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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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