I have the following minimization problem, where I want to find $ X $ $$ \min_{X \succeq 0} \mathrm{tr}(A X) ~~~ \mathrm{s.t.} ~ X \circ I = I $$ Assume that $ A $ is Hermitian positive definite, $ I $ is the identity matrix and $ \circ $ is the Hadamard product.
Solution: The Lagrangian is given by $ L = \mathrm{tr}(A X) - \mathrm{tr}(U \cdot (X \circ I - I)) - \mathrm{tr}(V X) $, where $ U,V \succeq 0 $ are the KKT multipliers. Computing the derivative w.r.t $ X $ yields:
- $ A - U \circ I - V = 0 $
In addition due to KKT complementary conditions:
$ U \cdot (X \circ I - I) = 0 $
$ V X = 0 $
Any hints how to continue? I do not see any useful relations.
The first KKT condition is called "dual feasibility". It represents the set $\{U : \min_{X \succeq 0} L(X,U) > -\infty\}$. Only if $X$ is unrestricted in sign you can simply set the derivative to $0$. Let me show it for LP duality on this problem: $$\min_x \{ c^Tx : Ax\geq b, x\geq 0\}.$$ If you use: $$L(x,y)=c^Tx + y^T(b-Ax)$$ the set $\{ y :\min_{x \geq 0} L(x,y) > -\infty\}$ is characterized by $c-A^Ty \geq 0$ (because if the derivative w.r.t. $x_i$ is negative, you can let $x_i$ go to $\infty$ and the value of $L(x,y)$ will go to $-\infty$).
On the other hand, if you use $$L(x,y,z)=c^Tx + y^T(b-Ax)-z^Tx$$ the set $\{ y :\min_{x \in \mathbb{R}^n} L(x,y) > -\infty\}$ is characterized by $c-A^Ty-z = 0$. Since $z\geq 0$, these conditions are the same.
Going back to your semidefinite problem, the first approach is difficult because elementwise reasoning no longer applies. If you consider the condition $X \succeq 0$ as a constraint (and add an extra term in the Lagrangian), then you can just set the derivative to $0$.