=(\Pi_i\sigma_i)^{1/n}a^{1/n}$$ (there is also the opposite solution when $$n$$ is even).

Finally, the candidate $$X$$ (to be a minimum) is a diagonal matrix: $$X=(\Pi_i\sigma_i)^{1/n}a^{1/n}\Sigma^{-1}-I$$. The associated value of $$f$$ is $$tr(X\Sigma)=(\Pi_i\sigma_i)^{1/n}a^{1/n}n-\sum_i\sigma_i$$.