Search Results

Suppose $\mathbf{v},\mathbf{w} \in \mathbb{R}^n$ (and if it helps, you can assume they each have non-negative entries), and let $\mathbf{v}^2,\mathbf{w}^2$ denote the vectors whose entries are the squ …

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 $ …

This equation always has a solution: $X = O$. I'll assume throughout this answer that you're interested in a non-zero solution. The equation $AX = XB$ is equivalent to $(A \otimes I - I \otimes B^T)\ …

One potential intuition for the trace norm is as a way of turning the rank of a matrix (which is very discontinuous) into a norm (which is continuous). Specifically, the trace norm is the unique norm …

The "canonical" example of a map that is $k$-positive but not $(k+1)$-positive is the map defined by $$ \Phi_k(X) = k\cdot\mathrm{Tr}(X)I_n - X. $$ Above, $n$ denotes the size of $X$ (i.e., $X \in M_ …

This is not true. For example, if $$ A = \frac{2}{1 + \sqrt{5}}\begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix} $$ then $\|A\| = 1$. Furthermore, $A$ has exactly $2$ square roots, which are $$ B_{\pm} = …

Here is a way to see that, for all $p$ (even if $p \nmid n$), you can reconstruct the absolute value of the matrix's determinant, which was suggested by Will Sawin's comment. The condition $p \mid n$ …

I don't know the answer, but I can at least provide some context that may convince others that this question is actually something that is studied and is indeed hard. First, let's rewrite the questio …

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 …

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 …

Many introductory books on quantum information theory go over the linear algebraic tools necessary to study the topic, including the tensor product (since it indeed models quantum entanglement). Takin …

As mentioned by Robert Bryant, in the $n = 3$ case, checking that the matrix is positive semidefinite and has all entries $\geq 0$ is both necessary and sufficient. In fact, the same is true when $n = …

The tensor product has a way of making easy problems into (NP-)hard problems. Rank of a 2-tensor (matrix)? Easy. Rank of a 3-tensor? NP-hard. Spectral norm of a matrix? Easy. Spectral norm of a 3-tens …

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( …

If the matrices are real and the function you have in mind is real-valued, then you indeed get the characterization you suggested. This was first shown in "I. J. Schoenberg. Positive definite function …