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To the best of my knowledge, already for $n=3$ case (which probably is close, if not identical, to $n=\dim(V)-3$ because of the duality argument) the situation is far from clear for $\dim(V)$ starting …

The coefficients are real if and only if (depending on whether or not $p_1x+p_2y+p_3z$ and $q_1x+q_2y+q_3z$ are proportional or not) such a quadratic form can be, by a coordinate change, brought to $\ …

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 …

Let us rewrite it using the commutators $[P,Q]=PQ-QP$, as follows: $$ tr(B)[A^2,B]=tr(A)[A,B^2]. $$ Now, for matrices $X$ of size~$2$, we have $X^2=tr(X)X-det(X)I$ (a particular case of Cayley--Hami …

The answer to your first question is - generally, it is not possible. For instance, let $V=\mathbb{C}$ viewed as a 2-dimensional vector space over $\mathbb{R}$, with the usual product of complex numbe …

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, …

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 …

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 …

A simple argument in the finite dimensional case: the commutator subgroup of $GL_n(\mathbb{C})$ is $SL_n(\mathbb{C})$, and the size of the center of $SL_n(\mathbb{C})$ is $n$, as scalar matrices with …

Here (see the very last paragraph) it is stated that every matrix with trace zero over a PID is a commutator. However, I can't come up with a proof right away; the only proof for matrices over a field …

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 …

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 $ …

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 …

If you want to find another $\sigma'_{V,W}\colon V\otimes W\to W\otimes V$ so that $\sigma'_{U,V\otimes W}=\sigma'_{U,V}\sigma'_{U,W}$ and assume that on homogeneous elements $\sigma'_{V,W}(v\otimes w …

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= …