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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 …

This isn't possible, even if you let $T$ be non-linear. You can see this in $\mathbb{R}^2$ as follows, but the intuition generalizes straightforwardly for larger $n$: Let $\{e_1,e_2\}$ be the standar …

Since you want an integer-valued function, you likely want the Schmidt rank (which is equal to the tensor rank mentioned by Qiaochu Yuan in the case of two Hilbert spaces) if you're dealing with vecto …

Begin by rewriting $\alpha_k(A_1,\ldots,A_k)$ as follows (just move integrals and traces around): $$ \alpha_k(A_1,\ldots,A_k) = \mathrm{tr}\Big( (A_1 \otimes \cdots \otimes A_k) \int_{S^{2n-1}}(vv^* \ …

I don't yet know of an explicit method of computing your desired infimum (although I'm fairly convinced that an explicit method exists), but here is MATLAB code that computes it efficiently via semide …

Here is another way of seeing that the answer is "yes". Recall that the orthogonal projection onto $\mathrm{Range}(A_k^T)$ is $P_k = A_k^T(A_kA_k^T)^{-1}A_k$. Since $A_k$ has full rank $n$ for each $k …

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 …

$\DeclareMathOperator\DNN{DNN}\DeclareMathOperator\CP{CP}$This question can be thought of as an offshoot of this MO question from a few months ago. Let $M_n(\mathbb{C})$ denote the set of $n \times n$ …

The best bound relating $m$, $n$, and $s$ (i.e., the best possible bound that does not take into account any structure of the $A_j$ matrices) is $$ s \leq \binom{n - m + 2}{2}. $$ To see that this bou …

Yes, this function is well-studied. It can be maximized by semidefinite programming (and is in fact one of the standard examples of a non-trivial function that can be maximized by SDP). The paper "M. …

I don't expect that you can get results that say too much beyond what is true in the general Hermitian case. You mentioned convexity of the (joint) numerical range in your question, so I'll address th …

The only result of this nature that I'm aware of (beyond the two cases involving $M$ and $N$ that you've listed) is Theorem 3.1 in "C.-K. Li and Y.-T. Poon. Convexity of the joint numerical range. SIA …

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 = …

Your equation is equivalent to $$ (B^{-1}A)X + X(B^{-1}A)^T = B^{-1}A^{-1}CB^{-T}B^{-T}, $$ where $X$ is the matricization of $u$. This is a Sylvester equation, which can be solved for $X$ in $O(d^3)$ …

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)\ …