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Let $X \in \mathbb{R}^{m \times n}$ be the matrix with columns $x_1$, $\ldots$, $x_n$. Then your matrix $A$ can be written as $$ A = X^TX(nI - J), $$ where $J$ is the $n \times n$ matrix with every en …

No. For example, let \begin{align*} p_1 = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \ \ p_2 = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}, \ \ S_1 = \begin{bmatrix} 1 & 1 \\ 1 & 2 \end{bmatrix}, …

Let $\mathbf{v}, \mathbf{w} \in \mathbb{R}^n$ and consider the following function $f : \mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}$: $$ f(\mathbf{v},\mathbf{w}) = \|\mathbf{v}\|\|\mathbf{w …

The correct value of $N$ is $$ N = \begin{cases} 2 & \text{if $D = 2$},\\ 3 & \text{if $D \geq 3$}. \end{cases} $$ The original question already proved that $N = 2$ when $D = 2$, and Michał's answer h …

I'll prove some bounds that disprove the conjectured $N = D^2 - 2$ when $D \geq 3$. Bound 1: $N \leq D(D-1)$. I'll illustrate this bound when $D = 3$ (so $N \leq 6$). Consider the following 7 of the …

As an alternative to LSpice's great answer: Every linear map $f$ acting on $M_n(\mathbb{C})$ has a Choi matrix defined by $$ C_f := \sum_{i,j}E_{i,j}\otimes f(E_{i,j}) \in M_n(\mathbb{C}) \otimes M_n( …

Let $1 \leq k \leq n$ be fixed integers. Let $\mathcal{M}_n^{\mathrm{H}}$ be the set of $n \times n$ complex Hermitian matrices (if it makes it easier to answer this question, you may instead use the …

I'll give a partial answer to your Question #2. If you know $k$ and also how many entries are equal to $1$ in each column, you can actually compute the absolute value of the determinant exactly. By th …

No, this is not true even if the matrices are $2 \times 2$ and $\alpha = 1/2$. For a concrete counter-example, consider $$A = \begin{bmatrix}1 & -\sqrt{3} \\ -\sqrt{3} & 3\end{bmatrix}, B = \begin{bma …

No. Here is an explicit counter-example for the $i = 3$ case: $$ A_3^\prime = \begin{bmatrix} 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 \\ 0 & 0 & 1 & 1 & 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1 & 0 & 1 & 0 & 1 \\ 1 & …

This is not true. For example, consider $H_{12}$: the $12 \times 12$ Hadamard matrix. Well, $H_{12}/\sqrt{12}$ is an orthonormal matrix with the "abs property", but the matrices $U$ and $V$ cannot pos …

Let $\mathbf{1}$ be the all-ones vector, and suppose $\mathbf{1}, \mathbf{v_1}, \mathbf{v_2}, \ldots, \mathbf{v_{n-1}} \in \{-1,0,1\}^n$ are mutually orthogonal non-zero vectors. Does it follow that $ …

The sum of the entrywise product of two vectors is just their dot product. So this question is asking how to find $x$ so that $$ (q+x)(\mathbf{W} \cdot \mathbf{T}) = q\mathbf{W} \cdot\mathbf{S}. $$ Th …

Iosif Pinelis already gave a nice solution to show that the answer is "no" for sets of infinitely many vectors, and thus for $N$ very large. I'll show that the answer is also "no" even for some set of …

You noted in your "Edit 2" that these $n \times n$ matrices $A$ are exactly those that can be written in the form $$ A = \sum_j \mathbf{v_j}\mathbf{v}_{\mathbf{j}}^*, $$ where each $\mathbf{v_j}$ has …