Optimizing over matrices in the Loewner ordering The so-called Loewner ordering introduces a partial ordering to the set of Hermitian matrices: $X\geq Y$ if $X - Y$ is positive semidefinite, $X>Y$ if $X-Y$ is positive definite.
Consider then the following two problems:
$$\max \left\{ X \,|\, X \text{ satisfies some conditions } \right\}\tag{1}\label{opt1}$$
$$\max \left\{ \operatorname{Tr}X \,|\, X \text{ satisfies some conditions } \right\}\tag{2}\label{opt2}$$
where $X$ is Hermitian, and the maximum in \eqref{opt1} is with respect to the Loewner ordering.
Are the problems equivalent?
 A: Firstly, notice that there might be matrices $X$ and $X'$ which are neither $X\geq X'$ nor $X' \geq X$ as the Loewner order is a partial order.
For problems of the type 
$$\max \{X: X\in \mathcal{Z}\},\tag{1}\label{eq:1}$$ 
where $\mathcal{Z}$ is a collection of symmetric matrices (also positive definite?), 
you should first check whether every chain in $\mathcal{Z}$ has an upper bound, that is, whether for every net $\{X_\alpha\}_{\alpha \in A}\subseteq \mathcal{Z}$ which is a chain (i.e., for $\alpha_1, \alpha_2\in A$, with $\alpha_1\leq \alpha_2$ it is $X_{\alpha_1}\leq X_{\alpha_2}$), there is an $\alpha^*\in A$ so that $X_\alpha \leq X_{\alpha^*}$ for all $\alpha \in A$. If so, by Zorn's lemma, there will be a maximal element in $\mathcal{Z}$, that is, an element $X^\star\in\mathcal{Z}$ so that if $X\in\mathcal{Z}$ and $X\geq X^\star$, then $X=X^\star$.
This said, problem \eqref{eq:1} should be interpreted as the problem of finding such a maximal element and then the problem is well posed.
Consider now the problem
$$\max \{\mathrm{trace}\, X: X\in\mathcal{Z}\}\tag{2}\label{eq:2}$$
and assume that $\mathcal{Z}$ is such that the problem has a solution, say $X^{\star\star}$. This means that for all $X\in\mathcal{Z}$, $\mathrm{trace}\, X \leq \mathrm{trace}\, X^{\star\star}$. Take $Y\in\mathcal{Z}$ with $Y\geq X^{\star\star}$. Then $\mathrm{trace}Y\leq \mathrm{trace} X^{\star\star}$. We have that $Y-X^{\star\star}$ is positive definite but it has a nonnegative trace, so it has zero trace, therefore, $Y=X^{\star\star}$. Therefore, the solutions of \eqref{eq:2} are also solutions of \eqref{eq:1}.
Conversely, let $X^\star$ be a solution of \eqref{eq:1}: for all $X\in\mathcal{Z}$ with $X\geq X^\star$ it is $X=X^\star$. Then, I'm not sure whether $X^\star$ is also a solution of \eqref{eq:2}. I would say that problem \eqref{eq:2} recovers some - but perhaps not all - solutions of problem \eqref{eq:1}.
As a side note, such problems may be solved with manifold optimization methods where $\mathcal{Z}$ is the manifold of symmetric (positive definite) matrices. You may have a look at software such as manopt.
A: The two problems are, in general, not equivalent.
For example, consider $2\times 2$ matrices. Assume "some conditions" means that the matrix is of the form $\begin{pmatrix}a&0\\0&-2a\end{pmatrix}$. Let $Z$ denote the set of all such matrices.
Now, for $M,N\in Z$ with $M\neq N$, we have that $M-N$ is of the form $\begin{pmatrix}a&0\\0&-2a\end{pmatrix}$ with $a\neq 0$, and thus $M-N$ is not positive semidefinit, and thus $M\nleq N$ and $M\ngeq N$. So every $M$ is maximal in the set $Z$.
Thus, every $M$ is a solution to (1).
But it is easy to see that (2) is maximized by $M=0$.
