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A Gram determinant is nonnegative and is positive exactly when the vector list is linearly independent

Statement

For every finite list $(v_{0}, \dots, v_{r - 1})$ in a real or complex inner product space, its Gram determinant is a nonnegative real number. It is positive if and only if the list is linearly independent, and it is zero if and only if the list is linearly dependent. The empty Gram determinant is $1$ .

Facts & Assumptions

Given: A finite list $(v_{0}, \dots, v_{r - 1})$ with Gram matrix $G$ .

[L1]

The Gram matrix has entries $G_{i j} = ⟨ v_{i}, v_{j} ⟩$ and the empty Gram determinant is $1$ (The Gram matrix $G (v_{0}, \dots, v_{r - 1}) = (⟨ v_{i}, v_{j} ⟩)_{i, j < r}$ and Gram determinant, with empty value $1$ ).

[L2]

Gram–Schmidt turns every independent finite list into an orthonormal list with the same successive spans (Gram–Schmidt turns every finite independent list into an orthonormal list with the same successive spans).

[L3]

A list is dependent exactly when some nonzero coefficient vector gives a vanishing linear combination (Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent).

[L4]

For $n \geq 1$ and same-sized $n \times n$ matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ and $\det (A^{T}) = \det (A)$ (For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ , For every square matrix over a commutative ring, $\det (A^{T}) = \det (A)$ ).

[L5]

Complex conjugation is a field automorphism, and $z \overline{z} = ∣ z ∣^{2} \geq 0$ with equality exactly when $z = 0$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[L6]

A square operator over a field is invertible exactly when its determinant is nonzero (A finite-dimensional linear operator over a field is invertible if and only if its determinant is nonzero).

[L7]

For $n \geq 1$ , the determinant of an $n \times n$ triangular matrix over a commutative ring is the product of its diagonal entries (The determinant of a triangular matrix is the product of its diagonal entries).

Proof

technique · direct

1.1L1L3L6algebra

If the list is dependent, choose nonzero coefficients $c_{j}$ with $\sum_{j < r} c_{j} v_{j} = 0$ by [L3]. Then $(G \overline{c})_{i} = \sum_{j} ⟨ v_{i}, v_{j} ⟩ \overline{c_{j}} = ⟨ v_{i}, \sum_{j} c_{j} v_{j} ⟩ = 0$ . Since $\overline{c} \neq 0$ , $G$ is singular and [L6] gives $\det G = 0$ .

1.2L2

Suppose the list is independent and $r \geq 1$ . Apply [L2], and write each $v_{j} = \sum_{i \leq j} R_{i j} e_{i}$ . The resulting $r \times r$ upper-triangular matrix $R$ has diagonal entries $R_{j j} = ∥ u_{j} ∥ > 0$ .

2.1step 1.2L4L5L7algebra

Orthonormality and the linear-first convention give $G = R^{T} \overline{R}$ . Because conjugation is a field automorphism by [L5], conjugating $R$ entrywise conjugates its determinant, so $\det \overline{R} = \overline{\det R}$ . Since $r \geq 1$ , [L4] and [L7] then give $\det G = \det (R^{T}) \det (\overline{R}) = \det R \overline{\det R} = ∣ \det R ∣^{2}$ , and $\det R = \prod_{j} R_{j j}$ is nonzero, so [L5] makes this positive.

3.1step 1.1step 2.1L1L3∎

If $r = 0$ , the empty list is independent by [L3] and [L1] gives determinant $1$ , which is positive, so all three assertions hold. If $r \geq 1$ , steps 1.1 and 2.1 exhaust the dependent and independent cases and give all three assertions.

Depends on

Used by

Dependency tree · two levels

43 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

Kenneth Hoffman and Ray Kunze, Linear Algebra, 2nd ed., p. 332, Theorem 7 (standard reference, not scraped)