Appendix A. Mathematical lemmas and proof sketches

Appendix A. Mathematical lemmas and proof: R is the field of real numbers and C is the field of complex numbers. Let V be a vector space over R or C. Thus b_j=0, so π(u_j) is a basis and dim(V/W)=n-k.

R is the field of real numbers and C is the field of complex numbers. Let V be a vector space over R or C. Thus b_j=0, so π(u_j) is a basis and dim(V/W)=n-k.

A.0 Notation and basic conventions (common)

(1) Numbers and vectors

R is the field of real numbers and C is the field of complex numbers. Let V be a vector space over R or C.

(2) Inner product and norm

In a complex inner-product space (V,⟨·,·⟩), the inner product satisfies the following properties.

  1. (Linearity) ⟨ ax+by, z⟩ = a⟨ x,z⟩ + b⟨ y,z⟩.
  2. (Conjugate symmetry) ⟨ x,y⟩ = ⟨ y,x⟩.
  3. (Positive definiteness) ⟨ x,x⟩ ≥ 0, and ⟨ x,x⟩=0 ⇔ x=0.

The norm is defined by |x|:=√(⟨ x,x⟩).

(3) Linear operators and adjoints

The adjoint A^dagger of a linear operator A:V→ V is defined by

\langle Ax,y\rangle=\langle x,A^\dagger y\rangle\quad(\forall x,y\in V)

as the operator satisfying the above. If A=A^dagger, then A is (in the mathematical sense) self-adjoint (Hermitian).

(4) Commutator

For operators A,B, the commutator is defined by [A,B]:=AB-BA.

A.1 Cauchy–Schwarz inequality

Theorem A.1 (Cauchy–Schwarz)

For any x,y∈ V,

\begin{equation} |\langle x,y\rangle|\le \|x\|\,\|y\| \end{equation}

holds.

Proof

If y=0 then the left-hand side is 0 and the claim holds. Now assume y≠ 0 and, for any λ∈C,

0\le \|x-\lambda y\|^2=\langle x-\lambda y, x-\lambda y\rangle

expand the expression.

\begin{align*} \|x-\lambda y\|^2 &=\langle x,x\rangle-\lambda\langle y,x\rangle-\overline{\lambda}\langle x,y\rangle+|\lambda|^2\langle y,y\rangle\\ &=\|x\|^2-\lambda\overline{\langle x,y\rangle}-\overline{\lambda}\langle x,y\rangle+|\lambda|^2\|y\|^2. \end{align*}

Choosing λ:=⟨ x,y⟩/|y|², we obtain

\begin{align*} 0\le \|x-\lambda y\|^2 &=\|x\|^2-\frac{\langle x,y\rangle\overline{\langle x,y\rangle}}{\|y\|^2} -\frac{\overline{\langle x,y\rangle}\langle x,y\rangle}{\|y\|^2} +\frac{|\langle x,y\rangle|^2}{\|y\|^4}\|y\|^2\\ &=\|x\|^2-\frac{|\langle x,y\rangle|^2}{\|y\|^2}. \end{align*}

Therefore |⟨ x,y⟩|²≤ |x|²|y|², and taking square roots yields (AppA_CS). square

A.2 Triangle inequality

Theorem A.2 (Triangle inequality)

For any x,y∈ V,

\begin{equation} \|x+y\|\le \|x\|+\|y\| \end{equation}

holds.

Proof

\|x+y\|^2=\langle x+y,x+y\rangle=\|x\|^2+\|y\|^2+2\operatorname{Re}\langle x,y\rangle.

From Cauchy–Schwarz (AppA_CS) we have |⟨ x,y⟩|≤ |x||y|, and since Re⟨ x,y⟩≤ |⟨ x,y⟩|,

\|x+y\|^2\le \|x\|^2+\|y\|^2+2\|x\|\|y\|=(\|x\|+\|y\|)^2.

taking square roots of both sides gives (AppA_triangle). square

A.3 Quotient-space dimension theorem (removing global reference degrees of freedom)

Theorem A.3 (Quotient-space dimension)

Let V be a finite-dimensional vector space and W⊂ V a subspace. Then

\begin{equation} \dim(V/W)=\dim(V)-\dim(W) \end{equation}

holds.

Proof

Extend a basis w₁,…,wₖ of W to a basis of V so that w₁,…,wₖ,u₁,…,uₙ₋ₖ becomes a basis of V (n=dim(V)). For the quotient map π:V→ V/W, show that π(u₁),…,π(uₙ₋ₖ) generates V/W. Any v∈ V can be written as

v=\sum_{i=1}^k a_i w_i + \sum_{j=1}^{n-k} b_j u_j

and since π(w_i)=0,

\pi(v)=\sum_{j=1}^{n-k} b_j \pi(u_j).

therefore

\sum_{j=1}^{n-k} b_j \pi(u_j)=0 \Rightarrow \pi\Bigl(\sum_{j=1}^{n-k} b_j u_j\Bigr)=0 \Rightarrow \sum_{j=1}^{n-k} b_j u_j \in W.

But since w_i,u_j is a basis, any vector in W must have all u_j components equal to 0. Thus b_j=0, so π(u_j) is a basis and dim(V/W)=n-k. square

Application A.3.1 (removing the global reference in a 6-face channel)

If V=R⁶, W=span1₆, and 1₆=(1,1,1,1,1,1), then dim(W)=1 and dim(V)=6, hence by (AppA_dim_quotient) we have dim(V/W)=5.

A.4 Discrete Fourier transform and Parseval's theorem (finite lattice)

Setup

For N∈N, let Z_N=0,1,…,N-1 be the cyclic group modulo N. Define the inner product of a complex sequence x:Z_N→C by

\langle x,y\rangle := \sum_{n=0}^{N-1} x_n \overline{y_n}

as above.

DFT

Define the discrete Fourier transform (DFT) by

\begin{equation} X_k := \sum_{n=0}^{N-1} x_n\,e^{-2\pi i kn/N}\qquad (k=0,\dots,N-1) \end{equation}

and define the inverse transform by

\begin{equation} x_n := \frac{1}{N}\sum_{k=0}^{N-1} X_k\,e^{2\pi i kn/N}\qquad (n=0,\dots,N-1) \end{equation}

as follows.

Theorem A.4.1 (Orthogonality)

\sum_{n=0}^{N-1} e^{2\pi i (k-\ell)n/N}=N\,\delta_{k\ell}.

Proof

If k=ℓ then the sum is N. If k≠ ℓ, it is a geometric series with ratio r=e^(2π i (k-ℓ)/N)≠ 1, hence

\sum_{n=0}^{N-1} r^n=\frac{1-r^N}{1-r}=\frac{1-1}{1-r}=0.

square

Theorem A.4.2 (Parseval)

\begin{equation} \sum_{n=0}^{N-1}|x_n|^2=\frac{1}{N}\sum_{k=0}^{N-1}|X_k|^2. \end{equation}

Proof

Using (AppA_IDFT),

x_n=\frac{1}{N}\sum_{k}X_k e^{2\pi i kn/N},\quad \overline{x_n}=\frac{1}{N}\sum_{\ell}\overline{X_\ell} e^{-2\pi i \ell n/N}.

therefore

\begin{align*} \sum_{n}|x_n|^2 &=\sum_{n}x_n\overline{x_n} =\sum_n \frac{1}{N^2}\sum_{k,\ell} X_k\overline{X_\ell} e^{2\pi i (k-\ell)n/N}\\ &=\frac{1}{N^2}\sum_{k,\ell}X_k\overline{X_\ell}\sum_n e^{2\pi i (k-\ell)n/N}. \end{align*}

By Theorem A.4.1, the inner sum equals Nδ_(kℓ), hence

\sum_{n}|x_n|^2=\frac{1}{N^2}\sum_{k,\ell}X_k\overline{X_\ell}\,N\delta_{k\ell} =\frac{1}{N}\sum_k |X_k|^2.

square

A.5 Commutator-based uncertainty inequality (pure algebra)

Theorem A.5 (Robertson form)

For self-adjoint A,B in an inner-product space and a normalized vector ψ (|ψ|=1),

\begin{equation} \Delta_\psi A\cdot \Delta_\psi B \ge \frac{1}{2}\left|\left\langle \psi,\;[A,B]\psi\right\rangle\right| \end{equation}

holds. Here

\Delta_\psi A:=\sqrt{\langle (A-\langle A\rangle)\psi,(A-\langle A\rangle)\psi\rangle}, \quad \langle A\rangle:=\langle \psi,A\psi\rangle

we define

Proof

Let A':=A-⟨ A⟩ I and B':=B-⟨ B⟩ I, and set u:=A'ψ and v:=B'ψ. Then

\|u\|=\Delta_\psi A,\quad \|v\|=\Delta_\psi B.

By Cauchy–Schwarz (AppA_CS),

|\langle u,v\rangle|\le \|u\|\,\|v\|=\Delta_\psi A\cdot \Delta_\psi B.

On the other hand,

\langle u,v\rangle=\langle A'\psi,B'\psi\rangle=\langle \psi,A'B'\psi\rangle.

For the complex number z:=⟨ ψ,A'B'ψ⟩, since Im(z)≤ |z|,

|z|\ge |\operatorname{Im}(z)|.

Also,

\begin{align*} z-\overline{z} &=\langle \psi,A'B'\psi\rangle-\overline{\langle \psi,A'B'\psi\rangle} =\langle \psi,A'B'\psi\rangle-\langle \psi,(A'B')^\dagger\psi\rangle\\ &=\langle \psi,A'B'\psi\rangle-\langle \psi,B'A'\psi\rangle =\langle \psi,[A',B']\psi\rangle. \end{align*}

Therefore

2i\,\operatorname{Im}(z)=\langle \psi,[A',B']\psi\rangle.

Taking absolute values gives

2|\operatorname{Im}(z)|=\left|\langle \psi,[A',B']\psi\rangle\right|.

But the constant term cancels in the commutator, so [A',B']=[A,B]. Therefore

|z|\ge |\operatorname{Im}(z)|=\frac{1}{2}\left|\langle \psi,[A,B]\psi\rangle\right|.

Finally, combining |z|=|⟨ u,v⟩|≤ Δ_ψ AΔ_ψ B yields (AppA_Robertson). square

A.6 Sensitivity (error propagation) upper bound

Theorem A.6 (first-order sensitivity bound)

Let f:Rⁿ→R be differentiable at x, and let the input error be δ x∈Rⁿ. Then

\begin{equation} |f(x+\delta x)-f(x)| \le \|\nabla f(x)\|_2\,\|\delta x\|_2 + o(\|\delta x\|_2). \end{equation}

Proof

By the definition of differentiability,

f(x+\delta x)-f(x)=\nabla f(x)\cdot \delta x + r(\delta x), \quad \frac{|r(\delta x)|}{\|\delta x\|_2}\to 0\ (\delta x\to 0).

By Cauchy–Schwarz, |∇ f(x)· δ x|≤ |∇ f(x)|₂|δ x|₂. Thus (AppA_sensitivity). square

A.7 Significant-digit rounding operator (reporting convention)

Definition A.7 (rounding operator)

For k∈Z, define Roundₖ:R→R by

\mathrm{Round}_k(x):=10^{-k}\cdot \mathrm{round}(10^k x)

as above. Here, round denotes rounding to the nearest integer (the tie-breaking rule is locked in analysis_lock).

Property A.7.1 (Error bound)

Regardless of the tie-breaking rule,

|\mathrm{Round}_k(x)-x|\le \frac{1}{2}\cdot 10^{-k}

holds (since the integer lattice spacing is 10^(-k), the worst case is half the spacing).