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Real symmetric quadratic forms tested on phasors

Statement

Let $V$ be a finite set and $B = (B_{x y})_{x, y \in V}$ a real symmetric matrix, meaning $B_{x y} = B_{y x} \in R$ . For $f = u + i v \in C^{V}$ with $u, v \in R^{V}$ , write $f^{*} B f = \sum_{x, y \in V} \overline{f_{x}} B_{x y} f_{y}$ and $u^{T} B u = \sum_{x, y \in V} u_{x} B_{x y} u_{y}$ . Then $f^{*} B f = u^{T} B u + v^{T} B v$ .

Thus $B$ is positive semidefinite on real vectors, meaning $u^{T} B u \geq 0$ for every $u \in R^{V}$ , if and only if $f^{*} B f \geq 0$ for every $f \in C^{V}$ . Likewise, $u^{T} B u > 0$ for every nonzero real $u$ if and only if $f^{*} B f > 0$ for every nonzero complex $f$ .

For any real symmetric array $b = (b_{x y})_{x, y \in V}$ , let $L = diag (\sum_{y} b_{x y}) - b$ , so that $L_{x y} = δ_{x y} \sum_{z} b_{x z} - b_{x y}$ , with $δ_{x y} = 1$ when $x = y$ and $0$ otherwise. Then $f^{*} L f = \frac{1}{2} \sum_{x, y \in V} b_{x y} ∣ f_{x} - f_{y} ∣^{2}$ . This identity allows signed coefficients $b_{x y}$ ; nonnegativity of the form requires a separate positive-semidefiniteness premise.

Statement references

Finite vertex Hilbert space in the article convention

Complex conjugation and modulus laws

Facts & Assumptions

[F1]

On a finite vertex set, matrix action and the second-linear inner product are given by finite sums, and a real symmetric matrix is self-adjoint (Finite vertex Hilbert space in the article convention).

[F2]

Complex conjugation preserves sums and products, fixes real numbers, and satisfies $z \overline{z} = ∣ z ∣^{2}$ (Conjugation is an involutive real-field automorphism, zz̄=|z|², and modulus is definite, multiplicative, and subadditive).

Proof

Given: A finite set $V$ , a real symmetric matrix $B$ , a vector $f = u + i v$ with real $u, v$ , and a real symmetric array $b$ .

Goal: Prove the quadratic-form decomposition, the two positivity equivalences, and the signed-weight identity.

1.1F1F2algebra

Expanding the finite sums using [F1] and [F2] gives $f^{*} B f = u^{T} B u + v^{T} B v + i (u^{T} B v - v^{T} B u)$ . Interchanging $x$ and $y$ in the finite sum and using $B_{x y} = B_{y x}$ gives $v^{T} B u = u^{T} B v$ . The imaginary term therefore vanishes, proving the decomposition.

2.1step 1.1algebra

If $u^{T} B u \geq 0$ for all real $u$ , the decomposition makes $f^{*} B f \geq 0$ for all complex $f$ ; the converse follows by taking $v = 0$ . If the real form is strictly positive on nonzero vectors, then a nonzero $f = u + i v$ has at least one nonzero component vector, and the decomposition is strictly positive because the other term is nonnegative. The converse again follows by taking $v = 0$ .

3.1F1F2step 2.1algebra∎

By [F2], $∣ f_{x} - f_{y} ∣^{2} = ∣ f_{x} ∣^{2} + ∣ f_{y} ∣^{2} - \overline{f_{x}} f_{y} - \overline{f_{y}} f_{x}$ . Since $b_{x y} = b_{y x}$ , interchanging the indices in the second diagonal sum and in the last cross term yields $\frac{1}{2} \sum_{x, y} b_{x y} ∣ f_{x} - f_{y} ∣^{2} = \sum_{x} (\sum_{y} b_{x y}) ∣ f_{x} ∣^{2} - \sum_{x, y} b_{x y} \overline{f_{x}} f_{y} = f^{*} L f$ , by the matrix action in [F1]. No sign condition on $b$ was used. If $L$ is positive semidefinite on real vectors, step 2.1 gives nonnegativity on complex vectors.

Depends on

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