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One reason linear algebra is so useful is that the basic notions, like rank, have so many equivalent definitions. Some are better for formulating problems, some for proving theorems, and some for doi …

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 …

Using the axiom of choice, one can show that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as additive groups. In particular, they are both vector spaces over $\mathbb{Q}$ and AC gives bases of thes …

The cone $C$ is called the cone of copositive matrices and its dual $C^*$ is called the cone of completely positive matrices. Here are some references. The paper most relevant to your question is pr …

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 …

This problem can be reformulated exactly as an SOS (sum of squares) program and then solved to any degree of accuracy efficiently as an SDP (semidefinite program). For lots of references I'd recommen …

Perhaps this should be a comment but it is too long. The classic result used for existence is (Farkas' Lemma), though this gives a non-existence condition rather than an existence condition. It says …

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 …

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.

There is such an $A$ if and only if $M\geq 5$. To see this, first note that the condition that $A$ be a convex combination of terms $yy^T$ each with trace $a$ is irrelevant. As long as $A$ is positi …

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 …

Not efficiently, at least not unless the problem has some additional structure which can be exploited. The set of mixed Nash equilibria of a two-player game can be written as the nonnegative solution …

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 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 " …

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 …