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The "good" matrices, those whose eigenvalues of largest absolute value are either a single simple real eigenvalue or a conjugate pair of simple eigenvalues, form a dense open set in the $n \times n$ r …

If there is no connection between the distributions for different $N$, it's not true. Suppose $w_{ij}$ and $v_i$ all have Bernoulli distributions with parameter $1 - 2^{-N}$. Then with high probabili …

So $A = Y X^{-1}$, where $X$ and $Y$ are the $n \times n$ matrices with these vectors as columns, and $A$ is the matrix with entries $a_{ij}$. Almost surely, $A$ has full rank because $Y$ and $X^{-1}$ …

$\text{tr}(AB+BA) = 2 \operatorname{tr}(A^{1/2} B A^{1/2}) > 0$, so that may produce some bias toward positive eigenvalues. In particular if you generate your "random" matrices in such a way that the …

Here's one way to get lots of examples. Start with a finite interval $[a,b]$ and a function $g$ on this interval such that $\int_a^b g(x)\; dx > 0$, and for $c > 0$ and $d > 0$ consider $f$ defined o …

Yes, if that is the distribution you want for the entries. This will depend on the distribution; in most cases (other than finite discrete distributions) where the answer is finite, there will be n …

As far as we know from what you told us, $A^T A$ is an arbitrary $K \times K$ positive definite matrix. Thus there is some $c > 0$ such that $z^T A^T A z \ge c z^T z$ for all $z$, and that's really …

Certainly not in terms of the eigenvalues of $A$, because this won't be invariant under similarity transformations on $A$. One thing I can say is that for any vector $b$, $K b = \sum_{j=0}^{n-1} b_{ …

You must assume $\sigma > 0$. Then it's just the fact that the if $D$ is a diagonal $n \times n$ matrix with minimum diagonal entry $m > \|B\|$, where $B$ is an $n \times n$ matrix, $\|D^{-1} B\| \le …

Let's try an easy case: $n=2$. If the matrix has entries $a,b,c,d$ we need $ab - cd$, $ac-bd$ and $ad-bc$ to all be nonzero. In particular if $b,c,d$ are all nonzero there are at most $3$ forbidden …

The trace is the sum of the diagonal elements, so (when $z$ is real) it's always true for at least one $i$, and the only way it can be true for all of them is that all diagonal elements are equal. F …

As soon as you have two stacks, each summing to the same all-1 vector, your matrix is going to be singular. So, unless I misunderstand you, the probability is $0$.