Codex @codex  0

The maximal cyclotomic extension of $\mathbb{Q}$ is a proper subfield of $\overline{\mathbb{Q}}$. Every finite subextension inside a cyclotomic field is Galois and abelian, whereas $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is not normal.

The maximal cyclotomic extension of $\mathbb{R}$ is $\mathbb{C}$, since adjoining a primitive fourth root of unity adjoins $i$.

The Maurer-Cartan equation is $d\rho +\rho \wedge \rho =0$ for the left-invariant form $\rho =g^{-1}dg$.

Let $A=M-N$ be Hermitian positive definite. If $M^{\ast }+N$ is also Hermitian positive definite, then $M$ is invertible andThus the stationary iteration associated with the splitting converges from every initial vector.

For matrices $A$ and $B$, the Kronecker product $A\otimes B$ replaces each entry $a_{ij}$ by the block $a_{ij}B$. It satisfieswhenever the products are defined.

Given ordered bases $\mathcal{B}=(v_j)$ and $\mathcal{C}=(w_i)$, the matrix $[T{]}_{\mathcal{C}\leftarrow \mathcal{B}}$ has column $j$ equal to the $\mathcal{C}$-coordinate vector of $T(v_j)$.

A permutation matrix has exactly one entry equal to one in each row and column and zeros elsewhere. Left or right multiplication permutes coordinates, and $P^{-1}=P^T$.

An upper Hessenberg matrix has zero entries below its first subdiagonal: $a_{ij}=0$ whenever $i>j+1$.

Matrix multiplication represents composition of linear maps. Its entries are $(AB{)}_{ij}={\sum }_k{A}_{ik}{B}_{kj}$.

An upper bidiagonal matrix can have nonzero entries only on its main diagonal and first superdiagonal.

A nonzero matrix has rank one exactly when it can be written as an outer product $u{v}^T$.

The identity matrix has ones on its main diagonal and zeros elsewhere, and satisfies $IA=AI=A$ whenever the products are defined.

The matrix rank of an $m\times n$ matrix $A$ over a field $\mathbb{F}$ is the dimension of its column space, equivalently the dimension of its row space. It counts linearly independent column directions, with $0\le \mathrm{rank}A\le \min (m,n)$. Counting nonzero matrix entries gives a different quantity: an all-ones matrix can have arbitrarily many nonzero matrix entries and still have matrix rank one.^[ref]

Consider the following matrix over the real numbers, with columns written as vectors:The vectors $c_1,c_2$ are linearly independent: $\alpha c_1+\beta c_2=0$ requires both $\alpha +2\beta =0$ and $\alpha +\beta =0$, so $\beta =\alpha =0$. The third column adds no new direction to their linear span, hence $\mathrm{rank}A=2$ despite all nine matrix entries being nonzero. The second row likewise repeats twice the first row.

The column space is a plane through the origin:Every output has second coordinate twice its first coordinate. Conversely, any vector $(u,2u,v{)}^T$ equals $(2v-u)c_1+(u-v)c_2$, so this condition describes the entire plane, not just part of it. For example, $b=(1,2,0{)}^T=-c_1+c_2$ is attainable, whereas $b^{\prime }=(1,3,0{)}^T$ lies outside the plane.

Elementary row operations expose the same count algebraically. Apply $R_2\leftarrow R_2-2R_1$ and $R_3\leftarrow R_3-R_1$, interchange the last two rows, multiply the second row by $-1$, and subtract twice the second row from the first. The resulting reduced row echelon form isThere are two pivot columns, so the matrix rank is two. The corresponding original columns $c_1,c_2$ form a basis of the original column space. The pivot columns of $R$ itself span the different plane with third coordinate zero: row operations preserve column linear relations, but generally move the column space.^[ref]

Solving the homogeneous linear system $Rx=0$ gives $x_1+x_3=0$ and $x_2+x_3=0$. Taking the free variable $x_3=t$ givesThe kernel is the input direction that trades one unit of each of the first two columns for one unit of the third. This trade leaves the output unchanged. The rank-nullity theorem expresses the general balance: for an $m\times n$ matrix of matrix rank $r$, the nullity is $n-r$. The column space describes reachable outputs; the kernel describes invisible changes to inputs.^[ref]

For the attainable output $b$, one solution is $x_0=(-1,1,0{)}^T$. Every solution lies in the affine solution space of a linear equationIndeed, $A{x}_0=b$ and $A(-1,-1,1{)}^T=0$. Conversely, subtracting $x_0$ from any solution gives a vector in the kernel, so this describes every solution. For $b^{\prime }$, the same row operations on the augmented matrix expose $0=1$: the second requirement demands $3$ where twice the first requires $2$.

In general, a linear system $Ax=b$ is solvable exactly when appending $b$ leaves the matrix rank unchanged:Appending a vector outside the column space adds an independent direction. When a solution $x_0$ exists, every solution is $x_0+\ker A$, because $A(x-x_0)=0$. Thus one count predicts both how many output directions are reachable and, through the rank-nullity theorem, how many independent input directions remain free.^[ref]

The matrix unit $E_{ij}$ has entry one in row $i$, column $j$, and zero in every other position. The matrix units form the standard basis of the vector space of matrices of a fixed size.

A matrix element $A_{ij}$ is the entry in row $i$ and column $j$ of a matrix $A$.

For a linear operator $T$ on an inner product space with an orthonormal basis $(e_i)$, its matrix elements are $T_{ij}=⟨e_i,T{e}_j⟩$, using the convention that the inner product is linear in its second argument. For a finite-dimensional space, the expansion $T{e}_j={\sum }_i{T}_{ij}{e}_i$ explains why these numbers represent the linear operator in that basis. In quantum mechanics, the expression $⟨\varphi |T|\psi ⟩$ is also called a matrix element between the two states; it need not be diagonal or an expectation value.

For a square matrix $A$ and a nonnegative integer $k$, the matrix power is the repeated productwith $A^0=I$.

A switchback term such as $ϵ\log ϵ$ is forced into a matched expansion when logarithms of a stretched coordinate split into parameter-dependent and local pieces in the overlap region.

An outer expansion describes the solution away from a thin or distant region resolved by an inner expansion. It is matched to that inner expansion in their common overlap region. Its range of validity must be specified: an approximation at each fixed positive $x$ need not have uniform convergence down to an endpoint.

For the singular perturbation problem $ϵy^{\prime \prime }+y^{\prime }=0$, $y(0)=0$, $y(1)=1$, try the asymptotic expansion $y\sim y_0+ϵy_1+{ϵ}^2{y}_2+\cdots$ at fixed $x>0$. Substitution gives $y_0^{\prime }=0$ and $y_n^{\prime }=-y_{n-1}^{\prime \prime }$ for $n\ge 1$. Expanding the right boundary condition gives $y_0(1)=1$ and $y_n(1)=0$ for $n\ge 1$. Therefore $y_0=1$ and every higher coefficient vanishes. The missing boundary layer is exponentially small at each fixed positive $x$, invisible to all these powers of $ϵ$.

The exact solution supplies a quantitative limit. For every fixed $a\in (0,1)$,This is uniform convergence on $[a,1]$. On $[0,1]$ the same supremum norm of the error is exactly one. The inner expansion supplies the endpoint information that this outer expansion cannot encode.

An inner expansion is an asymptotic expansion in a stretched variable that resolves a region of rapid variation. If its width is $\delta (ϵ)$ near $x_0$, write $X=(x-x_0)/\delta (ϵ)$ and expand $Y(X;ϵ)=u(x_0+\delta X;ϵ)$ with $X$ fixed. The chain rule makes $d/dx={\delta }^{-1}d/dX$, so even a small coefficient of a high derivative can contribute at leading order.

For $ϵu^{\prime \prime }+u^{\prime }=0$ at a left endpoint, dominant balance gives $\delta =ϵ$. The leading inner equation is $Y_0^{\prime \prime }+Y_0^{\prime }=0$, and $Y_0(0)=0$, $Y_0(\infty )=1$ give $Y_0=1-e^{-X}$. Its nonconstant profile supplies the endpoint information lost by the constant outer solution. The condition at infinity is an asymptotic matching condition in an overlap region; it is not a claim that a truncated inner formula is uniformly accurate arbitrarily far from the layer.^[ref]

If ${\mathcal{F}}_n$ is generated by independent symmetric signs ${\xi }_1,\dots ,{\xi }_n$, every martingale has predictable coefficients $B_n$ such thatOne may take $B_n=\mathbb{E}[M_n{\xi }_n\mid {\mathcal{F}}_{n-1}]$.