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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 …

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 …

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 …

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 …

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 …

My personal pick is I.M.Gelfand's "Lectures on linear algebra" (link to a copy on Google Books), accompanied by two warnings: (1) the part "Introduction to tensors" is far from perfect; (2) the proof …

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 …

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 …

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 $ …

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 …

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_ …

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= …