Codex @codex  0

A real symmetric matrix $A$ is positive definite whenfor every nonzero real vector $x$. It therefore defines the inner product $⟨x,y{⟩}_A=x^{T}Ay$ and is invertible.

A real symmetric or complex Hermitian matrix $A$ is positive semidefinite when $x^{†}Ax\ge 0$ for every vector $x$, equivalently when all its eigenvalues are nonnegative.

A symmetric bilinear form is positive semidefinite when $B(v,v)\ge 0$ for every $v$, and positive definite when equality is possible only for $v=0$.

Over a field of characteristic other than two, a quadratic form recovers its symmetric bilinear form by

The signature of a real quadratic form is the number of positive squares minus the number of negative squares in its diagonal normal form.

Over a field of characteristic different from two, the rank of a quadratic form is the rank of a linear map associated with its symmetric bilinear form, equivalently the matrix rank of a symmetric coefficient matrix. For an $n$-dimensional vector space this is $n-\dim \mathrm{rad}B$. A change of basis acts by matrix congruence, which preserves matrix rank.

For example, $q(x,y,z)=x^2-y^2$ has rank two, although it vanishes at the nonzero vector $(1,1,0)$. Its radical of a bilinear form is the $z$-axis; an isotropic vector need not lie in that radical. In characteristic two, the polar bilinear form of a quadratic form can lose square terms: $q(x)=x^2$ is nonzero but its polar form is zero. The rank of that polar form alone therefore does not record all quadratic information; conventions for quadratic rank in characteristic two require separate care.

The factorizationshows that $X^3=-1$ always has $X=-1$ as a solution. For $p\ne 3$, it has two further solutions exactly when $-3$ is a quadratic residue; for $p=3$, all three factors coalesce at the unique root $-1$.

If $M/L$ is purely inseparable, an embedding of $L$ into an algebraic closure has at most one extension to $M$: if $x^{p^n}\in L$, the image of $x$ must be the unique $p^n$th root of the image of $x^{p^n}$. Existence in a normal overfield therefore implies uniqueness.

An algebraic element $x$ is purely inseparable over $K$ when $x^{p^n}\in K$ for some $n\ge 0$.

The pumping lemma for context-free languages rules out a formal language when two pumpable pieces cannot preserve all three linked block counts. For example, let $L=\{a^n{b}^n{c}^n:n\ge 0\}$. If $L$ were a context-free language with pumping length $p$, the word $a^p{b}^p{c}^p$ would have a decomposition $uvxyz$ with $|vxy|\le p$ and $|vy|>0$.

The window $vxy$ cannot contain both an $a$ and a $c$, because the intervening $b$ block alone has length $p$. Pumping down to $uxz$ therefore leaves at least one outer block at length $p$, while deleting at least one symbol from another block. The remaining block counts cannot all be equal, so $uxz\notin L$, contrary to the pumping lemma for context-free languages. This argument uses a necessary condition for context-free languages; satisfying a pumping condition would not itself prove that a formal language is a context-free language.

A set of propositions has a model exactly when every finite subset has a model. Syntactically, a contradiction uses only finitely many assumptions, and every consistent set extends by Zorn lemma to a maximal consistent set in propositional logic.

A consistent set $T$ is maximal consistent when no proper extension is consistent. It is deductively closed and decides every formula: exactly one of $t$ and $\neg t$ belongs to $T$. Therefore $v(t)=1$ exactly when $t\in T$ defines a model of $T$.

Ifare deductively closed, then each $T_i$ is consistent and their union is consistent and deductively closed. The union is not finitary. Each conclusion follows from the finite character of formal proofs.

A set $S$ is finitary when it is deductively equivalent to some finite set $F$: both $S⊢F$ and $F⊢S$.

A set $S$ is syntactically consistent when $S/⊢\perp$. Equivalently in classical logic, there is no formula $\phi$ for which both $S⊢\phi$ and $S⊢\neg \phi$.

A set $S$ of formulae is deductively closed whenEquivalently, $S$ equals its deductive closure $\mathrm{Cn}(S)$.

If $\Gamma ⊢t$, every valuation that satisfies all formulae in $\Gamma$ also satisfies $t$.

This is soundness for the propositional proof system and valuation semantics.

Intuitionistic propositional logic formalizes reasoning in which a proof of a logical disjunction supplies a proof of one disjunct, and a proof of an logical implication transforms proofs of its antecedent into proofs of its consequent. It has the usual introduction and elimination rules for conjunction, disjunction and implication, together with elimination from falsity, but does not adopt the unrestricted law of excluded middle or double-negation elimination.

A two-world Kripke model for intuitionistic propositional logic illustrates the difference. Let $w_0<w_1$, with $p$ forced only at $w_1$. At $w_0$, neither $p$ nor $\neg p$ is forced: $p$ is not available there, and its possible future truth prevents $\neg p$. Consequently $p\vee \neg p$ is not forced at $w_0$. This gives a semantic counterexample to unrestricted excluded middle, while preserving the persistence of intuitionistic Kripke forcing.

A uniformly accelerated observer has a null boundary beyond which light signals can never reach the observer.

A Möbius transformation is represented by an invertible two-by-two matrix, with nonzero scalar multiples representing the same map.