Local convergence of the error-reduction algorithm for real-valued objects

A Derivation of (3.22)

We first observe that

The minimum point { θ 𝐤 } k ∈ 𝒦 ρ of the function r satisfies

which implies (3.22).

B Proof of Theorem 1

Denote

and

We have the following orthogonal decompositions for 𝐆 and 𝒫 ℳ ⁢ 𝐆 .

Decomposition for G :

where D 𝐆 = 𝒫 T ~ ℳ ⁢ ( π ⁢ ( 𝐆 ) ) ⁢ 𝐆 - π ⁢ ( 𝐆 ) ∈ T ℳ ⁢ ( π ⁢ ( 𝐆 ) ) and E 𝐆 = 𝐆 - 𝒫 T ~ ℳ ⁢ ( π ⁢ ( 𝐆 ) ) ⁢ 𝐆 ∈ ℰ .

Decomposition for P M ⁢ G :

for some D ∈ T ℳ ⁢ ( π ⁢ ( 𝐆 ) ) , E ∈ ℰ and F ∈ ℱ .

The orthogonality of T ℳ ⁢ ( π ⁢ ( 𝐆 ) ) , ℰ and ℱ implies

where the decomposition ( D , E , F ) has the following constraint.

The proof of Lemma 1 can be found in Appendix C. Because ($C_{DEF}$) is the necessary condition for ( D , E , F ) to be the orthogonal decomposition of 𝒫 ℳ ⁢ 𝐆 - π ⁢ ( 𝐆 ) , (B.1) implies that

By the method of Lagrange multipliers, the maximum in (B.2) is attained when F = 0 , so (B.2) implies

where ($C_{DE}$) is a constraint obtained by substituting F = 0 into the constraint ($C_{DEF}$), i.e.,

The constraint ($C_{DE}$) is depicted in Figure 2, which implies that the right-hand side of (B.3) can be rewritten as

In the following, we derive another constraint on ∥ E ∥ , which is motivated from [1, Proposition 2.4]. First of all, by the orthogonality property T ~ ℳ ⁢ ( π ⁢ ( 𝐆 ) ) ⊥ ( ℰ ⊕ ℱ ) , we have

where distance ⁡ ( 𝐔 , 𝐕 ) is defined by the Frobenius norm of 𝐔 - 𝐕 . Note that ∥ E ∥ → 0 when ∥ 𝐆 - π ⁢ ( 𝐆 ) ∥ → 0 . To get an effective constraint on ∥ E ∥ through (B.4), we need a more precise estimate for distance ⁡ ( 𝒫 ℳ ⁢ 𝐆 , T ~ ℳ ⁢ ( π ⁢ ( 𝐆 ) ) ) . Note that

We apply (3.21) and (3.22) to write down the vectorization of 𝒫 T ~ ℳ ⁢ ( π ⁢ ( 𝐆 ) ) ⁢ ( 𝒫 ℳ ⁢ 𝐆 ) as follows:

where

Because

we have

Therefore,

We start to estimate ∥ 𝒫 ℳ ⁢ ( 𝐆 ) - 𝒫 T ~ ℳ ⁢ ( π ⁢ ( 𝐆 ) ) ⁢ ( 𝒫 ℳ ⁢ 𝐆 ) ∥ (i.e., the right-hand side of (B.5)) as follows. By (3.9) and (B.6),

Because { 𝐟 𝐤 } 𝐤 ∈ 𝒦 ρ are orthonormal vectors, (B.7) implies

For every 𝐤 ∈ 𝒦 ρ , we define ν k , ω k ∈ ( - π , π ] as follows:

and define the difference | ν 𝐤 - ω 𝐤 | ⋆ between ν 𝐤 and ω 𝐤 by

The notation above helps us get the following estimation:

where the inequality above holds for any ν k and ω k . By substituting (B.9) into (B), we get

By the inequality

(B.10) implies that

We claim that

whose proof can be found in Appendix D. By substituting (B.12) into (B.11),

By denoting β = 2 3 ⁢ ( max 𝐤 ∈ 𝒦 ρ ⁡ b 𝐤 - 2 ) 1 2 , we have

By combining the estimation above with (B.4) and (B.5), we obtain another constraint as follows:

The entire constraints, including ($C_{DE}$) and (${C_{E}}$), on ( D , E ) are depicted in Figure 3.

Figure 3

Constraints ($C_{DE}$) and (${C_{E}}$) on D and E ( 𝒫 ℳ ⁢ 𝐆 belongs to the intersection of the shadow area andthe region { ( D , E ) : ∥ E ∥ ≤ β ⁢ ∥ 𝐆 - π ⁢ ( 𝐆 ) ∥ 2 } ).

(a)

(b)

Finally, we consider two cases as follows:

Case 1: 2 ⁢ ∥ E G ∥ + 1 2 ⁢ ∥ G - π ⁢ ( G ) ∥ 2 ⁢ ( ∑ k ∈ K ρ b k - 2 ) 1 2 ≥ β ⁢ ∥ G - π ⁢ ( G ) ∥ 2 . Using the notation described in Figure 3,

where r max = r 1 2 + r 2 2 and r 1 = β ⁢ ∥ 𝐆 - π ⁢ ( 𝐆 ) ∥ 2 . Both for Figures 3 (a) and 3 (b),

where the last inequality follows from ∥ E 𝐆 ∥ ≤ ∥ 𝐆 - π ⁢ ( 𝐆 ) ∥ and

By (B.14), we obtain

The statement of Theorem 1 follows by substituting (B.15) into (B.13).

Case 2: 2 ⁢ ∥ E G ∥ + 1 2 ⁢ ∥ G - π ⁢ ( G ) ∥ 2 ⁢ ( ∑ k ∈ K ρ b k - 2 ) 1 2 < β ⁢ ∥ G - π ⁢ ( G ) ∥ 2 . This case implies that the constraint C E cannot enhance the constraint ($C_{DE}$) (i.e., the dashed horizontal lines in Figures 3 (a) and 3 (b) do not have any intersection with the shadow areas), so we directly get

from the constraint ($C_{DE}$). Note that (B.16) implies the same conclusion as (B.15) by setting O 1 ⁢ ( ∥ 𝐆 - π ⁢ ( 𝐆 ) ∥ ) = 0 .

C Proof of Lemma 1

First of all, we show that Lemma 1 follows from the estimate

for any 𝐆 ∈ ℂ ⁢ ( 𝒩 ρ ) . Because ∥ 𝐆 - 𝒫 ℳ ⁢ 𝐆 ∥ is the shortest distance from 𝐆 to ℳ , we have the triangle inequality

where the last inequality is implied by (C.1). On the other hand, by the orthogonal decompositions of 𝐆 and 𝒫 ℳ ⁢ 𝐆 , we have

By combining (C.2) and (C.3), we get the desired result

Now we start to prove (C.1). Denote

where Φ is a matrix consisting of the column vectors { 𝐟 𝐤 } 𝐤 ∈ 𝒦 ρ . By (3.18), we know that ℳ π ⁢ ( 𝐆 ) ⊆ ℳ , which implies

By denoting k = k 1 ⁢ ( [ ρ ⁢ n 1 ] + 1 ) 2 + k 2 ⁢ ( [ ρ ⁢ n 2 ] + 1 ) for any 𝐤 = ( k 1 , k 2 ) ∈ 𝒦 ρ , 𝐟 k = 𝐟 𝐤 , b k = b 𝐤 , and 𝐠 = vec ⁢ ( 𝐆 ) , (3.21) and (3.22) imply that the vectorization of D 𝐆 can be expressed as

where

Hence, the term on the right-hand side of (C.4) can be calculated explicitly as follows:

where the minimum is taken over all diagonal matrices in ℝ | 𝒦 ρ | × | 𝒦 ρ | . Because μ ⁢ Φ is unitary,

which implies that

Applying the inequality ( 1 + δ ) 1 2 - 1 ≤ 1 2 ⁢ δ (for all δ ≥ 0 ) into (C.6), we get

The claim (C.1) follows by (C.4), (C.5) and (C.7).

D Proof of (B.12)

Because π ⁢ ( 𝐆 ) ∈ ℳ implies that b 𝐤 = | 𝐟 𝐤 * ⁢ μ * ⁢ vec ⁢ ( π ⁢ ( 𝐆 ) ) | for all 𝐤 ∈ 𝒦 ρ , the claim (B.12) is equivalent to

Let

be an orthonormal matrix such that 𝐕 * ⁢ vec ⁢ ( π ⁢ ( 𝐆 ) ) ∥ vec ⁢ ( π ⁢ ( 𝐆 ) ) ∥ = 𝐞 1 and 𝐕 * ⁢ 𝐠 = κ 1 ⁢ 𝐞 1 + κ 2 ⁢ 𝐞 2 , where 𝐞 1 and 𝐞 2 are the first and second columns in the standard basis of the real vector space ℝ | 𝒦 ρ | , and κ 1 , κ 2 ∈ ℂ . We have

Denote 𝐟 𝐤 * ⁢ μ * ⁢ 𝐕 = [ r 𝐤 , 1 ⁢ r 𝐤 , 2 ⁢ ⋯ ⁢ r 𝐤 , | 𝒦 ρ | ] ∈ ℂ 1 × | 𝒦 ρ | . Then (D.2) implies that

where Phase ⁡ ( y ) = y | y | for y ∈ ℂ ∖ { 0 } . For the function Phase ⁡ ( ⋅ ) , we have the following inequalities (see, for example, [26, Lemma A.7]):

By applying the inequality (D.4) with y = κ 1 ∥ π ⁢ ( 𝐆 ) ∥ + κ 2 ∥ π ⁢ ( 𝐆 ) ∥ ⁢ r 𝐤 , 2 r 𝐤 , 1 - 1 ∈ ℂ into (D.3), we get

Because | 𝐟 𝐤 * ⁢ μ * ⁢ vec ⁢ ( π ⁢ ( 𝐆 ) ) | = ∥ π ⁢ ( 𝐆 ) ∥ ⋅ | r 𝐤 , 1 | , (D.5) implies that

Because 𝐕 = [ V 1 ⁢ V 2 ⁢ ⋯ ⁢ V | 𝒦 ρ | ] is an orthonormal matrix and ∑ 𝐤 ∈ 𝒦 ρ 𝐟 𝐤 ⁢ 𝐟 𝐤 * = 𝐈 | 𝒦 ρ | × | 𝒦 ρ | , we have

which implies that

In the same way, we have

By substituting (D.7) and (D.8) into (D.6), we get

The claim (D.1) follows by (D.9) and noticing that | κ 1 - ∥ π ⁢ ( 𝐆 ) ∥ | 2 + | κ 2 | 2 = ∥ 𝐆 - π ⁢ ( 𝐆 ) ∥ 2 .