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Suppose real matrices $A,B$ satisfy $AP=PB$ for an invertible complex matrix $P=X+iY$. Then $AX=XB$ and $AY=YB$. The real polynomial $\det (X+tY)$ is not identically zero because its value at $t=i$ is nonzero, so some real $t$ makes $X+tY$ invertible. This real matrix intertwines $A$ and $B$.

Let $A$ be invertible over an algebraically closed field. If $A^m$ is diagonalizable for some $m>0$, then $A$ is diagonalizable. Indeed, a nontrivial Jordan block $A{|}_E=\lambda I+N$ has $\lambda \ne 0$ andThe first nonzero nilpotent term prevents this power from being diagonalizable.

For a compactly supported nonzero function $\psi$, the translates $f_n(x)=\psi (x-n)$ converge uniformly to zero on every compact subset of $\mathbb{R}$, while $\Vert f_n{\Vert }_{\infty }=\Vert \psi {\Vert }_{\infty }$ prevents uniform convergence on all of $\mathbb{R}$.

Every consistent recursively axiomatized extension of elementary arithmetic is incomplete. If it were complete, its theorem set would be decidable; representing that decision procedure and diagonalizing against it yields a contradiction.

Every recursively enumerable set many-one reduces to the diagonal halting set. If a program halts exactly on $X$, the S-m-n theorem produces from $x$ an index $g(x)$ for a unary program that ignores its input and performs that computation on $x$. ThenConversely, a computable preimage of a recursively enumerable set is recursively enumerable.

An affine algebraic group is solvable when its derived series reaches the identity. A connected solvable affine algebraic group can be conjugated into an upper triangular matrix group by the Lie-Kolchin theorem.

A Lie algebra is solvable when its derived series eventually becomes zero.

A random field is finite-range dependent if collections indexed by sets farther apart than a fixed distance are independent. Sparse subsets of long paths then restore enough independence for path-counting arguments.

For a random field $(Y_x)$, level-set percolation studies the random vertex set $\{x:Y_x\ge a\}$ as the threshold $a$ varies.

For an affine variety over a perfect field, the tangent-space codimension at a point is the rank of the Jacobian matrix of defining equations. The smooth locus of a pure $d$-dimensional variety in ${\mathbb{A}}^N$ is where this rank is $N-d$.

If an elliptic function $g$ has degree $d$ and $r$ distinct poles in a fundamental parallelogram, then each pole order increases by one under differentiation, soBecause $1\le r\le d$, this gives $d+1\le \mathrm{deg}{g}^{\prime }\le 2d$.

Let a nonconstant rational function $f$ have degree $d$, let $r$ be its number of distinct finite poles, and let $m_{\infty }$ be its pole order at infinity, taken as zero when infinity is not a pole. Differentiation raises each finite pole order by one and changes a pole of order $m_{\infty }>0$ at infinity into one of order $m_{\infty }-1$. HenceIn particular, $d-1\le \mathrm{deg}{f}^{\prime }\le 2d$.

If the base of a covering is path-connected, a path from $x_0$ to $x_1$ gives a bijectionby sending each initial lift point to the endpoint of its unique lifted path. Lifting the reversed path gives the inverse bijection.

Let three equal masses move on a circle of radius $r$, and let the potential be $U(\alpha )+U(\beta )+U(\gamma )$ with $\alpha +\beta +\gamma =2\pi$. At equal spacing $s_0=2\pi /3$, angular displacements have a zero-frequency rigid-rotation mode $(1,1,1)$ and a two-dimensional sum-zero eigenspace withThe two vibrational modes are degenerate by the permutation symmetry.

A Be star is a rapidly rotating B-type star showing Balmer emission from a circumstellar gaseous decretion disk.

For two polarizations, $n$ atoms in area $A$, and $\omega =C|k{|}^{\alpha }$, mode counting givesAt low temperature its energy scales as $T^{1+2/\alpha }$ and its constant-volume heat capacity as $T^{2/\alpha }$.

The Debye frequency is the upper frequency cutoff fixed by equating the continuum mode count to the crystal's number of vibrational degrees of freedom.

For $k>0$, the trajectory beginning at $(x,y)=(1,y_0)$ with $y_0>0$ remains in $x>1$, $y>0$ and obeysIt enters every strip $0<y<\epsilon$: otherwise $\dot{E}=-k{y}^2\le -k{\epsilon }^2$ would eventually make the nonnegative energy negative.

If $u(x,0)=0$, $u_t(x,0)=g(x)$, and the odd function $g$ is supported on $[-a,a]$, thenThe interval $[x-t,x+t]$ then contains the entire support, whose integral vanishes by oddness.

For the parametrizationthe Gaussian curvature and area element are