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There is not even an algorithm to test whether a solution exists. See e.g. Richardson's theorem. To have something sensible, you need at least to be able to restrict the domain to a bounded region. …

Newton's is not really a "minimizing method". If you're using Newton's method to find a root of $f'$, the root you find might be a local minimum, local maximum or neither. To remove the root $x=0$ fr …

If $B$ is one left inverse of $A$, then $B+X$ is a left inverse of $A$ (where $X$ is $n \times m$) iff $X A = 0$, i.e. the restriction of $X$ to $\text{Ran}(A)$ is $0$. Of course if $A$ is surjective, …

Note that since the objective is convex, there are optimal solutions that are extreme points of the feasible region, i.e. we can assume all $x_i \in \{0,1\}$. We can encode an Ising hamiltonian in …

Given a smooth function $g(x)$ on a convex compact $X \subset\mathbb R^n$, there is $K$ such that $K \|x\|^2 + g(x)$ is convex. Let $f(x,y) = g(x) + K \|x\|^2 - K \|y\|^2$, which is convex in $x$ and …

A local minimum of a quasi-convex function is a global minimum. At a critical point where the function is $C^2$, the Hessian matrix is positive semidefinite; if such a critical point is not a local …

You might try a local optimization approach. Guess values for the integer variables and solve the remaining continuous linear programming problem. Using sensitivity analysis, see if the solution can …

The gradient is $0$ iff $[x,y] = [1/D-1, 1/C-1]$, at which point the objective value is $1/(C+D-CD)$. If this is in the feasible region $[0, A] \times [0,B]$, it may be optimal. Compare to the optima …

No: maximizing the norm makes it a non-convex problem.

This can occur even in linear programming, in the presence of degeneracy. At an optimal basic solution, the slack variable for some binding constraint may be basic (but with value $0$ since it is bin …

If you make the objective to minimize the sum of the diagonal entries (i.e. the trace), your problem becomes a linear programming problem, solvable with readily available software (I think even Excel) …