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A triangle has zero Jordan content if and only if its vertices are collinear
Statement
A triangle has zero Jordan content if and only if its vertices are collinear.
Here collinear means that the displacement list is linearly dependent.
Facts & Assumptions
Given: Vertices .
Every triangle has content (A triangle has content , equal to half base times height when the chosen side is nonzero).
For a real square matrix, is the ordinary absolute value of its real determinant (For , the determinant over a commutative ring by the Leibniz formula, and for a real matrix).
A finite vector list is linearly dependent when a nonzero scalar list has zero linear combination (Linear independence: a finite list is independent when forces every , and a subset is independent when every injective finite list into is independent).
Proof
For the forward implication from collinearity to zero content, dependence in [L3] makes one of the two displacement vectors a scalar multiple of the other, including when either is zero; the two columns then have determinant zero, so [L1] and [L2] give content zero.
For the converse implication, suppose the content is zero. By [L1] and [L2], . If the list is dependent by [L3]; otherwise one coordinate of is nonzero, and the equation for shows by division in that nonzero coordinate that is a scalar multiple of . Thus [L3] gives collinearity.
Depends on
- A triangle has content $\tfrac12|\det[B-A\ C-A]|$, equal to half base times height when the chosen side is nonzero
- For $n\ge1$, the determinant over a commutative ring by the Leibniz formula, and $|\det A|$ for a real matrix
- Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i<n} \lambda_i v_i = 0_V$ forces every $\lambda_i = 0_F$, and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent
Used by
Dependency tree · two levels
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Sources
- W. F. Trench, Introduction to Real Analysis, §7.3 (standard reference, not scraped)