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The condition you want is exactly that the matrix multiplication map be locally open at the pair $(B,C)$. This is the topic of the recent paper Where is matrix multiplication locally open? by Behrend …

Update: I originally claimed to prove that $A$ is strictly positive definite, but there was a bug in the strictness part. I have revised the proof to show that $A$ is positive semidefinite. For an e …

Let $N=n$, $m=1$, $P_i = 1$ for all $i$, and let the $C_i$ be the standard unit vectors. Then the left hand side of $(\star\star)$ is the matrix whose diagonal entries are the inverses of the diagona …

One way to come at this is to try to stretch your intuition even farther, toward the Johnson-Lindenstrauss lemma, which says that while we can only fit $n$ orthogonal vectors into $\mathbb{R}^n$, we c …

Strictly speaking it doesn't make much sense to talk about defining a function on $S$ when you've explicitly said you don't know any of the elements of $S$. I'll assume rather that you have a sequenc …

It is a standard result that the matrices of the form $\mu^{\otimes 2}$ for nonzero $\mu$ are the extreme rays of the positive semidefinite cone. That is to say, your condition on the second moments …

This is false. In particular, $A^0$ need not be positive semidefinite. For example, take $n=3$, $K = \{1,2,3\}$, let $v(x)$ be the column vector with a $1$ at position $x$ and $-1$ elsewhere, and le …

The key phrase to google is "sparsest vector".

The given constraints are a system of linear inequalities, so you can find a feasible solution (or prove that none exists) by feeding these constraints to a linear program (LP) solver with some arbitr …

If you are willing to replace the vectors $c$, $x$, and $y$ by the spaces $U_1 = \text{span}(c)$ and $U_2 = \text{span}(x,y)$, which in some sense doesn't change the problem, then one generalization i …

This problem and various related problems are known to be NP-hard to solve exactly, but there has been a lot of work on efficient approximations. See this wikipedia page or try googling things like " …

For the purposes of this answer I will ignore the condition of constant column sums. You ask for a matrix $A$ with $A^TA = \Sigma$ and $A\geq 0$ element wise. Such a matrix need not exist. For exam …

One option for a convex relaxation is to search for a positive semidefinite hermitian $W$ with $\mathrm{Trace}(WC_i)\geq 1$ for all $i$. These conditions are equivalent to the ones you have written w …

In some sense you can view the singular value decomposition as a sharpening of this theorem (for real and complex matrices, anyway). This, in turn, is useful all over the place.

For the reasons discussed in the comments to Suvrit's answer, I will assume your friend would like to minimize $\det(\Lambda^{-1})$. The problem in question is a convex optimization problem: \begin{ …