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Your polynomial is precisely $$ \sum_{k_1+2k_2+\cdots+nk_n=n}\binom{k_1+\cdots+k_n}{k_1,\ldots,k_n}X_1^{k_1}\cdots X_n^{k_n}. $$ The proof is straightforward by induction: you have $$ p_n(X)=\sum_{i= …

There is an instance where the two formulas are very close: if $f_1,\ldots,f_n$ are fundamental solutions of a linear ODE with indeterminate (constant) coefficients. This is explained in the elegant p …

Suppose that $A,B\in M_{p,q}(\mathbb{Z})$ are two rectangular integer matrices of the same size. Suppose that one has a conjecture stating that the rank of the matrix $A+tB$ for Zariski generic values …

In addition to the other answer, a very good reference on that matter is the book of Yuri Ivanovich Manin "Gauge Field Theory and Complex Geometry", specifically Chapter 3 "Introduction to superalgebr …

One way to look at this question is via quiver representations. Two subspaces of a vector space form a representation of the quiver $A_3$ with orientations $\bullet \rightarrow \bullet \leftarrow \bul …

Wikipedia suggests "conformal orthogonal group" for the group of all such maps; see the articles https://en.wikipedia.org/wiki/Conformal_group https://en.wikipedia.org/wiki/Orthogonal_group#Conformal_ …

There is a general algebraic result which states that the abstractly defined associative $\mathbb{R}$-algebra with generators $X_1,\ldots,X_n$, $Y_1,\ldots,Y_n$ and relations $$ X_iX_j+X_jX_i=0, \qua …

I think your example is easily generalisable for any index. For example, let $$ Q_1=P_1\oplus(1), Q_2=P_2\oplus(1), Q_3=P_3\oplus(1)=(1)\oplus P_1, Q_4=(1)\oplus P_2, Q_5=(1)\oplus P_3. $$ Then $Q_5 …

This is an elaboration on the comment of Alexandre Eremenko. Algebraic multiplicity $n$ means that we have the equality of polynomials $$ \det(t I_n -a_1A_1+\cdots+a_nA_n)=(t-\lambda)^n $$ for some $ …

Multilinear equations are hardly easier than general equations. For instance, the multilinear equations $$ \begin{cases} x_0-x_1=0,\\ x_0x_1-x_2=0,\\ x_0x_2-x_3=0,\\ \ldots\\ x_0x_{n-1}-x_n=0 \end{c …

Of course. The eigenvalues of this matrix (over $\mathbb{C}$) may only be zeros and roots of unity (whose minimal polynomial is of degree at most $n$, as they are roots of the characteristic polynomia …

If you replace determinants by traces, then this inequality is just Cauchy-Schwarz for the inner product $(X,Y)=\mathop{\mathrm{tr}}(XY^T)$ on the space of matrices. Now, we just have to recall that d …

From your more general question I infer that you want to look at the coset of your matrix in the quotient (not at evaluation at specific $x_1,\ldots,x_n\in\mathbb{F}_q$). Without loss of generality, …

FWIW, for $n=d=2$, this polynomial is irreducible, as I just checked in Magma. The naive code for this (which even the online calculator http://magma.maths.usyd.edu.au/calc/ can handle) is S<x1,x2,x3 …

There is an improvement of the answer of Joseph Van Name which I feel is much more in the spirit in the question asked: Let $(X,d)$ be a compact metric space, and assume that the mapping $f\colon X\t …