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

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 …

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. …

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) …

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 …

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, …

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 …

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 …

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 …

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 …