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This problem is addressed in "On Fermat's problem in matrix rings and groups," by Z. Patay and A. Szakács, Publ. Math. Debrecen 61/3-4 (2002), 487–494, which summarizes previous work on the topic and gives some new results. It seems that the problem is not completely solved. When $k=2$, Khazonov showed that there are solutions in $SL_2(\mathbb ...


3

$\newcommand{\ep}{\varepsilon} \newcommand{\Ga}{\Gamma} \newcommand{\de}{\delta}$ For $\ep\in(0,1)$, let \begin{equation*} P_{m,n}:=P\Big(\bigcap_{1\le i<j\le m}\{|v_i\cdot v_j|\le\ep\}\Big) =1-Q_{m,n}, \end{equation*} where \begin{equation*} Q_{m,n}:=P\Big(\bigcup_{1\le i<j\le m}\{|v_i\cdot v_j|>\ep\}\Big). \end{equation*} By Bonferroni ...


2

I am not sure what's the exact meaning of "analytic" in "analytic methods". If you expand $|A-\lambda I|$ in the last row or column twice, you obtain a "three term recurrency" for the characteristic polynomials. Polynomials satisfying this type of recurrencies have been studied VERY much, and they have many remarkable properties. They are called orthogonal ...


2

$\newcommand{\la}{\lambda} \newcommand{\R}{\mathbb{R}}$Such a smooth field $(\la(t)v(t))$ does not exist in general. Indeed, let $$C=\begin{bmatrix}f&fg\\fg&fg^2\end{bmatrix},$$ where $f$ and $g$ are the (nonnegative) functions defined in this answer. The eigenvalues of $C$ are $\la:=f+fg^2$ and $0$. The eigenvectors of $C$ belonging to the ...


1

Here is a down-to-earth proof. Remark that for complex matrices, proving $A=B$ is equivalent to proving $\langle x,Ax\rangle=\langle x,Bx\rangle$ ; the superiority of the complex numbers over the real ones ! We must prove that $$\langle x,Ax\rangle=(n+1)\int_S|\langle h,x\rangle|^2\langle h,Ah\rangle d\lambda(h)-({\rm Tr}\,A)|x|^2.$$ By rotational ...


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