A Novel Framework for Wireless Digital Communication Signals via a Tensor Perspective

Abstract

In this paper, we introduce a novel signal characteristic, which is ubiquitous in the vast majority of communication signals. That is, a modulated signal is of an inherent low-rank structure following a reshaping operation. We first use a toy model to develop a framework for modelling this signal characteristic, and theoretically prove the impact on the signals’ rank structure by additive white Gaussian noise and inter-symbol interference, which is of great concern in the wireless communication field. Subsequently, the model is generalized to multi-input-multi-output signals, and tensor rank is taken into account. Using multi-linear algebra, we prove that the low-rankness of a reshaped signal only depends on the structure of its embedding subspace, and that its rank measure is upper bounded by the multi-rank of the basis tensor. As an application, we propose a novel adaptive sampling and reconstruction scheme for generic software-defined radio based on the low-rank structure. Numerical simulations demonstrate that the proposed method outperforms compressed sensing-based method, particularly when the modulated signal does not satisfy the sparsity assumption in the time and frequency domains. The results of practical experiments further demonstrate that many types of modulated signals can be effectively reconstructed from very limited observations using our proposed method.

Appendix: Proof of Theoretical Results

1.1 Proof of Theorem 2

To prove the theorem, we first introduce a lemma for a lower bound of the spectral norm of a product of two matrices.

Lemma 4

Let ${\mathbf {A}}\in {\mathcal {R}}^{m\times {}r}$ and ${\mathbf {B}}\in {\mathcal {R}}^{n\times {}r}$ be two full-rank matrices. Then,

$$\begin{aligned} \Vert {\mathbf {AB}}^{\mathrm {T}}\Vert _2\ge \min \left\{ \sigma _{\max }({\mathbf {A}})\sigma _{\min }({\mathbf {B}}),\sigma _{\max }({\mathbf {B}})\sigma _{\min }({\mathbf {A}})\right\} \end{aligned}$$

(32)

where $\sigma _{\max }(\cdot )$ and $\sigma _{\min }(\cdot )$ denote the largest and smallest singular of a matrix, respectively.

Proof

It is known that

$$\begin{aligned} \Vert {\mathbf {AB}}^{\mathrm {T}}\Vert _2=\sup _{\Vert {\mathbf {x}}\Vert =1} \Vert {\mathbf {AB}}^{\mathrm {T}}{\mathbf {x}}\Vert =\sup _{\Vert {\mathbf {x}}\Vert =1} \Vert {\mathbf {A}}{\mathbf {V}}_B{\mathbf {D}}_B{\mathbf {U}} _B^{\mathrm {T}}{\mathbf {x}}\Vert \end{aligned}$$

(33)

where ${\mathbf {B}}={\mathbf {U}}_B{\mathbf {D}}_B{\mathbf {V}}_B ^{\mathrm {T}}$ is the singular value decomposition (SVD) of ${\mathbf {B}}$. Hence, letting ${\mathbf {x}}={\mathbf {U}}_{\mathbf {B}}(:,1)$, which denotes the left singular vector corresponding the largest singular value, we have

$$\begin{aligned} \begin{aligned}&\Vert {\mathbf {AB}}^{\mathrm {T}}\Vert _2=\sup _{\Vert {\mathbf {x}}\Vert =1} \Vert {\mathbf {AB}}^{\mathrm {T}}{\mathbf {x}}\Vert \ge \sigma _{\hbox {max}}({\mathbf {B}})\Vert \mathbf {A}{\mathbf {V}}_ {\mathbf {B}}(:,1)\Vert \\&\quad \quad \quad \,\,\,\,\,\ge \sigma _{\hbox {max}}({\mathbf {B}})\sigma _{\hbox {min}}(\mathbf {A}) \end{aligned} \end{aligned}$$

(34)

Likewise, we have

$$\begin{aligned} \Vert {\mathbf {AB}}^{\mathrm {T}}\Vert _2\ge \sigma _{\hbox {max}}(\mathbf {A})\sigma _{\hbox {min}}({\mathbf {B}}) \end{aligned}$$

(35)

Combing inequality (34) and (35), the lemma is proved. $\square$

Next, the following lemma shows that the matrix nuclear norm would decrease if multiplying a diagonal matrix with the unit spectral norm.

Lemma 5

Consider that ${\mathbf {D}}\in {\mathcal {R}}^{n\times {}n}$ is a diagonal positive semi-definite matrix and $\Vert {\mathbf {D}}\Vert _2=1$. ${\mathbf {X}}\in {\mathcal {R}}^{n\times {}n}$ is any matrix. Then,

$$\begin{aligned} \Vert \mathbf {DX}\Vert _*\le {}\Vert {\mathbf {X}}\Vert _* \end{aligned}$$

(36)

Proof

$$\begin{aligned}&\Vert \mathbf {DX}\Vert _*\nonumber \\&\quad =\sup _{\Vert {\mathbf {Y}}\Vert _2=1}tr(\mathbf {DXY}^{{\mathrm {T}}})\nonumber \\&\quad =\sup _{\Vert {\mathbf {Y}}\Vert _2=1}\sum _{i=1}^n{\mathbf {D}}(i,i)tr\left( \mathbf {e}_i\mathbf {e}_i^{{\mathrm {T}}}\mathbf {XY}^{{\mathrm {T}}}\right) \end{aligned}$$

(37)

$$\begin{aligned}&\quad =\sup _{\Vert {\mathbf {Y}}\Vert _2=1}\sum _{i=1}^n{\mathbf {D}}(i,i)tr\left( \mathbf {e}_i\mathbf {e}_i^{{\mathrm {T}}}{\mathbf {X}} \sum _{j=1}^n\sigma _j\mathbf {v}_j\mathbf {u}_j^{{\mathrm {T}}}\right) \nonumber \\&\quad =\sup _{\Vert \mathbf {U\Sigma {}V}^{{\mathrm {T}}}\Vert _2=1}\sum _{i=1}^n\sum _{j=1}^n {\mathbf {D}}(i,i)\sigma _jtr\left( \mathbf {e}_i\mathbf {e}_i^ {{\mathrm {T}}}{\mathbf {X}}\mathbf {v}_j\mathbf {u}_j^{{\mathrm {T}}}\right) \nonumber \\&\quad \le \sup _{\Vert \mathbf {U\Sigma {}V}\Vert _2=1}\sum _{i=1}^n\sum _{j=1}^n {\mathbf {D}}(i,i)\sigma _j{}tr\left( \mathbf {e}_i^ {{\mathrm {T}}}{\mathbf {X}}\mathbf {v}_j\right) \nonumber \\&\quad \le \sup _{\Vert \mathbf {U\Sigma {}V}^{{\mathrm {T}}}\Vert _2=1}\sum _{i=1}^n\sum _{j=1}^n tr\left( \mathbf {e}_i^ {{\mathrm {T}}}{\mathbf {X}}\mathbf {v}_j\right) \nonumber \\&\quad =\sup _{\mathbf {VV}^{{\mathrm {T}}}={\mathbf {V}}^{{\mathrm {T}}} {\mathbf {V}}=\mathbf {I}}tr\left( \mathbf {XV}\right) \nonumber \\&\quad \le \sup _{\Vert {\mathbf {Z}}\Vert _2=1}tr(\mathbf {XZ}^{{\mathrm {T}}})\nonumber \\&\quad =\Vert {\mathbf {X}}\Vert _* \end{aligned}$$

(38)

In these formulas, $tr(\cdot )$ denotes calculating the trace of a matrix, $\left\{ \mathbf {e}_i,\,i=1,\ldots ,n\right\}$ denotes the standard basis for Euclidean space, and $\mathbf {Y=U\Sigma {}V}^{{\mathrm {T}}}$ denotes the SVD of ${\mathbf {Y}}$. Note that equation (37) holds because $tr(\cdot )$ is a linear function. Inequality (38) holds because $\Vert {\mathbf {D}}\Vert _2=\Vert {\mathbf {Y}}\Vert _2=1$. $\square$

Using Lemma 5, the following lemma provides an upper bound of the nuclear norm of a product of two matrices.

Lemma 6

Let ${\mathbf {A}}\in {\mathcal {R}}^{m\times {}r}$ and ${{\mathbf {B}}}\in {\mathcal {R}}^{n\times {}r}$ be any matrices. Then, it holds that

$$\begin{aligned} \Vert {{\mathbf {AB}}}^{{\mathrm {T}}}\Vert _*\le \Vert {\mathbf {A}}\Vert _2 \Vert {{\mathbf {B}}}\Vert _* \end{aligned}$$

(39)

Proof

First, let $k=\max \{m,n,r\}$, and square ${\mathbf {A}}$ and ${{\mathbf {B}}}$ into matrix $\bar{{\mathbf {A}}}\in {\mathcal {R}}^{k\times {}k}$ and $\bar{{{\mathbf {B}}}}\in {\mathcal {R}}^{k\times {}k}$, respectively, by using zero entries. Then, it is obvious that the spectral norm and nuclear norm are unchanged, namely,

$$\begin{aligned} \Vert {\mathbf {A}}\Vert _*=\Vert \bar{{\mathbf {A}}}\Vert _*, \Vert {\mathbf {A}}\Vert _2=\Vert \bar{{\mathbf {A}}}\Vert _2 \end{aligned}$$

(40)

and

$$\begin{aligned} \Vert {{\mathbf {B}}}\Vert _*=\Vert \bar{{{\mathbf {B}}}}\Vert _*, \Vert {{\mathbf {B}}}\Vert _2=\Vert \bar{{{\mathbf {B}}}}\Vert _2 \end{aligned}$$

(41)

Furthermore, it holds that

$$\begin{aligned} \Vert {{\mathbf {AB}}}^{{\mathrm {T}}}\Vert _*= \Vert \bar{{\mathbf {A}}}\bar{{{\mathbf {B}}}}^{{\mathrm {T}}}\Vert _* \end{aligned}$$

(42)

Hence,

$$\begin{aligned} \Vert {{\mathbf {AB}}}^{{\mathrm {T}}}\Vert _*= & {} \Vert \bar{{\mathbf {A}}}\bar{{{\mathbf {B}}}}^{{\mathrm {T}}}\Vert _*\\= & {} \sup _{\Vert {\mathbf {X}}\Vert _2=1}tr (\bar{{\mathbf {A}}}\bar{{{\mathbf {B}}}}^{{\mathrm {T}}} {\mathbf {X}}^{{\mathrm {T}}})\\\le & {} \sup _{\Vert {\mathbf {X}}\Vert _2=1}\sum _{i=1}^k \sigma _i(\bar{{\mathbf {A}}})\sigma _i ({\mathbf {X}}\bar{{{\mathbf {B}}}})\\&\quad \Vert {\mathbf {A}}\Vert _2 \sup _{\Vert {\mathbf {X}}\Vert _2=1}\Vert {\mathbf {X}}\bar{{{\mathbf {B}}}}\Vert _*\\= & {} \Vert {\mathbf {A}}\Vert _2 \sup _{\Vert \mathbf {UDV}^{{\mathrm {T}}}\Vert =1} \Vert \mathbf {DV}^{{\mathrm {T}}}\bar{{{\mathbf {B}}}}\Vert _*\\\le & {} \Vert {\mathbf {A}}\Vert _2\Vert {{\mathbf {B}}}\Vert _* \end{aligned}$$

Note that the last inequality holds by Lemma 5 and rotational invariance of the matrix nuclear norm. $\square$

To prove Theorem 2, we have the following equality by using Lemmas 4 and 6

$$\begin{aligned} \frac{\Vert {{\mathbf {AB}}}^{{\mathrm {T}}}\Vert _*}{\Vert {{\mathbf {AB}}}^{{\mathrm {T}}}\Vert _2}\le \frac{\Vert {\mathbf {A}}\Vert _*\sigma _{\max }\left( {{\mathbf {B}}}\right) }{\Vert {\mathbf {A}}\Vert _2\sigma _{\min }\left( {{\mathbf {B}}}\right) } =\frac{\Vert {\mathbf {A}}\Vert _*}{\Vert {\mathbf {A}}\Vert _2}cond({{\mathbf {B}}}) \end{aligned}$$

Likewise, we have

$$\begin{aligned} \frac{\Vert {{\mathbf {AB}}}^{{\mathrm {T}}}\Vert _*}{\Vert {{\mathbf {AB}}}^{{\mathrm {T}}}\Vert _2}\le =\frac{\Vert {{\mathbf {B}}}\Vert _*}{\Vert {{\mathbf {B}}}\Vert _2}cond({\mathbf {A}}) \end{aligned}$$

(43)

Combining the two inequalities, the theorem is proven. $\square$

1.2 Proof of Lemma 2

The probability

$$\begin{aligned}&{\mathbb {P}}\left( \Vert {\mathbf {X}}\Vert _2<\beta -\lambda \right) \\&\quad ={\mathbb {P}}\left( \sup _{\Vert {{\mathbf {x}}}\Vert =1}\Vert \beta {\mathbf {a}}<{\mathbf {p}},{{\mathbf {x}}}>+\sigma \mathbf {Nx}\Vert<\beta -\lambda \right) \\&\quad \le {\mathbb {P}}\left( \Vert \beta {\mathbf {a}}+\sigma \mathbf {Np}\Vert<\beta -\lambda \right) \\&\quad \le {\mathbb {P}}\left( \vert \beta -\sigma \Vert \mathbf {Np}\Vert \vert<\beta -\lambda \right) \\&\quad ={\mathbb {P}}\left( \frac{\lambda }{\sigma }<\Vert \mathbf {Np}\Vert <\frac{2\beta }{\sigma }-\frac{\lambda }{\sigma }\right) \\&\quad \le {\mathbb {P}}\left( \Vert \mathbf {Np}\Vert _2^2-K>\frac{\lambda ^2}{\sigma ^2}-K\right) \end{aligned}$$

Because $\Vert {\mathbf {Np}}\Vert ^2$ is ${\mathcal {X}}^2$ distributed with K degrees of freedom, then

$$\begin{aligned} {\mathbb {P}}\left( \Vert {\mathbf {Np}}\Vert ^2-K>\lambda _1\right) \le {}2e^{-4T} \end{aligned}$$

(44)

where $\lambda _1=2(\sqrt{4KT}+4T)$. Hence,

$$\begin{aligned} {\mathbb {P}}\left( \Vert {\mathbf {X}}\Vert _2 <\beta -\lambda \right) \le 2e^{-4T} \end{aligned}$$

(45)

where $\lambda =\sigma \sqrt{K+8T+4\sqrt{KT}}$. With some simple derivation, we can obtain

$$\begin{aligned} {\mathbb {P}}\left( \Vert {\mathbf {X}} \Vert _2\ge \beta -\sigma \sqrt{K+8T+4\sqrt{KT}}\right) >1-2e^{-4T} \end{aligned}$$

(46)

The lemma is proven. $\square$

1.3 Proof of Theorem 3

The probability

$$\begin{aligned} {\mathbb {P}}\left( \frac{\Vert {\mathbf {X}}\Vert _*}{\Vert {\mathbf {X}}\Vert _2}\ge \lambda \right)\approx & {} {\mathbb {P}}\left( \frac{\Vert {\mathbf {X}}\Vert _*}{\beta }\ge \lambda \right) \\= & {} {\mathbb {P}}\left( \frac{\Vert \beta \mathbf {ap}^{{\mathrm {T}}}+\sigma {\mathbf {N}}\Vert _*}{\beta }\ge \lambda \right) \\\le & {} {\mathbb {P}}\left( \frac{\beta +\sigma \Vert {\mathbf {N}}\Vert _*}{\beta }\ge \lambda \right) \\\le & {} {\mathbb {P}}\left( \frac{1+\sigma \min \left\{ K,T\right\} \Vert {\mathbf {N}}\Vert _2}{\beta }\ge \lambda \right) \\= & {} {\mathbb {P}}\left( \Vert {\mathbf {N}}\Vert _2\ge \frac{\beta \left( \lambda -1\right) }{\sigma \min \left\{ K,T\right\} }\right) \\\le & {} {}2e^{-T} \end{aligned}$$

where

$$\begin{aligned} \frac{\beta \left( \lambda -1\right) }{\sigma \min \left\{ K,T\right\} } =(4/3)\sqrt{K+8T+4\sqrt{KT}} \end{aligned}$$

(47)

namely,

$$\begin{aligned} \begin{aligned}&{\mathbb {P}}\left( \frac{\Vert {\mathbf {X}}\Vert _*}{\Vert {\mathbf {X}}\Vert _2}\ge {}1+\frac{4}{3}\min \left\{ K,T\right\} \left( \sigma /\beta \right) \sqrt{K+8T+4\sqrt{KT}}\right) \\&\le {}2e^{-T} \end{aligned} \end{aligned}$$

(48)

The theorem is proven. $\square$

1.4 Proof of Lemma 3

First, we define sets ${\mathcal {D}}_i, i=1,\ldots ,R$ as

$$\begin{aligned} {\mathcal {D}}_i=\left\{ x\Biggm |\,\vert {}x-{\mathbf {R}}_P(i,i)\vert \le \sum _{j\ne {}i}^R \vert {\mathbf {R}}_P(i,j)\vert \right\} ,\,i=1,\ldots ,R \end{aligned}$$

(49)

Using the assumption, we know

$$\begin{aligned} {\mathbf {R}}_P(1,1)-\sum _{i\ne {}1}\vert {\mathbf {R}}_P(1,i)\vert >\max _{j=2,\ldots ,R}\left\{ \sum _{i=1}^R \vert {\mathbf {R}}_P(j,i)\vert \right\} \end{aligned}$$

(50)

Hence, $\mathcal {D}_i$ is disjoint to other sets. It means that

$$\begin{aligned} {\mathcal {D}}_1\cap {\mathcal {D}}_i=\emptyset ,\,i=2,\ldots ,R \end{aligned}$$

(51)

Using Gershgorins Theorem, we can estimate the eigenvalues of ${\mathbf {R}}_P$. Because ${\mathcal {D}}_1$ is disjoint, then the largest eigenvalue of ${\mathbf {R}}_P$ must be contained in ${\mathcal {D}}_1$. Furthermore, it is known that the singular values of ${\mathbf {P}}$ are square roots of the corresponding eigenvalue of ${\mathbf {R}}_P$. Hence, the following inequalities hold

$$\begin{aligned} \begin{aligned}&\sqrt{\vert {\mathbf {R}}_P(1,1)\vert -\sum _{i\ne {}1}^R\vert {\mathbf {R}}_P(1,i)\vert }+(R-1)\sqrt{\min _{j\ne {}1} \left\{ \vert {\mathbf {R}}_P(j,j)\vert -\sum _{i\ne {}j}^R\vert {\mathbf {R}}_P(j,i)\vert \right\} }\\&\le \Vert {\mathbf {P}}\Vert _*=\sum _{i=1}^R \sigma _i\le \sqrt{\sum _{i=1}^R\vert {\mathbf {R}}_P(1,i)\vert }+(R-1)\sqrt{\max _{j\ne {}1}\left\{ \sum _{i=1}^R\vert {\mathbf {R}}_P(j,i)\vert \right\} } \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned}&\quad \sqrt{\vert {\mathbf {R}}_P(1,1)\vert -\sum _{i\ne {}1}^R\vert {\mathbf {R}}_P(1,i)\vert }\\&\le \Vert {\mathbf {P}}\Vert _2=\sigma _1\le \sqrt{\sum _{i=1}^R\vert {\mathbf {R}}_P(1,i)\vert } \end{aligned} \end{aligned}$$

(52)

By using these two inequalities, we can easily obtain the result of the lemma. $\square$

1.5 Proof of Theorem 5

Given flattening index $(\bar{a},\bar{b})$, let matrices ${\mathbf {X}}={\mathbf {X}}_{(\bar{a},\bar{b})}$ and ${\mathbf {P}}_k={\mathbf {P}}_{k,(\bar{a},\bar{b})},\,k=1,\ldots \,K$ for short. Moreover, assume that ${\mathbf {P}}_k$ can be decomposed as ${\mathbf {P}}_k={\mathbf {U}}_k{\mathbf {D}}_k{\mathbf {V}}_k^{{\mathrm {T}}}$ by SVD, where ${\mathbf {U}}_k\in {\mathcal {R}}^{M\times {}R_k}$, ${\mathbf {V}}_k\in {\mathcal {R}}^{T\times {}R_k}$, and diagonal matrix ${\mathbf {D}}_k\in {\mathcal {R}}^{R_k\times {}R_k}$ denotes left/right singular matrix and singular value matrix, respectively. By (24), ${\mathbf {X}}$ can be represented as

$$\begin{aligned} {\mathbf {X}}&= \left[ \begin{array}{ccc} {\mathbf {U}}_1&\ldots&{\mathbf {U}}_K \end{array} \right] \left[ \begin{array}{ccc} \alpha _1{\mathbf {D}}_1&{}\mathbf {0}&{}\mathbf {0}\\ \mathbf {0}&{}\ddots &{}\mathbf {0}\\ \mathbf {0}&{}\mathbf {0}&{}\alpha _K{\mathbf {D}}_K \end{array} \right] \\&\quad \cdot \left[ \begin{array}{ccc} {\mathbf {V}}_1&\ldots&{\mathbf {V}}_K \end{array} \right] ^{{\mathrm {T}}}\\& =\mathbf {UDV}^{{\mathrm {T}}} \end{aligned} $$

(53)

where operation $[{\mathbf {A}}_1\ldots {\mathbf {A}}_K]$ denotes the concatenation of ${\mathbf {A}}_k,\,k=1,\ldots ,K$. According the property of matrix rank, we know

$$\begin{aligned} rank({\mathbf {X}})\le \min \{{\mathbf {U}},{\mathbf {V}}\} \end{aligned}$$

(54)

Meanwhile, note that

$$\begin{aligned} & {\mathbf {P}}_{(\bar{a},\bar{b}\cup {}\{L+1\})}\\&\quad= \left[ \begin{array}{ccc} {\mathbf {P}}_1&\ldots&{\mathbf {P}}_K \end{array} \right] \\&\quad= \left[ \begin{array}{ccc} {\mathbf {U}}_1&\ldots&{\mathbf {U}}_K \end{array} \right] \left[ \begin{array}{ccc} {\mathbf {D}}_1{\mathbf {V}}_1^{{\mathrm {T}}}&{}\mathbf {0}&{}\mathbf {0}\\ \mathbf {0}&{}\ddots &{}\mathbf {0}\\ \mathbf {0}&{}\mathbf {0}&{}{\mathbf {D}}_K{\mathbf {V}}_K^{{\mathrm {T}}} \end{array} \right] \end{aligned}$$

(55)

and

$$\begin{aligned} \begin{aligned}&\quad {\mathbf {P}}_{(\bar{a}\cup {}\{L+1\},\bar{b})}\\&=\left[ \begin{array}{ccc} {\mathbf {P}}_1^{{\mathrm {T}}}&\ldots&{\mathbf {P}}_K^{{\mathrm {T}}} \end{array} \right] \\&=\left[ \begin{array}{ccc} {\mathbf {V}}_1&\ldots&{\mathbf {V}}_K \end{array} \right] \left[ \begin{array}{ccc} {\mathbf {D}}_1{\mathbf {U}}_1^{{\mathrm {T}}}&{}\mathbf {0}&{}\mathbf {0}\\ \mathbf {0}&{}\ddots &{}\mathbf {0}\\ \mathbf {0}&{}\mathbf {0}&{}{\mathbf {D}}_K{\mathbf {U}}_K^{{\mathrm {T}}} \end{array} \right] \end{aligned} \end{aligned}$$

(56)

It can be determined that

$$\begin{aligned} rank({\mathbf {P}}_{(\bar{a},\bar{b}\cup {}\{L+1\})})=rank({\mathbf {U}}) \end{aligned}$$

(57)

and

$$\begin{aligned} rank({\mathbf {P}}_{(\bar{a}\cup {}\{L+1\},\bar{b})})=rank({\mathbf {V}}) \end{aligned}$$

(58)

because the right block matrices are column full rank. Hence, the theorem is proven. $\square$

1.6 Proof of Theorem 7

From the assumptions used in the proof of Theorem 5, we have

$$\begin{aligned} \begin{aligned}&{\mathbf {X}}=\sum _{k=1}^K\alpha _k{\mathbf {P}}_k\\&=\left[ \begin{array}{ccc} {\mathbf {P}}_1&\ldots&{\mathbf {P}}_K \end{array} \right] \left[ \begin{array}{c} \alpha _1\mathbf {I}\\ \vdots \\ \alpha _K\mathbf {I} \end{array} \right] \\&={\mathbf {P}}_{(\bar{a},\bar{b}\cup {}\{L+1\})}{\mathbf {A}}^{{\mathrm {T}}} \end{aligned} \end{aligned}$$

(59)

It can be easily found that ${\mathbf {X}}$ is full rank and ${\mathbf {A}}$ can be decomposed using SVD such that

$$\begin{aligned} {\mathbf {A}}=\left( \left( \sum _{k=1}^K\alpha _i\right) ^{-\frac{1}{2}} \left[ \begin{array}{c} \alpha _1\mathbf {I}\\ \vdots \\ \alpha _K\mathbf {I} \end{array} \right] \right) \left( \left( \sum _{k=1}^K\alpha _i\right) ^{\frac{1}{2}}\mathbf {I}\right) \end{aligned}$$

(60)

Hence, we obtain that the condition number of ${\mathbf {A}}$ equals 1, namely, $cond({\mathbf {A}})=1$.

Using Theorem 2, we have

$$\begin{aligned} \frac{\Vert {\mathbf {X}}\Vert _*}{\Vert {\mathbf {X}}\Vert _2}\le \frac{\Vert {\mathbf {P}}_{(\bar{a},\bar{b}\cup {}\{L+1\})}\Vert _*}{\Vert {\mathbf {P}}_{(\bar{a},\bar{b}\cup {}\{L+1\})}\Vert _2}cond({\mathbf {A}}) \end{aligned}$$

(61)

Hence,

$$\begin{aligned} \frac{\Vert {\mathbf {X}}\Vert _*}{\Vert {\mathbf {X}}\Vert _2}\le \frac{\Vert {\mathbf {P}}_{(\bar{a},\bar{b}\cup {}\{L+1\})}\Vert _*}{\Vert {\mathbf {P}}_{(\bar{a},\bar{b}\cup {}\{L+1\})}\Vert _2} \end{aligned}$$

(62)

Likewise,

$$\begin{aligned} \frac{\Vert {\mathbf {X}}\Vert _*}{\Vert {\mathbf {X}}\Vert _2}\le \frac{\Vert {\mathbf {P}}_{(\bar{a}\cup {}\{L+1\},\bar{b})}\Vert _*}{\Vert {\mathbf {P}}_{(\bar{a}\cup {}\{L+1\},\bar{b})}\Vert _2} \end{aligned}$$

(63)

Combining these two inequalities, the theorem is proven.$\square$