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The squared cross-product norm is the Gram determinant of two vectors
Statement
For , , and this value is positive exactly when and are linearly independent.
Equivalently,
Facts & Assumptions
Given: Vectors .
The cross product has the displayed coordinate formula, and it is bilinear, alternating, and orthogonal to its factors (The cross product in , The cross product is bilinear, alternating, and orthogonal to both factors).
The two-vector Gram matrix has entries , and its determinant is positive exactly for a linearly independent pair and zero exactly for a dependent pair (The Gram matrix and Gram determinant, with empty value , A Gram determinant is nonnegative and is positive exactly when the vector list is linearly independent).
Proof
Expanding the three squared coordinates of and collecting terms gives .
The right side of step 1.1 is , which is by [L2].
The positivity and vanishing assertions follow from [L2], including the cases in which either vector is zero.
Depends on
- The cross product in $\mathbb R^3$
- The cross product is bilinear, alternating, and orthogonal to both factors
- The Gram matrix $G(v_0,\ldots,v_{r-1})=(\langle v_i,v_j\rangle)_{i,j<r}$ and Gram determinant, with empty value $1$
- A Gram determinant is nonnegative and is positive exactly when the vector list is linearly independent
Used by
Dependency tree · two levels
16 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- M. E. Taylor, Introduction to Analysis in Several Variables, formulas 3.2.18-3.2.20 (standard reference, not scraped)
- R. Sjamaar, Manifolds and Differential Forms, Theorem 8.4 (standard reference, not scraped)