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Using minimality can help. Without using the polynomial-root oracle you can find the minimal polynomial of $T$ by looking for the first linear dependence among $I, T, T^2, \dotsc, T^r$ (for the minim …

Change coordinates, or act by a linear transformation, so that $U$ is a general subspace and we are differentiating by $\partial = \frac{\partial}{\partial x_1}$. Since $U$ is general, it has a basis …

What is a linear operator---do you require $T_n(X_1 X_2) = T_n(X_1) T_n(X_2)$? If not, $T_n(X) = c^n X$ for a fixed scalar $c$ works. Even if yes, I think that $T_n(X) = X$ if $n$ odd, $0$ if $n$ even …

Say $M$ is $n \times n$. For $1 \leq k \leq n$, let $P_k(M)$ be the sum of all of the permanents of $k \times k$ minors of $M$; and let $P_0(M) = 1$. For example $P_n(M)$ is the permanent of $M$, whic …

In effect you are asking how to write monomials as sums of powers (with scalar coefficients; linear combinations of powers). The expression given by @abx is very elegant, but uses $2^k-1$ terms for $e …

Let's write $T \in \mathcal{A}^* \otimes \mathcal{B}^* \otimes \mathcal{C}$ for the tensor, and $L_T : \mathcal{A} \otimes \mathcal{B} \to \mathcal{C}$ for the linear map given by $A \otimes B \mapsto …

The variety $N$ of nilpotent matrices has codimension $n$ (e.g., this MSE answer). In projective space, generally, $A,B_1C_1,B_2C_2,\dotsc,B_nC_n$ span an $n$-plane. Because of the dimension, this pla …

The condition clearly implies $|A|=|B|$. (Fix a line $\ell$ and sum over all univariate polynomials on $\ell$.) Also, if $A$ and $B$ satisfy the condition and $p=p(x,y)$ is any bivariate polynomial of …

There is some recent work on tensors (and also the special cases of symmetric and alternating tensors) that admit orthogonal or unitary decompositions (resp., symmetric or alternating decompositions). …

Your condition (1) means: if $\hat{u}$, $\hat{v}$, and $\hat{w}$ are linearly dependent, then so are $\widehat{\varphi(u)}$, $\widehat{\varphi(v)}$, and $\widehat{\varphi(w)}$. So $\varphi$ preserves …

You know $\begin{pmatrix} A & 0 \\ 0 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} A^{-1} & 0 \\ 0 & 1 \end{pmatrix}$, and from there you can make two successive rank-$1$ modifications, first adding $b$ al …

A recent introduction is Carlini, et al, Four lectures on secant varieties. Adam mentioned Landsberg, Tensors: Geometry and Applications. In brief: 1(a). If $T$ is a symmetric tensor of tensor rank …

Consider the symmetric group action permuting the variables. The Vandermonde determinant $V$ is antisymmetric, meaning it spans an alternating representation—it's invariant under permutations, up to m …