1 Introduction

The low-rank matrix recovery problem has been a research hotpot recently [1, 2], and it has a range of applications in many fields such as signal or image processing [3, 4], subspace segmentation [5], collaborative filtering [6], and system identification [7]. Matrix rank minimization under affine equality constraints is generally formulated as follows:

$$ \min_{X} \operatorname{rank} ( X ) \quad\text{s.t.}\quad \mathrm{A} ( X ) = b, $$
(1.1)

where the linear map \(\mathrm{A}:R^{m \times n} \to R^{P}\) and the vector b are known.

Unfortunately, solving the above rank minimization problem (1.1) directly is an NP-hard problem [8], thus this problem is computationally infeasible. Therefore, the convex relations of these methods have been proposed and studied in the literature. For example, Recht et al. [8] proposed a nuclear norm minimization method for the matrix reconstruction. The tightest convex relaxation of problem (1.1) is the following nuclear norm minimization problem:

$$ \min_{X} \Vert X \Vert _{\bigstar} \quad\text{s.t.} \quad\mathrm{A} ( X ) = b, $$
(1.2)

where \(\Vert X \Vert _{\bigstar} = \sum_{i = 1}^{r} \sigma_{i} ( X )\) is the sum of all the singular values of \(X \in R^{m \times n}\) with \(\operatorname{rank} ( X ) = r\) (without loss of generality, \(n \le m\)). It has been shown that problem (1.2) shares common solutions with problem (1.1) under some sufficient conditions (see, e.g., [8, 9]).

However, the exact recovery of the low-rank matrix requires more measurements via nuclear norm minimization. Recently, some experimental observations and theoretical guarantees have shown the superiority of \(\ell_{p}\) quasi-norm minimization to \(\ell_{1}\) minimization in compressive sampling [10]. Therefore, the \(\ell_{p}\) quasi-norm minimization [11–13] was introduced instead of the nuclear norm minimization in order to give a better approximation to the original problem (1.1). Therefore, the \(\ell_{p}\) quasi-norm minimization can be formulated as

$$ \min_{X} \Vert X \Vert _{p}^{p}\quad \text{s.t.}\quad \mathrm{A} ( X ) = b, $$
(1.3)

where \(\Vert X \Vert _{p} = ( \sum_{i = 1}^{r} \sigma_{i}^{p} ( X ) )^{1/p}\) for some \(p \in ( 0,1 )\).

However, in practice, the observed data in the low-rank matrix recovery problem may be contaminated with noise, namely \(b = \mathrm{A}X + e\), where e contains measurement errors dominated by certain normal distribution. In order to recover the low-rank matrix robustly, problem (1.3) can be modified to

$$ \min_{X} \Vert X \Vert _{p}^{p} \quad\text{s.t.}\quad \bigl\Vert \mathrm{A} ( X ) - b \bigr\Vert _{2} \le \varepsilon, $$
(1.4)

where \(\Vert \cdot \Vert _{2}\) is the \(\ell_{2}\) norm of vector and \(\varepsilon\ge \Vert e \Vert _{2}\) is some constant.

Under some conditions, problems (1.3) and (1.4) can be rewritten as the following unconstrained model:

$$ \min_{X} \tau \Vert X \Vert _{p}^{p} + \frac{1}{2} \bigl\Vert \mathrm{A} ( X ) - b \bigr\Vert _{2}^{2}, $$
(1.5)

where \(\tau> 0\) is a given parameter. Since the above problem (1.5) is nonconvex and NP-hard, thus the researchers throughout the world proposed and analyzed some iterative reweighted algorithms [13–15]. The key idea of the iterative reweighted technique is to solve a convex problem with a given weight at each iteration and update the weight at every turn.

Different from previous studies, based on the weighted fixed point method, this paper puts forward a proximal iteratively reweighted algorithm to recover a low-rank matrix. Due to the special properties of nonconvex surrogate functions, the algorithm iteratively has a closed form solution to solve a weighted singular value thresholding problem. Also, in theory, this study has proved that the proximal iteratively reweighted algorithm decreases the objective function value monotonically, and any limit point is a stationary point.

The remainder of this paper is organized as follows. Section 2 introduces some notations and preliminary lemmas, and Section 3 describes the main results. The conclusion is followed in Section 4.

2 Preliminaries

Recently, Lai et al. [13] considered the following unconstrained problem:

$$ \min_{X} F ( X ) = \tau \operatorname{tr} \bigl( \bigl( X^{T}X + \varepsilon I \bigr)^{p/2} \bigr) + \frac{1}{2} \bigl\Vert \mathrm{A} ( X ) - b \bigr\Vert _{2}^{2}, $$
(2.1)

where I is the \(n \times n\) identity matrix and \(\varepsilon> 0\) is a smoothing parameter. By the definition in [13], we have

$$ \operatorname{tr} \bigl( \bigl( X^{T}X + \varepsilon I \bigr)^{p/2} \bigr) = \sum_{i = 1}^{n} \bigl( \sigma_{i} ( X )^{2} + \varepsilon \bigr)^{p/2}. $$
(2.2)

Lemma 2.1

([16])

Let \(\varphi ( X ) = \psi \circ\sigma ( X ) = \sum_{i = 1}^{n} ( \vert \sigma_{i} ( X ) \vert + \varepsilon )^{p}\), where the function \(\varphi: \mathbb{R}^{m \times n} \to [ - \infty, + \infty ]\) with \(n \le m\) is orthogonally invariant; \(\psi: \mathbb{R}^{m \times n} \to [ - \infty, + \infty ]\) is an absolutely symmetric function and \(p \in ( 0,1 )\), then \(\varphi= \psi\circ\sigma\) is subdifferentiable at matrix \(X \in\mathbb{R}^{m \times n}\) and

$$\partial\varphi ( X ) = pU\operatorname{Diag} \biggl\{ \frac{c_{i}}{ ( \sigma_{i} ( X ) + \varepsilon )^{1 - p}}:i \in\Omega \biggr\} V^{T} $$

with \(X = U\Sigma V^{T}\) being the SVD of X, and

$$c_{i} = \left \{ \textstyle\begin{array}{l@{\quad}l} 1, &\sigma_{i} ( X ) > 0, \\ {[ - 1,1 ]}, &\sigma_{i} ( X ) = 0, \end{array}\displaystyle \right . $$

is a constant depending only on the value of \(\sigma_{i} ( X )\) for each \(i \in\Omega\).

From Lemma 2.1, let \(m = n\) and the matrix Y be a semidefinite matrix, then \(Y = Y^{T}\) and the subdifferentiable of the function

$$ \varphi ( Y ) = \sum_{i = 1}^{n} \bigl( \bigl\vert \sigma _{i} ( Y ) \bigr\vert + \varepsilon \bigr)^{p/2} = \operatorname{tr} \bigl( ( Y + \varepsilon I )^{p/2} \bigr) $$
(2.3)

is

$$\partial\varphi ( Y ) = \frac{p}{2}U_{1}\operatorname{Diag} \biggl\{ \frac{c_{i}}{ ( \sigma_{i} ( Y ) + \varepsilon )^{1 - \frac{p}{2}}}:i \in\Omega \biggr\} U_{1}^{T}, $$

with \(Y = U_{1}\Sigma_{1}U_{1}^{T}\) being the SVD of Y, and \(\Omega= \{ 1,2, \ldots,n \}\).

From (2.3), it is easier to know exactly that \(\varphi ( Y )\) is concave, thus \(- \varphi ( Y )\) is convex. Besides, a vector \(Y^{*}\) is said to be a subgradient of a convex function f at a point Y if \(f(z) \ge f(Y) + \langle Y^{*},Y - x \rangle\), for any Z. Therefore, based on the definition of subgradient of the convex function, we have

$$ - \varphi ( Y ) \ge- \varphi ( Y_{k} ) + \langle- G_{k},Y - Y_{k} \rangle, $$
(2.4)

where \(- G_{k}\) is the subgradient of \(- \varphi ( Y )\) at \(Y_{k}\), i.e., \(- G_{k} \in\partial ( - \varphi ( Y_{k} ) )\). The inequality of (2.4) is equivalent to

$$ \varphi ( Y ) \le\varphi ( Y_{k} ) + \langle G_{k},Y - Y_{k} \rangle. $$
(2.5)

Then \(\varphi ( Y_{k} ) + \langle G_{k},Y - Y_{k} \rangle\) is used as a surrogate function of \(\varphi ( Y )\).

3 Main results

Let \(Y = X^{T}X\), then \(Y = V\Sigma^{2}V^{T}\) can be obtained, where \(X = U\Sigma V^{T}\) with \(U \in\mathbb{R}^{m \times n}\), \(V \in\mathbb{R}^{n \times n}\), and \(\Sigma= \operatorname{Diag} \{ \sigma_{i} ( X ) \} \in\mathbb{R}^{n \times n}\), then \(\sigma_{i} ( Y ) = ( \sigma_{i} ( X ) )^{2}\). From (2.2), (2.3), and (2.5),

$$ \operatorname{tr} \bigl( \bigl( X^{T}X + \varepsilon I \bigr)^{p/2} \bigr) \le \operatorname{tr} \bigl( \bigl( X_{k}^{T}X_{k} + \varepsilon I \bigr)^{p/2} \bigr) + \bigl\langle W_{k},X^{T}X - X_{k}^{T}X_{k} \bigr\rangle $$
(3.1)

can be obtained, where\(W_{k} \in\frac{p}{2}V\operatorname{Diag} \{ \frac{c_{i}}{ ( ( \sigma_{i} ( X_{k} ) )^{2} + \varepsilon )^{1 - \frac{p}{2}}}:i \in\Omega \}V^{T}\).

In order to introduce the following lemma, the definitions of Lipschitz continuous of a function and the norm \(\| \cdot\|_{F}\) are given, namely a function is Lipschitz continuous with constant L if, for any x, y, \(|f(x) - f(y)| \le L\|x - y\|\); and the \(\| \cdot\|_{F}\) of a matrix X is defined as \(\|X\|_{F}: = \sqrt{\sum_{i = 1}^{m} \sum_{j = 1}^{n} x_{ij}^{2}} \).

Lemma 3.1

([17])

Let \(f:\mathbb{R}^{m \times n} \to \mathbb{R}\) be a continuously differentiable function with Lipschitz continuous gradient and the Lipschitz constant \(L ( f )\). Then, for any \(L \ge L ( f )\),

$$ f ( X ) \le f ( Y ) + \bigl\langle \nabla f ( Y ),X - Y \bigr\rangle + \frac{L}{2} \Vert X - Y \Vert _{F}^{2},\quad \forall X,Y \in\mathbb{R}^{m \times n}. $$
(3.2)

Now let \(f ( X ) = \frac{1}{2} \Vert \mathrm{A} ( X ) - b \Vert _{2}^{2}\), thus the Lipschitz constant of the gradient \(\nabla f ( X ) = \mathrm{A}^{\bigstar} ( \mathrm{A} ( X ) - b )\) is \(L ( f ) = \lambda ( \mathrm{A}^{\bigstar} \mathrm{A} )\), where \(\lambda ( \mathrm{A}^{\bigstar} \mathrm{A} )\) is the maximum eigenvalue of \(\mathrm{A}^{\bigstar} \mathrm{A}\).

By using (2.1), (2.3), (3.1), and (3.2), we update \(X_{k + 1}\) by minimizing the sum of these two surrogate functions

$$ \begin{aligned}[b] X_{k + 1} ={}& \arg\min\varphi \bigl( X_{k}^{T}X_{k} \bigr) + \bigl\langle W_{k},X^{T}X - X_{k}^{T}X_{k} \bigr\rangle + f ( X_{k} ) + \bigl\langle \nabla f ( X_{k} ),X - X_{k} \bigr\rangle \\ &+ \frac{L ( f )}{2} \Vert X - X_{k} \Vert _{F}^{2} \\ ={}& \arg\min\tau \bigl\langle W_{k},X^{T}X \bigr\rangle + \frac{\rho}{ 2} \biggl\Vert X - \biggl( X_{k} - \frac{1}{\rho} \nabla f ( X_{k} ) \biggr) \biggr\Vert _{F}^{2}, \end{aligned} $$
(3.3)

where \(\rho\ge\frac{L ( f )}{2}\).

Lemma 3.2

If the function \(g ( X ) = \langle Q,X^{T}X \rangle\) with \(X \in\mathbb{R}^{m \times n}\) and \(Q \in\mathbb{R}^{n \times n}\), then the gradient of \(g ( X )\) is \(\nabla g ( X ) = 2XQ\).

Proof

Consider the auxiliary function \(\theta:\mathbb{R} \to \mathbb{R}\), given by \(\theta ( t ) = g ( X + tY )\), for any arbitrary matrix \(Y \in\mathbb{R}^{m \times n}\). From the basic calculus, it can be known that \(\theta' ( 0 ) = \langle \nabla g ( X ),Y \rangle\). By the definition of the derivative of function, it follows that

$$ \begin{aligned}[b] \theta' ( 0 )& = \lim _{t \to0}\frac{\theta ( t ) - \theta ( 0 )}{t} = \lim_{t \to 0} \frac{g ( X + tY ) - g ( X )}{t} \\ &= \lim_{t \to0}\frac{ \langle Q, ( X + tY )^{T} ( X + tY ) \rangle- \langle Q,X^{T}X \rangle}{t} \\ &= \lim_{t \to0}\frac{t \langle Q,X^{T}Y \rangle+ t \langle Q,Y^{T}X \rangle+ t^{2} \langle Q,Y^{T}Y \rangle}{t} \\ &= \bigl\langle Q,X^{T}Y \bigr\rangle + \bigl\langle Q,Y^{T}X \bigr\rangle = \operatorname{tr} \bigl( Q^{T}X^{T}Y \bigr) + \operatorname{tr} \bigl( Y^{T}XQ \bigr) = \langle2XQ,Y \rangle \end{aligned} $$
(3.4)

thus the gradient of \(g ( X )\) is \(\nabla g ( X ) = 2XQ\). □

Based on the above analysis, this paper proposes the following algorithm.

Algorithm 1

(Proximal iteratively reweighted algorithm to solve problem (2.1))

1::

Input: \(\rho\ge\frac{L ( f )}{2}\), where \(L ( f )\) is the Lipschitz constant of \(f ( x )\).

2::

Initialization: \(k = 0\), \(W_{k}\).

3::

Update \(X^{k + 1}\) by solving the following problem:

$$X_{k + 1} = \arg\min\tau \bigl\langle X^{T}X,W_{k} \bigr\rangle + \frac{\rho}{2} \bigg\Vert X - \biggl( X_{k} - \frac{1}{\rho} \mathrm {A}^{\bigstar} \bigl( \mathrm{A} ( X_{k} ) - b \bigr) \biggr) \bigg\Vert _{F}^{2}.$$
4::

Update the weight \(W_{k + 1}\) by

$$W_{k + 1} \in- \partial \bigl( - \varphi ( Y_{k + 1} ) \bigr),\quad\text{where } Y_{k + 1} = X_{k}^{T}X_{k}.$$
5::

Output low-rank matrix \(X^{k}\).

Theorem 3.3

Let \(\rho\ge\frac{L ( f )}{2}\), where \(L ( f )\) is the Lipschitz constant of \(f ( x )\). The sequence \(\{ X_{k} \}\) generated in Algorithm 1 satisfies

  1. (1)
    $$F ( X_{k} ) - F ( X_{k + 1} ) \ge \biggl( \rho- \frac{L ( f )}{2} \biggr) \Vert X_{k} - X_{k + 1} \Vert _{F}^{2}. $$
  2. (2)

    The sequence \(\{ X_{k} \}\) is bounded.

  3. (3)

    \(\sum_{k = 1}^{\infty} \Vert X_{k} - X_{k + 1} \Vert _{F}^{2} < \frac{2\theta}{2\rho- L ( f )}\). In particular, \(\lim_{k \to\infty} \Vert X_{k} - X_{k + 1} \Vert _{F} = 0\).

Proof

Since \(X_{k + 1}\) is the globally optimal solution of problem (3.3), and the zero matrix is contained in the subgradient with respect to X. That is, there exists a matrix \(X_{k + 1} \in\) such that

$$ 2\tau X_{k + 1}W_{k} + \nabla f ( X_{k} ) + \rho ( X_{k + 1} - X_{k} ) = 0. $$
(3.5)

By using the above equality of (3.4) and (3.5), we get

$$ 2\tau \langle X_{k + 1}W_{k},X_{k + 1} - X_{k} \rangle+ \bigl\langle \nabla f ( X_{k} ),X_{k + 1} - X_{k} \bigr\rangle + \rho \Vert X_{k + 1} - X_{k} \Vert _{F}^{2} = 0. $$
(3.6)

Since the function \(\langle W_{k},X^{T}X \rangle\) is a convex function on X, thus

$$\bigl\langle W_{k},X_{k}^{T}X_{k} \bigr\rangle - \bigl\langle W_{k},X_{k + 1}^{T}X_{k + 1} \bigr\rangle \ge2 \langle X_{k + 1}W_{k},X_{k} - X_{k + 1} \rangle, $$

and the above equality also can be rewritten as

$$ \bigl\langle W_{k},X_{k}^{T}X_{k} - X_{k + 1}^{T}X_{k + 1} \bigr\rangle \ge2 \langle X_{k + 1}W_{k},X_{k} - X_{k + 1} \rangle. $$
(3.7)

Then it follows from (3.6) and (3.7) that

$$ \tau \bigl\langle W_{k},X_{k}^{T}X_{k} - X_{k + 1}^{T}X_{k + 1} \bigr\rangle \ge- \bigl\langle \nabla f ( X_{k} ),X_{k} - X_{k + 1} \bigr\rangle + \rho \Vert X_{k + 1} - X_{k} \Vert _{F}^{2}. $$
(3.8)

Let \(f ( X ) = \frac{1}{2} \Vert \mathrm{A} ( X ) - b \Vert _{2}^{2}\), and according to Lemma 3.1,

$$ f ( X_{k} ) - f ( X_{k + 1} ) \ge \bigl\langle \nabla f ( X_{k} ),X_{k} - X_{k + 1} \bigr\rangle - \frac{L ( f )}{2} \Vert X_{k} - X_{k + 1} \Vert _{F}^{2} $$
(3.9)

can be obtained. Since the function \(\operatorname{tr} ( ( X^{T}X + \varepsilon I )^{p/2} )\) is concave, and just like (3.1), then it can be obtained

$$ \operatorname{tr} \bigl( \bigl( X_{k}^{T}X_{k} + \varepsilon I \bigr)^{p/2} \bigr) - \operatorname{tr} \bigl( \bigl( X_{k + 1}^{T}X_{k + 1} + \varepsilon I \bigr)^{p/2} \bigr) \ge \bigl\langle W_{k},X_{k}^{T}X_{k} - X_{k + 1}^{T}X_{k + 1} \bigr\rangle . $$
(3.10)

Now, combining (3.8), (3.9), and (3.10), we get

$$\begin{aligned} F ( X_{k} ) - F ( X_{k + 1} ) &= \tau \operatorname{tr} \bigl( \bigl( X_{k}^{T}X_{k} + \varepsilon I \bigr)^{p/2} \bigr) + f ( X_{k} ) - \tau \operatorname{tr} \bigl( \bigl( X_{k + 1}^{T}X_{k + 1} + \varepsilon I \bigr)^{p/2} \bigr) - f ( X_{k + 1} ) \\ &\ge \biggl( \rho- \frac{L ( f )}{2} \biggr) \Vert X_{k + 1} - X_{k} \Vert _{F}^{2} \ge0. \end{aligned} $$

Thus, \(F ( X_{k} )\) is monotonically decreasing. Given the facts of all inequalities above for \(k \ge1\), it can be obtained

$$ F ( X_{1} ) - F ( X_{k + 1} ) \ge \biggl( \rho- \frac{L ( f )}{2} \biggr)\sum_{i = 1}^{k} \Vert X_{i + 1} - X_{i} \Vert _{F}^{2}, $$
(3.11)

and from (3.11) it follows that

$$ \biggl( \rho- \frac{L ( f )}{2} \biggr)\sum_{i = 1}^{k} \Vert X_{i + 1} - X_{i} \Vert _{F}^{2} \le F ( X_{1} ) < + \infty. $$
(3.12)

Then, for \(k \to\infty\), (3.12) implies that

$$\lim_{k \to\infty} \Vert X_{k + 1} - X_{k} \Vert _{F} = 0. $$

Since the objective function \(F ( X )\) in problem (2.1) is nonnegative and satisfies

$$F ( X ) \to\infty,\quad \text{as }\Vert X \Vert _{F} \to\infty, $$

then \(X_{k} \in \{ X:0 \le F ( X ) \le F ( X_{1} ) \}\) and the sequence \(\{ X_{k} \}\) is bounded.

Therefore, the proof has been completed. □

Theorem 3.4

Let \(\{ X_{k} \}\) be the sequence generated in Algorithm 1. Then any accumulation point of \(\{ X_{k} \}\) is a stationary point \(X^{\bigstar} \) of the problem. Moreover, for \(k = 1,2, \ldots,N\), there always exists

$$\min_{1 \le k \le N} \Vert X_{k + 1} - X_{k} \Vert _{F}^{2} \le \frac{F ( X_{1} ) - F ( X^{\bigstar} )}{n ( \rho- \frac{L ( f )}{2} )}. $$

Proof

Since the sequence \(\{ X_{k} \}\) generated in Algorithm 1 is bounded, there exist an accumulation point \(X^{\bigstar} \) and a subsequence \(\{ X_{kj} \}\) such that \(\lim_{j \to\infty} X_{kj} = X^{\bigstar} \). Assume that \(X_{kj}\) is the solution of problem (3.3), it can be obtained

$$2\tau X_{kj + 1}W_{kj} + \nabla f ( X_{kj} ) + \rho ( X_{kj + 1} - X_{kj} ) = 0. $$

Let \(j \to\infty\), according to Theorem 3.3, \(\lim_{j \to\infty} \Vert X_{kj + 1} - X_{kj} \Vert _{F} = 0\) can be obtained. Hence, there exists the matrix

$$W^{\bigstar} = \frac{p}{2}V_{2}\operatorname{Diag} \biggl\{ \frac{1}{ ( ( \sigma_{i} ( X^{\bigstar} ) )^{2} + \varepsilon )^{1 - \frac{p}{2}}} \biggr\} V_{2}^{T} = \frac{p}{2} \bigl( \bigl( X^{\bigstar} \bigr)^{T}X^{\bigstar} + \varepsilon I \bigr)^{p/2 - 1}, $$

where \(X^{\bigstar} = U_{2}\Sigma_{2}V_{2}^{T}\) with \(U_{2} \in\mathbb{R}^{m \times n}\), \(V_{2} \in\mathbb{R}^{n \times n}\), and \(\sum= \operatorname{Diag} \{ \frac{1}{ ( ( \sigma_{i} ( X^{\bigstar} ) )^{2} + \varepsilon )^{1 - \frac{p}{2}}} \}\).

By the above analysis, it can be known that

$$\tau\rho X^{\bigstar} \bigl( \bigl( X^{\bigstar} \bigr)^{T}X^{\bigstar } + \varepsilon I \bigr)^{p/2 - 1} + \nabla f ( X_{\bigstar} ) = 0, $$

then \(X^{\bigstar} \) is a stationary point of problem (2.1).

Moreover, by using (3.11), for \(k = 1,2, \ldots,N\), it can be obtained

$$\begin{aligned} F ( X_{1} ) - F ( X_{N + 1} ) &\ge \biggl( \rho- \frac{L ( f )}{2} \biggr)\sum_{k = 1}^{N} \Vert X_{k + 1} - X_{k} \Vert _{F}^{2} \\ &\ge N \biggl( \rho- \frac{L ( f )}{2} \biggr)\min_{1 \le k \le N} \Vert X_{k + 1} - X_{k} \Vert _{F}^{2}. \end{aligned} $$

Thus

$$\begin{aligned} \min_{1 \le k \le N} \Vert X_{k + 1} - X_{k} \Vert _{F}^{2} &\le\frac{F ( X_{1} ) - F ( X_{N + 1} )}{n ( \rho- \frac{L ( f )}{2} )} \\ &\le\frac{F ( X_{1} ) - F ( X^{\bigstar} )}{n ( \rho- \frac{L ( f )}{2} )} \end{aligned} $$

can be obtained, which completes the proof. □

4 Conclusion

A proximal iteratively reweighted algorithm based on the weighted fixed point method for recovering a low-rank matrix problem has been presented in this paper. Due to the special properties of the nonconvex surrogate function, the algorithm in this study iteratively has a closed form solution to solving a weighted singular value thresholding problem. Finally, it has been proved that the algorithm can decrease the objective function value monotonically and any limit point is a stationary point.