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Matrix Recipes for Hard Thresholding Methods

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Abstract

In this paper, we present and analyze a new set of low-rank recovery algorithms for linear inverse problems within the class of hard thresholding methods. We provide strategies on how to set up these algorithms via basic ingredients for different configurations to achieve complexity vs. accuracy tradeoffs. Moreover, we study acceleration schemes via memory-based techniques and randomized, ϵ-approximate matrix projections to decrease the computational costs in the recovery process. For most of the configurations, we present theoretical analysis that guarantees convergence under mild problem conditions. Simulation results demonstrate notable performance improvements as compared to state-of-the-art algorithms both in terms of reconstruction accuracy and computational complexity.

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Notes

  1. The distinction between \(\mathcal{P}_{\mathcal{S}} \) and \(\mathcal {P}_{k} \) for k positive integer is apparent from context.

  2. In the case of multiple identical singular values, any ties are lexicographically dissolved.

  3. From a different perspective and for a different problem case, similar ideas have been used in [18].

  4. We can move between these two cases by a simple transpose of the problem.

  5. While such operation has O(max{m 2 n,mn 2}) complexity, each application of \(\mathcal{P}_{\mathcal{S}} \boldsymbol {X}\) requires three matrix-matrix multiplications. To reduce such computational cost, we relax this operation in Sect. 10 where in practice we use only \(\mathcal{P}_{\mathcal{U}}\) that needs one matrix-matrix multiplication.

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Acknowledgements

This work was supported in part by the European Commission under Grant MIRG-268398, ERC Future Proof, SNF 200021-132548 and DARPA KeCoM program #11-DARPA-1055. VC also would like to acknowledge Rice University for his Faculty Fellowship.

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Correspondence to Anastasios Kyrillidis.

Appendix

Appendix

Remark 1

Let \(\boldsymbol {X}\in\mathbb{R}^{m \times n}\) with SVD: X=UΣV T, and \(\boldsymbol{Y} \in\mathbb{R}^{m \times n}\) with SVD: \(\boldsymbol {Y} = \widetilde{\boldsymbol{U}} \widetilde{\boldsymbol{\varSigma}} \widetilde{\boldsymbol{V}}^{T}\). Assume two sets: (i) \(\mathcal{S}_{1} = \lbrace\boldsymbol{u}_{i}\boldsymbol{u}_{i}^{T}: i \in\mathcal {I}_{1} \rbrace\) where u i is the i-th singular vector of X and \(\mathcal{I}_{1} \subseteq\lbrace1, \dots,\allowbreak \operatorname{rank}(\boldsymbol {X}) \rbrace\) and, (ii) \(\mathcal{S}_{2} = \lbrace \boldsymbol{u}_{i}\boldsymbol{u}_{i}^{T}, \tilde{\boldsymbol{u}_{j}} \tilde {\boldsymbol{u}_{j}}^{T}: i \in\mathcal{I}_{2},~j \in\mathcal{I}_{3} \rbrace\) where \(\tilde{\boldsymbol{u}_{i}} \) is the i-th singular vector of Y, \(\mathcal{I}_{1} \subseteq\mathcal{I}_{2} \subseteq\lbrace1, \dots, \operatorname{rank}(\boldsymbol {X}) \rbrace\) and, \(\mathcal{I}_{3} \subseteq\lbrace1, \dots, \operatorname{rank}(\boldsymbol {Y}) \rbrace\). We observe that the subspaces defined by \(\boldsymbol {u}_{i}\boldsymbol{u}_{i}^{T}\) and \(\tilde{\boldsymbol{u}_{j}} \tilde {\boldsymbol{u}_{j}}^{T}\) are not necessarily orthogonal.

To this end, let \(\widehat{\mathcal{S}}_{2} = \text{ortho}(\mathcal {S}_{2})\); this operation can be easily computed via SVD. Then, the following commutativity property holds true for any matrix \(\boldsymbol {W} \in\mathbb{R}^{m \times n} \):

(36)

1.1 A.1 Proof of Lemma 6

Given \(\mathcal{X}^{\ast}\leftarrow\mathcal{P}_{k}(\boldsymbol {X}^{\ast}) \) using SVD factorization, we define the following quantities: \(\mathcal{S}_{i} \leftarrow\mathcal{X}_{i} \cup\mathcal{D}_{i}\), \(\mathcal {S}^{\ast}_{i} \leftarrow\text{ortho} (\mathcal{X}_{i} \cup\mathcal {X}^{\ast}) \). Then, given the structure of the sets \(\mathcal {S}_{i}\) and \(\mathcal{S}_{i}^{\ast}\)

(37)

and

(38)

Since the subspace defined in \(\mathcal{D}_{i} \) is the best rank-k subspace, orthogonal to the subspace spanned by \(\mathcal {X}_{i} \), the following holds true:

Removing the common subspaces in \(\mathcal{S}_{i} \) and \(\mathcal {S}_{i}^{\ast}\) by the commutativity property of the projection operation and using the shortcut \(\mathcal{P}_{\mathcal{A} \setminus\mathcal {B}} \equiv\mathcal{P}_{\mathcal{A}} \mathcal{P}_{\mathcal {B}^{\bot}} \) for sets \(\mathcal{A}\), \(\mathcal{B}\), we get:

(39)

Next, we assume that \(\mathcal{P}_{(\mathcal{A} \setminus\mathcal {B})^{\bot}}\) denotes the orthogonal projection onto the subspace spanned by \(\mathcal{P}_{\mathcal{A}} \mathcal{P}_{\mathcal {B}^{\bot}}\). Then, on the left hand side of (39), we have:

(40)

where (i) due to triangle inequality over Frobenius metric norm, (ii) since \(\mathcal{P}_{\mathcal{S}_{i} \setminus\mathcal {S}_{i}^{\ast}} (\boldsymbol {X}(i) - \boldsymbol{X}^{\ast}) = \mathbf{0} \), (iii) by using the fact that \(\boldsymbol {X}(i) - \boldsymbol{X}^{\ast}:= \mathcal {P}_{\mathcal{S}_{i} \setminus\mathcal{S}_{i}^{\ast}}(\boldsymbol {X}(i) - \boldsymbol{X}^{\ast}) + \mathcal{P}_{(\mathcal{S}_{i} \setminus \mathcal {S}_{i}^{\ast})^{\bot}}(\boldsymbol {X}(i) - \boldsymbol{X}^{\ast}) \), (iv) due to Lemma 4, (v) due to Lemma 5 and (vi) since \(\Vert\mathcal{P}_{(\mathcal{S}_{i} \setminus\mathcal {S}_{i}^{\ast})^{\bot}} (\boldsymbol{X}^{\ast}- \boldsymbol {X}(i)) \Vert_{F} \leq \Vert \boldsymbol {X}(i) - \boldsymbol{X}^{\ast}\Vert_{F} \).

For the right hand side of (39), we calculate:

(41)

by using Lemmas 4 and 5. Combining (40) and (41) in (39), we get:

1.2 A.2 Proof of Theorem 1

Let \(\mathcal{X}^{\ast}\leftarrow\mathcal{P}_{k}(\boldsymbol {X}^{\ast}) \) be a set of orthonormal, rank-1 matrices that span the range of X . In Algorithm 1, \(\boldsymbol{W}(i) \leftarrow \mathcal{P}_{k}(\boldsymbol{V}(i)) \). Thus:

(42)

From Algorithm 1, (i) \(\boldsymbol{V}(i) \in\text {span}(\mathcal {S}_{i}) \), (ii) \(\boldsymbol {X}(i) \in\operatorname{span}( \mathcal{S}_{i}) \) and (iii) \(\boldsymbol{W}(i) \in\operatorname{span}(\mathcal{S}_{i}) \). We define \(\mathcal{E} \leftarrow \text{ortho}(\mathcal{S}_{i} \cup \mathcal{X}^{\ast}) \) where \(\operatorname{rank}(\operatorname{span}(\mathcal {E})) \leq3k\) and let \(\mathcal{P}_{\mathcal{E}} \) be the orthogonal projection onto the subspace defined by \(\mathcal{E} \).

Since \(\boldsymbol{W}(i) - \boldsymbol{X}^{\ast}\in\text {span}(\mathcal{E})\) and \(\boldsymbol{V}(i) - \boldsymbol{X}^{\ast}\in\text {span}(\mathcal{E})\), the following hold true:

Then, (42) can be written as:

(43)

In B, we observe:

(44)

where (i) holds since \(\mathcal{P}_{\mathcal{S}_{i}} \mathcal {P}_{\mathcal{E}} = \mathcal{P}_{\mathcal{E}}\mathcal{P}_{\mathcal {S}_{i}} = \mathcal{P}_{\mathcal{S}_{i}} \) for \(\operatorname{span}(\mathcal {S}_{i}) \in\operatorname{span}(\mathcal{E}) \), (ii) is due to Cauchy-Schwarz inequality and, (iii) is easily derived using Lemma 2.

In A, we perform the following motions:

(45)

where (i) is due to \(\mathcal{P}_{\mathcal{E}}(\boldsymbol {X}(i) - \boldsymbol{X}^{\ast}) := \mathcal{P}_{\mathcal{S}_{i}} \mathcal {P}_{\mathcal{E}}(\boldsymbol {X}(i) - \boldsymbol{X}^{\ast}) + \mathcal {P}_{\mathcal {S}_{i}^{\bot}} \mathcal{P}_{\mathcal{E}}(\boldsymbol {X}(i) - \boldsymbol {X}^{\ast}) \) and (ii) follows from Cauchy-Schwarz inequality. Since \(\frac{1}{1+\delta_{2k}} \leq\mu_{i} \leq\frac {1}{1-\delta_{2k}} \), Lemma 4 implies:

and thus:

Furthermore, according to Lemma 5:

since \(\operatorname{rank}(\mathcal{P}_{\mathcal{K}}\boldsymbol {X}) \leq3k\), \(\forall \boldsymbol {X}\in\mathbb{R}^{m \times n} \) for \(\mathcal{K} \leftarrow\text{ortho}(\mathcal{E} \cup\mathcal{S}_{i})\). Since \(\mathcal{P}_{\mathcal{S}_{i}^{\bot}}\mathcal{P}_{\mathcal {E}}(\boldsymbol {X}(i) - \boldsymbol{X}^{\ast}) = \mathcal{P}_{\mathcal {X}^{\ast}\setminus(\mathcal{D}_{i} \cup\mathcal{X}_{i})}\boldsymbol{X}^{\ast}\) where

then:

using Lemma 6. Combining the above in (45), we compute:

(46)

Combining (44) and (46) in (43), we get:

(47)

Focusing on steps 5 and 6 of Algorithm 1, we perform similar motions to obtain:

(48)

Combining the recursions in (47) and (48), we finally compute:

for \(\rho:= ( \frac{1 + 2\delta_{2k}}{1-\delta_{2k}} ) (\frac{4\delta_{2k}}{1-\delta_{2k}} + (2\delta_{2k} + 2\delta_{3k})\frac{2\delta_{3k}}{1-\delta_{2k}} )\) and

For the convergence parameter ρ, further compute:

(49)

for δ k δ 2k δ 3k . Calculating the roots of this expression, we easily observe that \(\rho < \hat{\rho} < 1 \) for δ 3k <0.1235.

1.3 A.3 Proof of Theorem 2

Before we present the proof of Theorem 2, we list a series of lemmas that correspond to the motions Algorithm 2 performs.

Lemma 9

[Error norm reduction via least-squares optimization]

Let \(\mathcal{S}_{i} \) be a set of orthonormal, rank-1 matrices that span a rank-2k subspace in \(\mathbb{R}^{m \times n} \). Then, the least squares solution V(i) given by:

(50)

satisfies:

(51)

Proof

We observe that \(\Vert\boldsymbol{V}(i) - \boldsymbol{X}^{\ast}\Vert_{F}^{2} \) is decomposed as follows:

(52)

In (50), V(i) is the minimizer over the low-rank subspace spanned by \(\mathcal{S}_{i} \) with \(\text {rank}(\operatorname{span}(\mathcal{S}_{i})) \leq2k\). Using the optimality condition (Lemma 1) over the convex set \(\varTheta= \lbrace \boldsymbol {X}: \operatorname{span}(\boldsymbol {X}) \in\mathcal{S}_{i} \rbrace\), we have:

(53)

for \(\mathcal{P}_{\mathcal{S}_{i}}\boldsymbol{X}^{\ast}\in\text {span}(\mathcal{S}_{i}) \). Given condition (53), the first term on the right hand side of (52) becomes:

(54)

Focusing on the term \(| \langle\boldsymbol{V}(i) - \boldsymbol {X}^{\ast}, (\mathbf {I}- \boldsymbol {\mathcal {A}}^{\ast} \boldsymbol {\mathcal {A}})\mathcal{P}_{\mathcal {S}_{i}}(\boldsymbol{V}(i) - \boldsymbol{X}^{\ast}) \rangle| \), we derive the following:

where (i) follows from the facts that \(\boldsymbol{V}(i) - \boldsymbol{X}^{\ast}\in \operatorname{span}(\text{ortho}(\mathcal{S}_{i} \cup \mathcal {X}^{\ast})) \) and thus \(\mathcal{P}_{\mathcal{S}_{i} \cup\mathcal {X}^{\ast}}(\boldsymbol{V}(i) - \boldsymbol{X}^{\ast}) = \boldsymbol {V}(i) - \boldsymbol{X}^{\ast}\) and (ii) is due to \(\mathcal{P}_{\mathcal{S}_{i} \cup\mathcal{X}^{\ast}} \mathcal{P}_{\mathcal{S}_{i}} = \mathcal {P}_{\mathcal{S}_{i}} \) since \(\operatorname{span}(\mathcal{S}_{i}) \subseteq \operatorname{span}(\text{ortho}(\mathcal{S}_{i} \cup\mathcal{X}^{\ast})) \). Then, (54) becomes:

(55)

where (i) comes from Cauchy-Swartz inequality and (ii) is due to Lemmas 2 and 4. Simplifying the above quadratic expression, we obtain:

(56)

As a consequence, (52) can be upper bounded by:

(57)

We form the quadratic polynomial for this inequality assuming as unknown variable the quantity ∥V(i)−X F . Bounding by the largest root of the resulting polynomial, we get:

(58)

 □

The following lemma characterizes how subspace pruning affects the recovered energy:

Lemma 10

[Best rank-k subspace selection]

Let \(\boldsymbol{V}(i) \in\mathbb {R}^{m \times n} \) be a rank-2k proxy matrix in the subspace spanned by \(\mathcal{S}_{i} \) and let \(\boldsymbol {X}(i+1) \leftarrow \mathcal{P}_{k}(\boldsymbol{V}(i)) \) denote the best rank-k approximation to V(i), according to (5). Then:

(59)

Proof

Since X(i+1) denotes the best rank-k approximation to V(i), the following inequality holds for any rank-k matrix \(\boldsymbol {X}\in\mathbb{R}^{m \times n} \) in the subspace spanned by \(\mathcal{S}_{i} \), i.e. \(\forall \boldsymbol {X}\in \operatorname{span}(\mathcal{S}_{i}) \):

(60)

Since \(\mathcal{P}_{\mathcal{S}_{i}} \boldsymbol{V}(i) = \boldsymbol {V}(i) \), the left inequality in (59) is satisfied for \(\boldsymbol {X}:= \mathcal{P}_{\mathcal{S}_{i}} \boldsymbol{X}^{\ast}\) in (60). □

Lemma 11

Let V(i) be the least squares solution in Step 2 of the ADMiRA algorithm and let X(i+1) be a proxy, rank-k matrix to V(i) according to: \(\boldsymbol {X}(i+1) \leftarrow\mathcal {P}_{k}(\boldsymbol{V}(i))\). Then, ∥X(i+1)−X F can be expressed in terms of the distance from V(i) to X as follows:

(61)

Proof

We observe the following

(62)

Focusing on the right hand side of expression (62), \(\langle \boldsymbol{V}(i) - \boldsymbol{X}^{\ast}, \boldsymbol{V}(i) - \boldsymbol {X}(i+1) \rangle= \langle\boldsymbol{V}(i) - \boldsymbol {X}^{\ast}, \mathcal{P}_{\mathcal{S}_{i}}(\boldsymbol{V}(i) - \boldsymbol {X}(i+1) ) \rangle\) can be similarly analysed as in Lemma 10 where we obtain the following expression:

(63)

Now, expression (62) can be further transformed as:

(64)

where (i) is due to (63). Using Lemma 10, we further have:

(65)

Furthermore, replacing \(\Vert \mathcal{P}_{\mathcal {S}_{i}}(\boldsymbol{X}^{\ast}- \boldsymbol{V}(i))\Vert _{F} \) with its upper bound defined in (56), we get:

(66)

where (i) is obtained by completing the squares and eliminating negative terms. □

Applying basic algebra tools in (61) and (51), we get:

Since \(\boldsymbol{V}(i) \in\operatorname{span}(\mathcal{S}_{i}) \), we observe \(\mathcal{P}_{\mathcal{S}_{i}^{\bot}}(\boldsymbol{V}(i) - \boldsymbol{X}^{\ast}) = -\mathcal{P}_{\mathcal{S}_{i}^{\bot}} \boldsymbol{X}^{\ast}= -\mathcal{P}_{\mathcal{X}^{\ast}\setminus (\mathcal{D}_{i} \cup\mathcal{X}_{i})} \boldsymbol{X}^{\ast}\). Then, using Lemma 6, we obtain:

(67)

Given δ 2k δ 3k , ρ is upper bounded by \(\rho\,{<}\, 4\delta_{3k}\sqrt{\frac{1+3\delta_{3k}}{1-\delta_{3k}^{2}}} \). Then, \(4\delta_{3k}\sqrt{\frac {1+3\delta_{3k}}{1-\delta_{3k}^{2}}} < 1 \Leftrightarrow \delta_{3k} < 0.2267\).

1.4 A.4 Proof of Theorem 3

Let \(\mathcal{X}^{\ast}\leftarrow\mathcal{P}_{k}(\boldsymbol {X}^{\ast}) \) be a set of orthonormal, rank-1 matrices that span the range of X . In Algorithm 3, X(i+1) is the best rank-k approximation of V(i). Thus:

(68)

From Algorithm 3, (i) \(\boldsymbol{V}(i) \in\operatorname{span}(\mathcal {S}_{i}) \), (ii) \(\boldsymbol{Q}_{i} \in \operatorname{span}( \mathcal{S}_{i}) \) and (iii) \(\boldsymbol{W}(i) \in\operatorname{span}(\mathcal{S}_{i}) \). We define \(\mathcal{E} \leftarrow\text{ortho}(\mathcal{S}_{i} \cup \mathcal{X}^{\ast}) \) where we observe \(\operatorname{rank}(\text {span}(\mathcal{E})) \leq4k\) and let \(\mathcal{P}_{\mathcal {E}} \) be the orthogonal projection onto the subspace defined by \(\mathcal{E} \).

Since \(\boldsymbol {X}(i+1) - \boldsymbol{X}^{\ast}\in\operatorname{span}(\mathcal {E}) \) and \(\boldsymbol{V}(i) - \boldsymbol{X}^{\ast}\in \operatorname{span}(\mathcal{E}) \), the following hold true:

and,

(69)

Then, (68) can be written as:

(70)

where (i) is due to \(\mathcal{P}_{\mathcal{E}}(\boldsymbol{Q}_{i} - \boldsymbol{X}^{\ast}) := \mathcal{P}_{\mathcal{S}_{i}} \mathcal {P}_{\mathcal{E}}(\boldsymbol{Q}_{i} - \boldsymbol{X}^{\ast}) + \mathcal{P}_{\mathcal{S}_{i}^{\bot}} \mathcal{P}_{\mathcal {E}}(\boldsymbol{Q}_{i} - \boldsymbol{X}^{\ast}) \) and (ii) follows from Cauchy-Schwarz inequality. Since \(\frac{1}{1+\delta_{3k}} \leq\mu_{i} \leq\frac {1}{1-\delta_{3k}} \), Lemma 4 implies:

and thus:

Furthermore, according to Lemma 5:

since \(\operatorname{rank}(\mathcal{P}_{\mathcal{K}}\boldsymbol{Q}) \leq 4k\), \(\forall\boldsymbol{Q} \in\mathbb{R}^{m \times n} \) where \(\mathcal{K} \leftarrow\text{ortho}(\mathcal{E} \cup\mathcal{S}_{i})\). Since \(\mathcal{P}_{\mathcal{S}_{i}^{\bot}}\mathcal{P}_{\mathcal{E}} (\boldsymbol{Q}_{i} - \boldsymbol{X}^{\ast}) = \mathcal{P}_{\mathcal{X}^{\ast}\setminus(\mathcal{D}_{i} \cup\mathcal {X}_{i})}\boldsymbol{X}^{\ast}\) where

then:

(71)

using Lemma 6. Using the above in (70), we compute:

(72)

Furthermore:

(73)

Combining (72) and (73), we get:

(74)

Let \(\alpha:= \frac{4\delta_{3k}}{1-\delta_{3k}} + (2\delta_{3k} + 2\delta_{4k})\frac{2\delta_{3k}}{1-\delta_{3k}} \) and g(i):=∥X(i+1)−X F . Then, (74) defines the following homogeneous recurrence:

(75)

Using the method of characteristic roots to solve the above recurrence, we assume that the homogeneous linear recursion has solution of the form g(i)=r i for \(r \in\mathbb{R} \). Thus, replacing g(i)=r i in (75) and factoring out r (i−2), we form the following characteristic polynomial:

(76)

Focusing on the worst case where (76) is satisfied with equality, we compute the roots r 1,2 of the quadratic characteristic polynomial as:

Then, as a general solution, we combine the above roots with unknown coefficients b 1,b 2 to obtain (69). Using the initial condition \(g(0) := \Vert \boldsymbol {X}(0) - \boldsymbol{X}^{\ast} \Vert _{F} \stackrel{\boldsymbol {X}(0) = \mathbf{0}}{=} \Vert \boldsymbol{X}^{\ast} \Vert _{F} = 1 \), we get b 1+b 2=1. Thus, we conclude to the following recurrence:

1.5 A.5 Proof of Lemma 7

Let \(\mathcal{D}_{i}^{\epsilon} \leftarrow\mathcal{P}_{k}^{\epsilon }(\mathcal{P}_{\mathcal{X}_{i}^{\bot}} \nabla f(\boldsymbol {X}(i))) \) and \(\mathcal{D}_{i} \leftarrow \mathcal{P}_{k}(\mathcal{P}_{\mathcal {X}_{i}^{\bot}} \nabla f(\boldsymbol {X}(i)))\). Using Definition 4, the following holds true:

(77)

Furthermore, we observe:

(78)

Here, we use the notation defined in the proof of Lemma 6. Since \(\mathcal{P}_{\mathcal{D}_{i}} \nabla f(\boldsymbol {X}(i)) \) is the best rank-k approximation to ∇f(X(i)), we have:

(79)

where \(\operatorname{rank}(\operatorname{span}(\text{ortho}(\mathcal{X}^{\ast}\setminus\mathcal{X}_{i}))) \leq k\). Using (77) in (79), the following series of inequalities are observed:

(80)

Now, in (78), we compute the series of inequalities in (81)-(82).

(81)
(82)

Focusing on \(\Vert \mathcal{P}_{\mathcal{X}^{\ast}\setminus \mathcal{X}_{i}}^{\bot} \boldsymbol {\mathcal {A}}^{\ast}(\boldsymbol {y}- \boldsymbol {\mathcal {A}}\boldsymbol {X}(i))\Vert _{F} \), we observe:

(83)

Moreover, we know the following hold true from Lemma 6:

(84)

and

(85)

Combining (83)–(85) in (82), we obtain:

1.6 A.6 Proof of Theorem 4

To prove Theorem 4, we combine the following series of lemmas for each step of Algorithm 1.

Lemma 12

[Error norm reduction via gradient descent]

Let \(\mathcal{S}_{i} \leftarrow\text{\textit{ortho}}(\mathcal{X}_{i} \cup\mathcal{D}_{i}^{\epsilon}) \) be a set of orthonormal, rank-1 matrices that span a rank-2k subspace in \(\mathbb{R}^{m \times n} \). Then (86) holds.

(86)

Proof

We observe the following:

(87)

The following equations hold true:

Furthermore, we compute:

(88)

where (i) is due to Lemmas 2, 4, 5 and \(\frac{1}{1+\delta_{2k}} \leq\mu_{i} \leq\frac {1}{1-\delta_{2k}} \).

Using the subadditivity property of the square root in (87), (88), Lemma 7 and the fact that \(\Vert \mathcal {P}_{\mathcal{S}_{i}}(\boldsymbol {X}(i) - \boldsymbol{X}^{\ast})\Vert _{F} \leq \Vert \boldsymbol {X}(i) - \boldsymbol{X}^{\ast} \Vert _{F} \), we obtain:

(89)

where \(\hat{\rho} := ( 1+ \frac{\delta_{3k}}{1-\delta _{2k}} ) (2\delta_{2k} + 2\delta_{3k} ) + \frac{2\delta_{2k}}{1-\delta_{2k}}\). □

We exploit Lemma 8 to obtain the following inequalities:

(90)

where the last inequality holds since W(i) is the best rank-k matrix estimate of V(i) and, thus, ∥W(i)−V(i)∥ F ≤∥V(i)−X F .

Following similar motions for steps 6 and 7 in Matrix ALPS I, we obtain:

(91)

Combining (91), (90) and (89), we obtain the desired inequality.

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Kyrillidis, A., Cevher, V. Matrix Recipes for Hard Thresholding Methods. J Math Imaging Vis 48, 235–265 (2014). https://doi.org/10.1007/s10851-013-0434-7

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