DefinitionDefinition: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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Finite compatible matrix system

Physical scope

Pure mathematics; no physical premise.

Definition

Let $n_{e}, n_{b}, n_{c}, n_{g}$ be nonnegative integers. For dimension zero, $R^{0}$ has its single empty vector and its unique square matrix acts as the identity on that space. Choose real matrices $M_{e} \in R^{n_{e} \times n_{e}}$ and $M_{b} \in R^{n_{b} \times n_{b}}$ that are symmetric positive definite, meaning $M_{e}^{T} = M_{e}$ , $M_{b}^{T} = M_{b}$ , $x^{T} M_{e} x > 0$ for every nonzero $x \in R^{n_{e}}$ , and $y^{T} M_{b} y > 0$ for every nonzero $y \in R^{n_{b}}$ ; these positivity conditions are vacuous on $R^{0}$ . Choose $C \in R^{n_{b} \times n_{e}}$ , a symmetric positive semidefinite $S \in R^{n_{e} \times n_{e}}$ , meaning $S^{T} = S$ and $x^{T} S x \geq 0$ for every $x \in R^{n_{e}}$ , and matrices $D \in R^{n_{c} \times n_{b}}$ and $G \in R^{n_{e} \times n_{g}}$ satisfying $D C = 0$ and $C G = 0$ . These are finite real matrix data; no mesh or continuum field is assumed.

Let $I \subseteq R$ be an open interval, let $j : I \to R^{n_{e}}$ be continuous, and let candidate states $e : I \to R^{n_{e}}$ and $b : I \to R^{n_{b}}$ be of class $C^{1}$ . The finite compatible matrix system is

$M_{e} \dot{e} (t) = C^{T} M_{b} b (t) - S e (t) - j (t), \dot{b} (t) = - C e (t) (t \in I) .$

Its quadratic energy is $W (e, b) : = \frac{1}{2} (e^{T} M_{e} e + b^{T} M_{b} b)$ . Its magnetic constraint is the vector $D b \in R^{n_{c}}$ . For the chosen $G$ , its weak-divergence coordinate is $q : = - G^{T} M_{e} e \in R^{n_{g}}$ . The term “weak-divergence” names this finite-dimensional quantity only. If the columns of $G$ form a basis of $\ker C : = {v \in R^{n_{e}} : C v = 0}$ , then $q = 0$ exactly when $M_{e} e \in im C^{T} : = {C^{T} u : u \in R^{n_{b}}}$ .

Definition references

For a subspace $W$ of a finite-dimensional inner product space, $V = W \oplus W^{⊥}$

Facts & Assumptions

Given: The finite real matrices and the interval and maps specified in the Definition; for the final characterization, assume additionally that $im G = \ker C$ .

[F1]

Every subspace $U$ of a finite-dimensional real inner-product space $V$ satisfies $V = U \oplus U^{⊥}$ (For a subspace $W$ of a finite-dimensional inner product space, $V = W \oplus W^{⊥}$ ).

Proof

1.1givenalgebra

In positive dimension, if $M_{e} x = 0$ , then $x^{T} M_{e} x = 0$ , so positive definiteness gives $x = 0$ ; the same argument applies to $M_{b}$ . In dimension zero the unique square matrix is the identity map on the singleton space and is invertible. Thus both matrices are injective and hence invertible in their finite-dimensional domains, so their inverse operations used by consumers are defined in every allowed dimension.

1.2givenF1algebra

For $u \in R^{n_{b}}$ and $v \in \ker C$ , $⟨ C^{T} u, v ⟩ = u^{T} C v = 0$ , hence $im C^{T} \subseteq (\ker C)^{⊥}$ . Also $z \in (im C^{T})^{⊥}$ exactly when $u^{T} C z = 0$ for every $u \in R^{n_{b}}$ , which is exactly $C z = 0$ ; therefore $(im C^{T})^{⊥} = \ker C$ . By [F1], decompose any $w \in (\ker C)^{⊥}$ as $w = u + v$ with $u \in im C^{T}$ and $v \in (im C^{T})^{⊥} = \ker C$ . Both $w$ and $u$ are orthogonal to $v$ , so $∥ v ∥^{2} = ⟨ w - u, v ⟩ = 0$ and $v = 0$ . Hence $(\ker C)^{⊥} = im C^{T}$ .

2.1givenstep 1.2algebra

Assume $im G = \ker C$ . If $q = 0$ , then $G^{T} M_{e} e = 0$ , so $M_{e} e$ is orthogonal to every column of $G$ and hence to $\ker C$ . By step 1.2, $M_{e} e \in im C^{T}$ .

3.1givenstep 1.2algebra∎

Conversely, if $M_{e} e \in im C^{T}$ , then step 1.2 gives $M_{e} e \in (\ker C)^{⊥}$ . Every column of $G$ lies in $\ker C$ , so $G^{T} M_{e} e = 0$ and therefore $q = 0$ . The two implications prove the stated characterization.

Depends on

For a subspace $W$ of a finite-dimensional inner product space, $V = W \oplus W^{⊥}$