On the Secrecy Capacity of a MIMO Gaussian Wiretap Channel with a Cooperative Jammer
aa r X i v : . [ c s . I T ] S e p On the Secrecy Capacity of a MIMO GaussianWiretap Channel with a Cooperative Jammer
Lingxiang Li, Zhi Chen, Jun Fang,
Member, IEEE , and Athina P. Petropulu,
Fellow, IEEE
Abstract —We study the secrecy capacity of a helper-assistedGaussian wiretap channel with a source, a legitimate receiver,an eavesdropper and an external helper, where each terminalis equipped with multiple antennas. Determining the secrecycapacity in this scenario generally requires solving a nonconvexsecrecy rate maximization (SRM) problem. To deal with thisissue, we first reformulate the original SRM problem into asequence of convex subproblems. For the special case of single-antenna legitimate receiver, we obtain the secrecy capacity via acombination of convex optimization and one-dimensional search,while for the general case of multi-antenna legitimate receiver, wepropose an iterative solution. To gain more insight into how thesecrecy capacity of a helper-assisted Gaussian wiretap channelbehaves, we examine the achievable secure degrees of freedom(s.d.o.f.) and obtain the maximal achievable s.d.o.f. in closed-form. We also derive a closed-form solution to the original SRMproblem which achieves the maximal s.d.o.f.. Numerical resultsare presented to illustrate the efficacy of the proposed schemes.
I. I
NTRODUCTION
The area of physical (PHY) layer security has been pio-neered by Wyner [1], who introduced the wiretap channel andquantified security with the maximal achievable secrecy rate(also known as the secrecy capacity) at which the legitimatereceiver can correctly decode the source message, while theeavesdropper can retrieve almost no information. Results in[2] further show that for the classical source-destination-eavesdropper Gaussian wiretap channel, the secrecy capacityis zero when the quality of the legitimate channel is worsethan that of the eavesdropper’s channel. One way to achievenon-zero secrecy rate in the latter case is to introduce externalhelpers which act as cooperative jammers [3]. By transmittingjamming signals the external helpers degrade the eavesdrop-per’s channel without hurting the legitimate channel, thusallowing secret communication even when the eavesdropper’schannel has a much better quality. Works along these linesinclude [4]–[7] which consider one external helper and [8]–[14] which consider the case of multiple external helpers. Morecomplex relaying scenarios are considered in [15]–[18] wherethe jamming signal is sent in the relaying phase, or in boththe broadcasting phase and the relaying phase.
Lingxiang Li, Zhi Chen, and Jun Fang are with the National KeyLaboratory of Science and Technology on Communications, University ofElectronic Science and Technology of China, Chengdu 610054, China (e-mails:lingxiang li [email protected]; { chenzhi, JunFang } @uestc.edu.cn)A. P. Petropulu is with the Department of Electrical and Computer Engi-neering, Rutgers–The State University of New Jersey, New Brunswick, NJ08854 USA (e-mail: [email protected]).This work was supported in part by the Important National Science andTechnology Specific Projects of China under Grant 2014ZX03004003, andby the Sichuan Province Project under Grant 2012FZ0119. Although cooperative jamming approaches improve the se-crecy rate, their advantage comes from optimally designedinput covariance matrices, which are difficult to obtain dueto the nonlinear fractional nature of the problem. To ad-dress this issue, for the single-antenna eavesdropper case,[10]–[12] propose a suboptimal but cost efficient null-spacejamming scheme that spreads the jamming signal within thenull-space of the legitimate receiver’s channel, while [6]–[9], [18] design algorithms to get the optimal solution usinga combination of convex optimization and one-dimensionalsearch. For the multi-antenna eavesdropper case, [13], [14]design the jamming signals so that they align into a pre-specified jamming subspace at the legitimate receiver, whilespanning the whole received signal space at the eavesdropper.This approach allows the legitimate receiver to completelyremove the interference by projecting the received signal tothe secrecy subspace, while confounding the eavesdropper.Still for the multi-antenna eavesdropper case, the work of[4] provides a closed-form expression for the structure of thejamming signal covariance matrix that guarantees secrecy ratelarger than the secrecy capacity of the wiretap channel withno jamming signals. The results of [4] are obtained under thepower covariance constraint.In this paper, we consider a multi-input multi-output(MIMO) Gaussian wiretap channel with one external multi-antenna helper as in [4]. Different from [4], we investigate thesecrecy rate maximization (SRM) problem under an averagepower constraint. To the best of the authors’ knowledge,determining the exact secrecy capacity of a helper-assistedMIMO Gaussian wiretap channel has not been previouslyaddressed. We first address the problem for the special caseof single-antenna legitimate receiver. By decomposing theoriginal nonconvex SRM problem into a sequence of convexsubproblems, we obtain the optimal solution to the originalSRM problem via a combination of convex optimization andone-dimensional search. For the general case of multi-antennalegitimate receiver, we propose an iterative algorithm to solvethe original SRM problem via employing the Gauss-Seidel ap-proach, which successively optimizes each variable while theother variables are kept fixed. Specifically, each subproblem isconvex and admits an optimal solution. Though the proposediterative algorithm provides no guarantee of finding the globaloptimal solution, it constitutes an efficient way for attaining ameaningful achievable secrecy rate.In order to gain more insight into how the secrecy capacityof a helper-assisted MIMO Gaussian wiretap channel behaves,we examine the rate at which the secrecy capacity scales with log( P ) , i.e., the maximal achievable secure degrees of freedom (s.d.o.f.) [19]. To this end, we first introduce an alternativeoptimization problem, i.e., maximizing the dimension of thesubspace spanned by the message signal received at thelegitimate receiver, under the constraints that the messageand jamming signals lie in different subspaces at the legiti-mate receiver, but are aligned into the same subspace at theeavesdropper. We then give a critical lemma, proving that themaximal achievable objective value of the newly introducedoptimization problem equals the maximal achievable s.d.o.f..Consequently, the original s.d.o.f. maximization reduces tothe newly introduced optimization problem. Subsequently, wesolve analytically the newly introduced optimization problem,thus obtaining the maximal achievable s.d.o.f. of the helper-assisted MIMO Gaussian wiretap channel in closed-form.Further, we derive an analytical solution to the original SRMproblem, which achieves the maximal s.d.o.f.. Our analyticalresults prove that for the special case of single-antenna legit-imate receiver, a s.d.o.f. of 1 can be achieved if and only if N e < N a + N j − ; for the case of multi-antenna legitimatereceiver, the maximal achievable s.d.o.f. is zero if and only if N e ≥ N a + N j .We should note that the s.d.o.f. for the helper-assistedGaussian wiretap channel has also been investigated in [14],[20]–[22]. Different from our work, the work of [14] studies ascenario in which a large number of helpers is available, andexploits multiuser diversity via opportunistic helper selectionto enhance security. The works of [20], [21] consider ascenario in which each terminal is equipped with one antenna,while the work of [22] considers the special scenario inwhich the source, the legitimate receiver and the eavesdropperare equipped with the same number of antennas. Further,the works of [20]–[22] examine the s.d.o.f. based on realinterference alignment, while our work is based on spatialinterference alignment.The rest of this paper is organized as follows. In Section II,we describe the system model for the MIMO Gaussian wiretapchannel with one external multi-antenna helper, and formulatethe secrecy rate maximization problem. In Section III, weconsider the special case of single-antenna legitimate receiver.We investigate the secrecy rate maximization problem, andexamine the conditions under which a secure degrees offreedom equal to 1 can be achieved. In Section IV, we considerthe general case of multi-antenna legitimate receiver, inves-tigate the secrecy rate maximization problem, and examinethe maximal achievable secure degrees of freedom. Numericalresults are provided in Section V and conclusions are drawnin Section VI. Notation: A H , tr { A } and rank { A } stand for the hermitiantranspose, trace and rank of the matrix A , respectively; A (: , j ) indicates the j -th column of A while and A (: , i : j ) denotesthe columns from i to j of A . span( A ) and span( A ) ⊥ arethe subspace spanned by the columns of A and its orthogonalcomplement, respectively; null( A ) denotes the null space of A ; span( A ) / span( B ) , { x | x ∈ span( A ) , x / ∈ span( B ) } . A (cid:23) B denotes that A − B is a hermitian positive semidefinitematrix. C N × M indicates a N × M complex matrix set. i ∈ Z denotes that i is a positive integer. I represents an identitymatrix with appropriate size. Besides, a + , max( a, ; | a | is the magnitude of a ; x ∼ CN (0 , Σ) means x is a randomvariable following a complex circular Gaussian distributionwith mean zero and covariance Σ .II. S YSTEM M ODEL AND P ROBLEM S TATEMENT
We consider the MIMO Gaussian wiretap channel with acooperative jammer (see Fig.1) where the source, the legit-imate receiver, the eavesdropper and the external helper areequipped with N a , N b , N e and N j antennas, respectively. Thesource wishes to send its message, x , to the legitimate receiver,without being eavesdropped by the eavesdropper. Towards thatobjective, the source is aided by a cooperative terminal, whichsimultaneously transmits jamming signal, z , to confuse theeavesdropper. The signals received at the legitimate receiverand the eavesdropper can be respectively expressed as y d = H Vx + G Wz + n d (1a) y e = G Vx + H Wz + n e , (1b)where V and W are the precoding matrices at the source andthe helper, respectively; n d ∼ CN ( , I ) and n e ∼ CN ( , I ) represent noise at the legitimate receiver and the eavesdropper,respectively; G ∈ C N b × N j and H ∈ C N e × N j representthe helper to legitimate receiver and the helper to eaves-dropper channel matrices, respectively; H ∈ C N b × N a and G ∈ C N e × N a denote the channel matrix from the sourceto the legitimate receiver and the source to the eavesdropper,respectively. All channels are assumed to be flat fading.We assume that global channel state information (CSI) isavailable, including the CSI for the eavesdropper. This ispossible in situations in which the eavesdropper is normally anactive member of the network, communicating nonconfidentialinformation with the other parties in other time slots [4]. Aminimum-Mean-Square-Error (MMSE) receiver is consideredat the legitimate receiver and the eavesdropper. The rate at thelegitimate receiver and the eavesdropper can be respectivelyexpressed as R d = log | I + ( I + G Q j G H ) − H Q a H H | (2a) R e = log | I + ( I + H Q j H H ) − G Q a G H | , (2b)where Q a , VV H and Q j , WW H are the transmitcovariance matrices at the source and the helper, respectively.In the paper, we focus on the SRM problem [23], i.e., C s , max { Q a (cid:23) , Q j (cid:23) } R d − R e s.t. tr { Q a + Q j } ≤ P , (3)where P is a given total transmit power budget and C s denotesthe maximal achievable secrecy rate, also known as the secrecycapacity.Generally, the optimization problem of (3) is nonconvex.It is challenging and still an open problem to determine theexact secrecy capacity. In this paper, we propose to solve theproblem of (3) by reformulating it into a sequence of convexproblems. Also, we study the rate at which the secrecy capacity For a given point { Q a , Q j } , the achieved secrecy rate is max( R d − R e , . For ease of exposition, the trivial case with zero achievable secrecyrate is omitted. Source Legitimate receiverEavesdropper G H Helper •• •• •• a N j N e N •• b N H G Fig. 1: MIMO wiretap channel with an external helperscales with log( P ) , i.e., the maximal achievable s.d.o.f. [19],which equals s.d.o.f , lim P →∞ C s log P . (4)We compute s.d.o.f analytically and determine its connectionto system parameters, i.e., the number of antennas at eachterminal.In the following sections, we will begin with the simple casewhere the legitimate receiver is equipped with one antenna.Then, a more complicated scenario, in which each terminal isequipped with multiple antennas will be investigated.III. H
ELPER -A SSISTED
MISOME W
IRETAP C HANNEL
In this section, we consider the helper-assisted multi-inputsingle-output multi-antenna-eavesdropper (MISOME) wiretapchannel where the legitimate receiver is equipped with a singleantenna ( N b = 1 ). In such case, the legitimate receiver canreceive at most one data stream. Thus, the source transmitsone data stream x . Let v denote the precoding vector at thesource. The signals received at the legitimate receiver and theeavesdropper can be respectively expressed as y d = h v x + g Wz + n d (5) y e = G v x + H Wz + n e , (6)where h ∈ C × N a denotes the channel vector from the sourceto the legitimate receiver, and g ∈ C × N j represents thechannel vector from the helper to the legitimate receiver. Therate at the legitimate receiver and the eavesdropper can besimplified as, R d = log(1 + (1 + g Q j g H ) − h vv H h H ) (7a) R e = log(1 + v H G H ( I + H Q j H H ) − G v ) . (7b)Correspondingly, the secrecy capacity equals C s = max { v , Q j (cid:23) } R d − R e s.t. tr { vv H + Q j } ≤ P. (8)Due to the presence of the multi-antenna eavesdropper,the SRM problem becomes more complex as compared with the problem considered in [6]–[8], [10]. To cope with thisissue, we resort to theorems on partial ordering of hermi-tian matrix [24]. Based on these theorems, we transformthe original matrix inverse constraint into a convex linearmatrix inequality (LMI) constraint, which, together with thesemidefinite relaxation (SDR) technique, enables us to recastthe original nonconvex optimization problem into a sequenceof semidefinite programmes (SDPs). Further, we prove thatthe optimal solutions to the relaxed optimization problemare also optimal solutions to the original SRM problem.Consequently, we obtain the optimal solution to the originalSRM problem with a combination of convex optimizationand one-dimensional search. On the other hand, to gain moreinsight into how the secrecy capacity of the helper-assistedMISOME Gaussian wiretap channel behaves, we examine theconditions under which a s.d.o.f. equal to 1 can be achieved.To this end, we first introduce an alternative optimizationproblem which keeps the message signal and the jammingsignal into different subspaces at the legitimate receiver, butaligns them into the same subspace at the eavesdropper. Wethen give two key lemmas. Lemma 1 proves that, a s.d.o.f.equal to 1 can be achieved if and only if the newly introducedoptimization problem returns a nonempty set. Lemma 2 givesthe conditions under which the newly introduced optimizationproblem returns a nonempty set. Combining the two lemmas,we finally show that a s.d.o.f. equal to 1 can be achieved ifand only if N e < N a + N j − . A. Secrecy rate maximization
To solve the SRM problem in (8), the Two-Layer idea of [8]is adopted. The key insight is to recast the original optimiza-tion problem in (8) as a two-level optimization problem. Theinner-level part is dealt with the SDR technique, and the outer-level part is handled by one-dimensional search. Specifically,the outer-level part is max τ ∈ [ τ lb ,τ ub ] log(1 + g ( τ )) − log(1 + τ ) , (9)where g ( τ ) is obtained by solving the following inner-levelpart optimization problem for a fixed τ : g ( τ ) = max { v , Q j (cid:23) } h vv H h H / (1 + g Q j g H ) (10a) s . t . v H G H ( I + H Q j H H ) − G v ≤ τ (10b) tr { vv H + Q j } ≤ P. (10c)By performing one-dimensional search on τ , the optimal τ ⋆ maximizing the objective function in (9) can be found.Correspondingly, the optimal solution { v ⋆ , Q ⋆j } to the originaloptimization problem of (8) can be obtained.In (9), τ lb and τ ub denote the lower and upper bound on γ e , respectively. Firstly, it is obvious that γ e is no less than0. Thus, we have τ lb = 0 . Secondly, according to the securityrequirement, γ e should be no more than γ d . Further, γ d isupper bounded by the maximal received signal-to-noise ratio(SNR) value of P | h | at the legitimate receiver. Therefore, τ ub = P | h | .So far, τ lb and τ ub have been determined. In the following,we focus on solving the optimization problem of (10), which is still nonconvex. To solve it, we resort to the SDR techniqueof [25]. On denoting Q a = vv H and dropping the rank-oneconstraint, the optimization problem of (10) can be rewrittenas f ( τ ) = max { Q a (cid:23) , Q j (cid:23) } h Q a h H / (1 + g Q j g H ) (11a) s . t . G Q a G H (cid:22) τ ( I + H Q j H H ) (11b) tr { Q a + Q j } ≤ P , (11c)where the replacement of the constraint (10b) with (11b) canbe proven using basic theorems on partial ordering [24] asfollows:(10b) ⇔ λ max (( I + H Q j H H ) − G vv H G H ) ≤ τ ⇔ ( I + H Q j H H ) − G vv H G H (cid:22) τ I ⇔ G vv H G H (cid:22) τ ( I + H Q j H H ) ⇔ (11b) . In the above, λ max ( A ) denotes the maximum eigenvalue of A .Letting ξ = (1 + g Q j g H ) − > , ˜Q a = ξ Q a , ˜Q j = ξ Q j ,and using the Charnes-Cooper transformation [26], we canrecast the optimization problem of (11) as f ( τ ) = max { ˜Q a (cid:23) , ˜Q j (cid:23) ,ξ> } h ˜Q a h H s . t . ξ + g ˜Q j g H = 1 G ˜Q a G H (cid:22) τ ( ξ I + H ˜Q j H H )tr { ˜Q a + ˜Q j } ≤ ξP , (12)which is a SDP and can be efficiently solved using availablesoftware packages, e.g., CVX [26].Let us consider the power minimization problem associatedwith (11), which can be formulated as follows: min { Q a , Q j } tr { Q a + Q j } s . t . h Q a h H / (1 + g Q j g H ) ≥ f ( τ ) G Q a G H (cid:22) τ ( I + H Q j H H ) Q a (cid:23) , Q j (cid:23) , (13)where f ( τ ) is obtained by solving the optimization problemof (12). We have the following two propositions. Proposition 1:
Denote the optimal solution to (13) as { ˆ Q a , ˆ Q j } . Then, ˆ Q a is rank-one provided that a positivesecrecy rate is achieved. Proof:
See Appendix A.
Proposition 2:
Denote the optimal solution to (13) as { ˆ Q a , ˆ Q j } . Then, { ˆ Q a , ˆ Q j } is also the optimal solution tothe problem of (11). Proof:
See Appendix B.Let Q oa = ˆ Q a and Q oj = ˆ Q j . Combining Proposition 1 withProposition 2, we get that { Q oa , Q oj } is the optimal solution tothe problem of (11), such that rank { Q oa } = 1 . Therefore, theoptimization problem of (11) is indeed a tight approximationof the optimization problem of (10). Moreover, { Q oa , Q oj } isalso the optimal solution to the problem of (10) and g ( τ ) = f ( τ ) . B. Conditions to ensure s.d.o.f. equal to 1
As stated in the preceding sections, it is difficult to obtain ananalytical expression for the secrecy capacity for the helper-assisted Gaussian wiretap channel. Instead, in this subsection,we investigate the conditions under which a s.d.o.f. equal to 1can be achieved. To this end, we first introduce an alternativeoptimization problem as follows: find { v , W } (14a) s . t . span( G v ) ⊂ span( H W ) (14b) span( g W ) ∩ span( h v ) = { } (14c) | h v | > . (14d)Specifically, we aim to find the point at which the subspacespanned by the message signal and that spanned by thejamming signal have no intersection at the legitimate receiver,such that R d scales with log( P ) . Simultaneously, the sub-space spanned by the message signal belongs to the subspacespanned by the jamming signal at the eavesdropper, such that R e converges to a constant as P approaches to infinity.In the sequel, we first give two key lemmas. Lemma 1proves that s.d.o.f. equal to 1 can be achieved if and only if theoptimization problem of (14) returns a nonempty set. Lemma2 gives the conditions under which the optimization problemof (14) returns a nonempty set. Combining the two lemmas,we finally obtain the conditions to ensure s.d.o.f equal to 1 inthe helper-assisted MISOME Gaussian wiretap channel. Lemma 1: The secure degrees of freedom equal to 1 canbe achieved if and only if the optimization problem of (14)returns a nonempty set.Proof:
Clearly, if the optimization problem of (14) returnsa nonempty set, then s.d.o.f. equal to 1 can be achieved. Sothe sufficiency holds true.We now turn to prove the necessity by contradiction. If theoptimization problem of (14) returns an empty set, then atleast one of the constraints in (14) does not hold true. We test(14b)-(14d) one by one:1) If (14b) does not hold true, then there exists a directionalong which the eavesdropper can extract the messagesignal without interference, so the rate at which R e scales with log( P ) is 1. Together with (4),(8) and thefact that the rate at which R d scales with log( P ) is atmost 1 for the multi-input single-output (MISO) source-receiver channel, we arrive at s.d.o.f. = 0 .2) If (14c) does not hold true, then the message signal isaligned in the subspace spanned by the jamming signal,so R d converges to a constant when P approaches toinfinity, which indicates that s.d.o.f. = 0 .3) If (14d) does not hold true, then | h v | = 0 , whichindicates that R d = 0 , thus s.d.o.f. = 0 .Summarizing, if the optimization problem of (14) returns anempty set, s.d.o.f. = 0 . Therefore, if s.d.o.f. = 1 , theoptimization problem of (14) returns a nonempty set. Thiscompletes the proof.Before proceeding to Lemma 2, we first introduce thegeneralized singular value decomposition (GSVD) transform,which provides the basis for the proof of Lemma 2 to follow. Definition 1 (GSVD Transform) : Given two matrices H ∈ C N × M and G ∈ C N × K , let k , rank { [ H H , G H ] T } (15a) p , dim { span( H ) ⊥ ∩ span( G ) } (15b) r , dim { span( H ) ∩ span( G ) ⊥ } (15c) s , dim { span( H ) ∩ span( G ) } , (15d)then we have k = min { M + K, N } (16a) p = k − min { M, N } (16b) r = k − min { K, N } (16c) s = min { M, N } + min { K, N } − k. (16d)The proof is given in Appendix C. According to [27], theGSVD of ( H H , G H ) returns unitary matrices Ψ ∈ C M × M and Ψ ∈ C K × K , non-negative diagonal matrices D ∈ C M × k and D ∈ C K × k , and a matrix X ∈ C N × k with rank { X } = k , such that HΨ = XD H (17a) GΨ = XD H , (17b)in which D = I r
00 0 0 , D =
00 0 I p ,where the diagonal entries of S ∈ R s × s and S ∈ R s × s aregreater than 0, and D H D + D H D = I .For simplicity, in the following text of this paper, we denotethe above GSVD Transform as ( Ψ , Ψ , D , D , X , k, r, s, p ) = gsvd( H H , G H ) . Lemma 2: The optimization problem of (14) returns anonempty set if and only if N e < N a + N j − . Proof:
We start with the constraint of (14c). With the
GSVD Transform of ( h H , g H ) , we get s , dim { span( h ) ∩ span( g ) } = 1 . Thus to satisfy (14c), we should have | h v | =0 or | g W | = 0 . However, | h v | = 0 contradicts (14d), sowe should have | g W | = 0 .Without loss of generality, let W = ΓW where Γ =null { g } ∈ C N j × ( N j − , W ∈ C ( N j − × ( N j − . Substitut-ing W = ΓW into (14b), we arrive at span( G v ) ⊂ span( H Γ W ) , (18)in which G ∈ C N e × N a and ¯ H , H Γ ∈ C N e × ( N j − .Invoking the GSVD Transform of ( ¯ H H , G H ) , we obtain ( ¯ Ψ , ¯ Ψ , ¯ D , ¯ D , ¯ X , k , r , s , p ) = gsvd( ¯ H H , G H ) ,such that ¯ H ¯ Ψ = ¯ X ¯ D H (19a) G ¯ Ψ = ¯ X ¯ D H , (19b)where k = min { N a + N j − , N e } , p = k − min { N j − , N e } , r = k − min { N a , N e } and s = min { N a , N e } +min { N j − , N e } − k . To satisfy (14d), we should have v = , which, togetherwith (19a) and (19b), indicates that (18) holds true if andonly if p < N a .1) For the case of N e ≤ N j − , p = N e − N e = 0 . So p < N a .2) For the case of N j − < N e < N a + N j − , p = N e − ( N j − < N a .3) For the case of N e ≥ N a + N j − , p = N a + N j − − ( N j −
1) = N a .Summarizing, p < N a if and only if N e < N a + N j − .Therefore, if N e < N a + N j − , (18) holds true, thusthe optimization problem of (14) returns a nonempty set.Otherwise, if N e ≥ N a + N j − , (18) does not hold true,so the optimization problem of (14) returns an empty set. Ina word, the optimization problem of (14) returns a nonemptyset if and only if N e < N a + N j − . This completes the proof. Throrem 1: The secure degrees of freedom equal to 1 canbe achieved if and only if N e < N a + N j − . Proof:
Combining Lemma 1 with Lemma 2, it is clearthat the secure degrees of freedom equal to 1 can be achievedif and only if N e < N a + N j − . This completes the proof. Remark:
It is worthwhile to note that Lemma 2 also providesus with a way to determine the precoding matrices at thesource and the helper to achieve s.d.o.f. of 1. In the remainingtext of this subsection, we first give the precoding matricesto achieve s.d.o.f. of 1 in closed-form. We then substitute thederived precoding matrices into (8) and solve for the optimalpower allocation between the message signal and the jammingsignal. Consequently, we obtain an analytical lower bound onthe secrecy capacity.Revisiting the proof of Lemma 2, for the case of N e
MIMOME W
IRETAP C HANNEL
In this section, we consider the helper-assisted MIMOMEwiretap channel where each terminal is equipped with multipleantennas. In such MIMO case, the SRM problem becomesmore complex as compared with the problem considered inSection III, and actually, it is still an open problem. To dealwith this issue, we first reformulate the SRM problem in (3) toa form that can be processed with the Gauss-Seidel approach,which successively optimizes each variable given that theother variables are fixed, thus giving an iterative algorithmto solve (3). We then examine the maximal achievable s.d.o.f.and reveal its connection to the number of antennas at eachterminal. We obtain both the maximal achievable s.d.o.f. andthe solution that achieves the maximal s.d.o.f. in closed-form.
A. Secrecy rate maximization
In order to reformulate the SRM problem in (3) to a formthat can be processed with the Gauss-Seidel approach, we needthe following lemma.
Lemma 3: Given a positive definite matrix E ∈ C N × N , itholds that ln | E − | = max S ∈ N × N , S (cid:23) ϕ ( S ) , (26) where ϕ ( S ) = − tr( SE )+ln | S | + N . Moreover, for the optimalsolution to the right-hand side of (26), it holds that S ⋆ = E − .Proof: Please refer to [28]. Applying Lemma 3, we arrive at, respectively, ln (cid:12)(cid:12) ( I + G Q j G H ) − (cid:12)(cid:12) = max S (cid:23) ϕ b ( S ) (27a) ln (cid:12)(cid:12)(cid:12) ( I + H Q j H H + G Q a G H ) − (cid:12)(cid:12)(cid:12) = max S (cid:23) ϕ e ( S ) , (27b)where ϕ b ( S ) = − tr { S ( I + G Q j G H ) } + ln | S | + N b , and ϕ e ( S ) = − tr { S ( I + H Q j H H + G Q a G H ) } +ln | S | + N e .Substituting (27a) and (27b) into (2a) and (2b), respectively,we arrive at R d = max S (cid:23) ϕ b ( S ) + ln (cid:12)(cid:12) I + H Q a H H + G Q j G H (cid:12)(cid:12) (28a) R e = − max S (cid:23) ϕ e ( S ) − ln (cid:12)(cid:12) I + H Q j H H (cid:12)(cid:12) . (28b)Further, substituting (28a)(28b) into (3), we arrive at C s = max { Q a (cid:23) , Q j (cid:23) , S (cid:23) , S (cid:23) } θ ( S , S , Q a , Q j ) s.t. tr { Q a + Q j } ≤ P , (29)where θ ( S , S , Q a , Q j ) = ϕ ( S ) + ϕ e ( S ) + ω ( Q a , Q j ) in which ω ( Q a , Q j ) = ln (cid:12)(cid:12) I + H Q a H H + G Q j G H (cid:12)(cid:12) +ln (cid:12)(cid:12) I + H Q j H H (cid:12)(cid:12) .Although the optimization problem of (29) is still not con-vex, we observe that if we fix either { Q a , Q j } or S i ( i = 1 , ,the remaining problem is convex and can thus be solvedefficiently. Hence, we turn to a two-stage iterative method(Gauss-Seidel approach), and approximately solve the opti-mization problem of (29) via iterations between the followingtwo subproblems.1) Fix { Q a , Q j } , and maximize θ ( S , S , Q a , Q j ) over { S , S } .2) Fix { S , S } , and maximize θ ( S , S , Q a , Q j ) over { Q a , Q j } .Specifically, when { Q a , Q j } is fixed, let { S ⋆ , S ⋆ } = arg max { S , S } θ ( S , S , Q a , Q j ) . Applying Lemma 3, we arrive at S ⋆ = ( I + G Q j G H ) − , S ⋆ = ( I + H Q j H H + G Q a G H ) − . Besides, when { S , S } is fixed, the maximization of θ ( S , S , Q a , Q j ) over { Q a , Q j } is a convex optimizationproblem and can be efficiently solved by available softwarepackages, e.g., CVX [26].One can easily verify that the above iterative process leadsto a monotonically non-descending objective function valueof θ ( S , S , Q a , Q j ) . Moreover, for a given limited transmitpower, the achievable secrecy rate is upper bounded. Thus, theabove iterative algorithm is convergent. Remark:
Although the above iterative algorithm provides noguarantee of finding the global optimal solution to the problemof (29), our numerical results in the following section showthat it attains a fairly good secrecy rate performance.
B. Maximal achievable secure degrees of freedom
In this subsection, we examine the maximal achievables.d.o.f. and determine its connection to the number of antennasat each terminal. Similar to Section III, we first introduce analternative optimization problem as follows: d , max { V , W } rank { H V } (30a) s . t . span( G V ) ⊂ span( H W ) (30b) span( G W ) ∩ span( H V ) = { } . (30c)Specifically, we find the feasible points at which the subspacespanned by the message signal and that spanned by thejamming signal have no intersection at the legitimate receiver.Simultaneously, the subspace spanned by the message signalbelongs to the subspace spanned by the jamming signal at theeavesdropper. Among these feasible points, we determine theone at which the rank of H V is maximized. Lemma 4: The maximal achievable secure degrees of free-dom, defined in (4), equal to d . That is, s.d.o.f = d .Proof: See Appendix EFrom Lemma 4, we observe that to obtain the s.d.o.f , weneed only to focus on solving (30). To this end, in the sequel,we first give a heuristic method which gives a closed-formfeasible point { ˆ V , ˆ W } to (30), followed by the derivation of d ⋆ , rank { H ˆ V } in closed-form. Subsequently, in Lemma 5,we prove that d = d ⋆ . Combining Lemma 4 and Lemma 5, wefinally obtain s.d.o.f = d ⋆ . Further, we prove that { ˆ V , ˆ W } achieves the maximal s.d.o.f., i.e., { ˆ V , ˆ W } constituting thes.d.o.f.-optimal solution to the original SRM problem.The aforementioned heuristic method to obtain { ˆ V , ˆ W } isshown in Table I. Notice that in Table I, null { G } returns anempty matrix when N a ≤ N e . In the following text, we provethat { ˆ V , ˆ W } is a feasible solution for (30), and derive theclosed-form expression for d ⋆ . As in Table I, four cases arediscussed.In Case I and
Case II , it is clear that { ˆ V , ˆ W } is feasibleto (30) and d ⋆ = rank { H ˆ V } = min { N a , N b } .In Case III , for the subcase of d + d ≥ N b , ˆ V =[ V , V ] and ˆ W = W . According to (31), we get span( G W ) = { } and span( H W ) = span( G V ) .In addition, G V = . So, span( H ˆ W ) = span( G ˆ V ) and span( H ˆ V ) T span( G ˆ W ) = { } , which indicate that { ˆ V , ˆ W } is feasible to (30). Furthermore, V is orthogonalwith V by definition, thus d ⋆ =rank { [ V , V ] } =rank { V } + rank { V } = N b . For the subcase of d + d < N b , ˆ V = [ V , V , V ] and ˆ W =[ W , W ] . As in the subcase of d + d ≥ N b , G V = , span( G W ) = { } and span( H W ) = span( G V ) . Inaddition, according to (32), span( H W ) = span( G V ) and d = min { s , ⌊ N b − ( d + d )2 ⌋} . Thus span( H ˆ W ) =span( G ˆ V ) and span( H ˆ V ) T span( G ˆ W ) = { } . There-fore { ˆ V , ˆ W } is feasible to (30). Furthermore, [ V , V ] is TABLE I: A heuristic method to obtain { ˆ V , ˆ W } which isfeasible to (30) Case I: N a ≥ N e + N b . Let ˆ V = null { G } and ˆ W = . Case II: N j ≥ N b + N e . Let ˆ W = null { G } ∈ C N a × ( N j − N b ) and ˆ V be the right singular matrix of H . Case III: N a < N e + N b and N b < N j < N e + N b . For a start, let V = null { G } and d = ( N a − N e ) + . Secondly, denote ¯ H = H Γ ∈ C N e × ( N j − N b ) where Γ = null { G } ∈ C N a × ( N j − N b ) .Denote ¯ G = G V c where V c = null { V H } ∈ C N a × N e . Invokingthe GSVD Transform of ( ¯ H H , ¯ G H ) yields ( Ψ , Ψ , D , D , X , k , r , s ) = gsvd( ¯ H H , ¯ G H ) . (31)Subsequently, let d = s , c = r + N e − k , and check1) If d + d ≥ N b , let W = ΓΨ (: , r + 1 : r + N b − d ) and V = V c Ψ (: , c +1 : c + N b − d ) . Lastly, let ˆ V = [ V , V ] and ˆ W = W .2) Otherwise, let W = ΓΨ (: , r + 1 : r + s ) and V = V c Ψ (: , c + 1 : c + s ) . Thirdly, denote V c =null { [ V , V ] H } ∈ C N a × ( N a − d − d ) and ˜ G = G V c .Invoking the GSVD Transform of ( H H , ˜ G H ) yields ( ˜ Ψ , ˜ Ψ , ˜ D , ˜ D , ˜ X , k , r , s ) = gsvd( H H , ˜ G H ) . (32)Then let W = ˜ Ψ (: , r + 1 : r + d ) and V = V c ˜ Ψ (: , c + 1 : c + d ) in which d = min { s , ⌊ N b − ( d + d )2 ⌋} and c = r + ( N a − d − d ) − k . Lastly, let ˆ V = [ V , V , V ] and ˆ W = [ W , W ] . Case IV: N a < N e + N b and N j ≤ N b . For a start, let V = null { G } and d = ( N a − N e ) + . Secondly, denote V c = null { V H } ∈ C N a × N e and ¯ G = G V c . Invoking the GSVD Transform of ( H H , ¯ G H ) yields ( Ψ , Ψ , D , D , X , k , r , s ) = gsvd( H H , ¯ G H ) . (33)Then let W = Ψ (: , r + 1 : r + d ) and V = V c Ψ (: , c + 1 : c + d ) in which d = min { s , ⌊ N b − d ⌋} and c = r + N a − k .Lastly, let ˆ V = [ V , V ] and ˆ W = W . orthogonal with V by definition, thus d ⋆ =rank { [ V , V , V ] } =rank { [ V , V ] } + rank { V } =rank { V } + rank { V } + rank { V } = min { d + d + d , N a } . In Case IV , ˆ V = [ V , V ] and ˆ W = W . Accord-ing to (33), span( H W ) = span( G V ) , which, togetherwith G V = , gives span( H ˆ W ) = span( G ˆ V ) . Inaddition, span( H ˆ V ) ∩ span( G ˆ W ) = { } due to d =min { s , ⌊ N b − d ⌋} . Thus, { ˆ V , ˆ W } is feasible to (30). Fur-thermore, V is orthogonal with V by definition, therefore d ⋆ =rank { [ V , V ] } =rank { V } + rank { V } = min { d + d , N a } . Summarizing the above four cases, we can rewrite d ⋆ intoa more compact form as follows: d ⋆ = min { d ⋆ + d ⋆ + d ⋆ , N a , N b } , (34) TABLE II: Summary of the closed-form results on d ⋆ ( s.d.o.f. ) Inequalities on the number of antennas at terminals d ⋆ ( s.d.o.f. ) N a ≥ N e + N b N j ≥ N e + N b min { N a , N b } N b + N e − N j ≤ N a < N e + N b N b < N j < N e + N b N b + N e − N j < N a < N b + N e − N j N a + N j − ( N b + N e ) + min { s, ⌊ N b + N e − N a − N j ⌋} N b < N j < N e + N b s = min { N b + N e − N j , N e } + min { N j , N e } − N e N e < N a < N e + N b , N j ≤ N b N a − N e + min { s, ⌊ N b + N e − N a ⌋} , s = min { N j , N e } N a ≤ N b + N e − N j , N b < N j < N e + N b min { s, ⌊ N b ⌋} N a ≤ N e , N j ≤ N b s = min { N a , N e } + min { N j , N e } − min { N a + N j , N e } in which d ⋆ = ( N a − N e ) + (35a) d ⋆ = (min { N a , N e } + ( N j − N b ) + − N e ) + (35b) d ⋆ = min { s, ( ⌊ N b − ( d ⋆ + d ⋆ )2 ⌋ ) + } , (35c)where s = min { N a − ( d ⋆ + d ⋆ ) , N e } + min { N j , N e } − min { N a − ( d ⋆ + d ⋆ ) + N j , N e } .To gain more insight into d ⋆ , we give Table II whichclarifies the connection of d ⋆ to the number of antennas ateach terminal. Lemma 5: On d defined in (30), we have d = d ⋆ where d ⋆ is given in (34).Proof: See Appendix F
Remark:
According to Lemma 5, it is straight-forward thatthe feasible solution { ˆ V , ˆ W } given in Table I is also theoptimal solution to (30). Throrem 2: Consider a helper-assisted MIMO Gaussianwiretap channel, as depicted in Fig.1, where the source, thelegitimate receiver, the eavesdropper and an external helperare equipped with N a , N b , N e and N j antennas, respectively.The maximal achievable secure degrees of freedom s.d.o.f. = d ⋆ , (36) where d ⋆ is given in (34).Proof: Combining Lemma 4 and Lemma 5, it is clear that s.d.o.f. = d ⋆ . This completes the proof. Corollary 1:
The feasible point { ˆ V , ˆ W } for the optimiza-tion problem of (30), given in Table I, serves as a s.d.o.f.-optimal solution to the original SRM problem in (3). Itachieves the maximal s.d.o.f.. Moreover, Table II clarifiesthe maximal achievable s.d.o.f. of a helper-assisted MIMOGaussian wiretap channel, and reveals its specific connectionto the number of antennas at each terminal. Proof:
With Theorem 2, it is straight-forward to arrive atthese conclusions.
Corollary 2:
When N b > , the maximal achievable s.d.o.f.of a helper-assisted MIMO Gaussian wiretap channel is zeroif and only if N e ≥ N a + N j . When N b = 1 , the maximalachievable s.d.o.f. of a helper-assisted MIMO Gaussian wire-tap channel is zero if and only if N e ≥ N a + N j − . Proof:
See Appendix G A v er ag e S ecrec y R a t e ( b i t / s / H z ) ZF SchemeAlignment SchemeOptimal Scheme
Fig. 2: Average secrecy rate versus SNR, N a = N e = 3 , N j = 2 , N b = 1 V. N
UMERICAL R ESULTS
In this section, we give numerical results to show the secrecyrate performance of the proposed schemes and validate ourtheoretical findings. All channels are assumed to be quasi-static flat Rayleigh fading and independent of each other,with entries distributed as CN (0 , . The noise vector at eachreceiver is assumed to be AWGN, with i.i.d. entries distributedas CN (0 , . In each figure, details on the number of antennasat each terminal will be depicted. Results are averaged over1000 independent channel trials.We first test the secrecy rate performance of our proposedschemes and compare them with the existing method.Fig.2 illustrates the secrecy rate performance of the single-antenna legitimate receiver case. The lines labeled as “OptimalScheme” and “Alignment Scheme” illustrate the secrecy rateperformance of C s and C sub s in (8) and (20), respectively.The line labeled as “ZF Scheme” [10] gives the secrecyrate achieved by the scheme which completely nulls out thejamming signal at the legitimate receiver and matches themessage signal with the source-receiver (legitimate) channel.It shows that both C s and C sub s increase linearly with SNR.In contrast, there exists a performance ceiling on the secrecyrate achieved by the ZF Scheme.Fig. 3 illustrates the secrecy rate performance of the multi-antenna legitimate receiver case. The bar labeled as “Align- ment Scheme” shows the secrecy rate result of our proposedheuristic method. In such case, closed-form precoding matrices { ˆ V , ˆ W } are given in Table I and the total power is equallydistributed over all message signal streams and jammingsignal streams. The bar labeled as “Gauss-Seidel Approach”shows the secrecy rate performance of our proposed iterativealgorithm in Section IV. A . As stated in Corollary 1, { ˆ V , ˆ W } given in Table I, actually acts as a s.d.o.f.-optimal solution tothe original SRM problem in (3). So the initial point is setas the closed-form solutions in the Alignment Scheme. Forcomparison, Fig. 3 also plots the secrecy rate performanceof the method proposed in [4], wherein the secrecy ratemaximization method is derived under a power covarianceconstraint. Thus, to find the maximal achievable secrecy rateunder an average power constraint, we have to solve [5,equation (41)] [29] R s ( P ) = max S (cid:23) , tr { S }≤ P R s ( S ) . That is, numerical search over the power covariance matrix S is performed to compute R s ( P ) . Since such numerical searchis based on random choices of S , it is difficult to decide whento stop it. To deal with this issue, we first determine the runtime of our Gauss-Seidel based algorithm of Section IV. A ,which terminates when the relative secrecy rate improvementbetween two adjacent iterations is less than − . We thenrun the algorithm proposed in [4] for the same run time. Itis worthwhile to note that our proposed Alignment Schemehas closed-form solutions, so it is the most computationallyinexpensive scheme. Fig. 3 shows that, with the same runtime, our proposed Gauss-Seidel based algorithm achieveshigher secrecy rate than the algorithm proposed in [4]. Moreencouragingly, it can be seen that the Alignment Schemeachieves nearly the same secrecy rate as the Gauss-SeidelApproach based algorithm in the high SNR regime. This isconsistent with the fact that the Alignment Scheme is s.d.o.f.-optimal, thus near-optimal at high SNR. Besides, it can be seenthat the Alignment Scheme achieves higher secrecy rate thanthe algorithm proposed in [4] in the high and medium SNRregimes. To gain more insight into the Gauss-Seidel basedalgorithm, Fig. 4 plots the convergence of it. Results showthat our proposed Gauss-Seidel Approach converges very fastand stabilizes after several loops.We then test the achievable s.d.o.f. performance for thehelper-assisted MIMO wiretap channel and validate the theo-retical results of Section IV. B .In Fig. 5, the stem labeled as “Theoretical Results” showsthe theoretical maximal achievable s.d.o.f. according to TableII. The stem labeled as “Numerical Results” shows the s.d.o.f.achieved by the proposed Alignment Scheme. In the proposedAlignment Scheme, closed-form precoding matrices { ˆ V , ˆ W } are given in Table I and the total power is equally distributedover all message signal streams and jamming signal streams.The total power P is set as 50dB. For each channel trial, wesubstitute the closed-from solution into (3) and compute thesecrecy rate C o s . We then compute the s.d.o.f. as the rate atwhich the secrecy rate C o s scales with log P , i.e., C o s / log P .It can be seen that the theoretical results almost coincide with A v er ag e S ecrec y R a t e ( b i t \ s \ H z ) Gauss-Seidel ApproachAlgorithm Proposed in [5]Alignment Scheme
Fig. 3: Average secrecy rate versus SNR, N a = N b = 3 , N j = 4 , N e = 3 A v er ag e S ecrec y R a t e ( b i t / s / H z ) Fig. 4: Average secrecy rate versus the number of iterations, N j = 4 , N a = N b = 3 , N e = 3 S ec u re D e g ree s o f F ree d o m Theoretical ResultsNumerical Results a N Fig. 5: s.d.o.f. versus N a , N j = 3 , N b = 3 , N e = 4 the numerical results.Fig. 6 and Fig. 7 plot the maximal achievable s.d.o.f. forthe helper-assisted MIMO Gaussian wiretap channel undervarious antenna configurations, according to Table II. Resultsshow that for the case of single-antenna legitimate receiver,the maximal achievable s.d.o.f. is zero if and only if N e ≥ N a + N j − , while for the case of multi-antenna legitimatereceiver, the maximal achievable s.d.o.f. is zero if and only if S ec u re D e g ree s o f F ree d o m b N = b N = a N a N Fig. 6: s.d.o.f. versus N a , N j = 2 , N e = 4 S ec u re D e g ree s o f F ree d o m b N = b N = a N a N Fig. 7: s.d.o.f. versus N a , N j = 4 , N e = 4 N e ≥ N a + N j . Further, it is illustrated that the maximalachievable s.d.o.f. benefits from the increasing number ofantennas at any of the three terminals, i.e., the source, thelegitimate receiver and the external helper. These results areconsistent with the theoretical findings of Section IV. B .VI. C ONCLUSION
We have studied the secrecy capacity of the MIMO Gaus-sian wiretap channel, where a multi-antenna external helperis available. For the special case of single-antenna legiti-mate receiver, we have obtained the secrecy capacity usinga combination of convex optimization and one-dimensionalsearch. For the case of multi-antenna legitimate receiver, wehave reformulated the original nonconvex SRM problem intoseveral convex subproblems. By doing so, we have been ableto provide an iterative algorithm, which attains a fairly goodsecrecy rate performance. In addition, we have addressedthe s.d.o.f. maximization analytically. Specifically, we haveobtained an analytical s.d.o.f.-optimal solution to the originalSRM problem, based on which, we have obtained the maximalachievable s.d.o.f. in closed-form. These results uncovered theconnection between the maximal achievable s.d.o.f. and thesystem parameters, thus shedding light on how the secrecycapacity of a helper-assisted MIMO Gaussian wiretap chan-nel behaves. Numerical results have validated the theoreticalfindings and confirmed the efficacy of our proposed schemes. A
PPENDIX AP ROOF OF P ROPOSITION L =tr { Q a + Q j } + µ [ f ( τ )(1 + g Q j g H ) − h Q a h H ]+ tr { Z [ G Q a G H − τ ( I + H Q j H H )] }− tr { Z a Q a } − tr { Z j Q j } , (37)where Z , Z a , Z j and µ are dual variables associated with theinequalities in (13). The optimization problem of (13) is con-vex with part of the Karush-Kuhn-Tucker (KKT) conditionsas follows: Z a = I − µ h H h + G H Z G (38a) Z a Q a = (38b) Z (cid:23) , Z a (cid:23) , µ ≥ . (38c)Substituting (38a) into (38b), we arrive at ( I + G H Z G ) Q a = µ h H h Q a . (39)Since I + G H Z G ≻ , so rank { Q a } = rank { ( I + G H Z G ) Q a } = rank { µ h H h Q a } ≤ . In addition, rank { Q a } = 0 implies that Q a = , which contradicts thepositive secrecy rate requirement. Thus, rank { Q a } = 1 . Thiscompletes the proof. A PPENDIX BP ROOF OF P ROPOSITION Φ( Q a , Q j ) = h Q a h H / (1 + g Q j g H ) . Denote the optimal solution to (11) as { ¯ Q a , ¯ Q j } . Apparently,the point { ¯ Q a , ¯ Q j } is also feasible to (13). Thus we havetr { ˆ Q a + ˆ Q j } ≤ tr { ¯ Q a + ¯ Q j } ≤ P ,which indicates the point { ˆ Q a , ˆ Q j } is feasible to (11). Ac-cording to the definition of f ( τ ) , we see Φ( ˆ Q a , ˆ Q j ) ≤ f ( τ ) .Moreover, from (13), Φ( ˆ Q a , ˆ Q j ) ≥ f ( τ ) . Thus, Φ( ˆ Q a , ˆ Q j ) = f ( τ ) ,which implies that the point { ˆ Q a , ˆ Q j } is also the optimalsolution to (11). This completes the proof.A PPENDIX CP ROOF OF THE EQUALITIES IN (16)Apparently, by the definition of (15a), (16a) holds true.According to [24], for any given matrix of A , it holds that span( A H ) = null( A ) ⊥ . (40)Applying (40) to (15b), we get span( H ) ⊥ ∩ span( G ) = null( H H ) ∩ null( G H ) ⊥ = null( H H ) / [null( H H ) ∩ null( G H )]= null( H H ) / null([ H H , G H ] T ) . (41)In addition, null([ H H , G H ] T ) ⊂ null( H H ) by definition, sowe have p = dim { null( H H ) } − dim { null([ H H , G H ] T ) } = N − min { M, N } − ( N − k ) = k − min { M, N } .Similarly, we can prove (16b)-(16d). This completes theproof. A PPENDIX DP ROOF OF THE LOWER BOUND ON C s IN (24)The optimization problem of (22) can be solved by resortingto carefully mathematical deductions. Let y = 1+ cP +( b − c ) x , α = min { b, c } and β = max ( b, c ) . Then, y ∈ [1+ αP, βP ] .Substituting x = ( y − (1 + cP )) / ( b − c ) into (23) and makingsome mathematical transformations yield η ( y ) = κ − ( Ay + B/y ) , (42)in which κ = a (1 + cP ) + c − bc − b + [2 a (1 + cP ) + c − b ] b ( c − b ) , A = ac ( c − b ) and B = b (1 + cP )[ a (1 + cP ) + c − b ]( c − b ) .Resorting to (42), the optimization problem of (22) can betransformed into a new optimization problem that searches for y as follows: η max , max αP ≤ y ≤ βP κ − ( Ay + B/y ) . (43)1) For the case of a (1 + cP ) > b − c , B > . Thus, η ( y ) is increasing in y when y < y , and decreas-ing in y when y > y . Herein, y = p B/A = p b (1 + cP )[ a (1 + cP ) − b + c ]/( ac ) . Therefore, if αP ≤ y ≤ βP , η max = κ − √ AB . Otherwise, η ( y ) achieves its maximal value at the two endpoints ( y = 1 + αP or y = 1 + βP ) , and η max = max { (1 + aP ) / (1 + bP ) , } .2) For the case of a (1 + cP ) ≤ b − c , B ≤ . Thus, η ( y ) decreases monotonically with respect to y . It achievesits maximal value at the two endpoints, and η max = max { (1 + aP ) / (1 + bP ) , } .Summarizing, if αP ≤ y ≤ βP , then η max = κ − √ AB and C sub s =log( c (1 + aP )[ c + b − √ bc p c − b ) / ( a + acP )]( c − b ) + c ( a − b ) + 2( c − a ) √ bc p c − b ) / ( a + acP )( c − b ) ) ,(44)where the optimal solution x ⋆ = ( y − (1 + cP )) / ( b − c )= p b (1 + cP )[ a (1 + cP ) − b + c ] / ( ac ) − (1 + cP ) b − c . (45)Otherwise, when a > b , η max = (1 + aP ) / (1 + bP ) and C sub s = log(1 + aP ) / (1 + bP ) , where the optimal solution x ⋆ = P ; when a ≤ b , η max = 1 and C sub s = 0 , where theoptimal solution x ⋆ = 0 . In the sequel, we refer to these twosolutions x ⋆ = 0 and x ⋆ = P as the trivial solutions.Consider the nontrivial solution of (45). When the totaltransmit power P is big enough, C sub s ≈ log( aP ) − p b/c ) . (46)This completes the proof. A PPENDIX EP ROOF OF L EMMA A and B , if B is invertible, then span( A ) = span( AB ) (47) rank { A } = rank { AB } . (48)Firstly, span( A ) = span( ABB − ) ⊂ span( AB ) . Secondly, span( A ) ⊃ span( AB ) . Therefore, the equality (47) holdstrue. With (47), it is clear that the equality (48) holds true.Given an arbitrary point of { V , W } , with the definition in(4), we can re-express the achieved s.d.o.f. as follows: h ( V , W ) = rank { H V } − m ( V , W ) − n ( V , W ) , (49)in which m ( V , W ) = dim { span( G V ) / span( H W ) } and n ( V , W ) = dim { span( G W ) ∩ span( H V ) } .Assume that { ¯ V , ¯ W } is the optimal solution to (30), thenwe have span( G ¯ V ) ⊂ span( H ¯ W ) and span( G ¯ W ) ∩ span( H ¯ V ) = { } . The achieved s.d.o.f. h ( ¯ V , ¯ W ) =rank { H ¯ V } = d . In addition, s.d.o.f. ≥ h ( ¯ V , ¯ W ) bydefinition. Thus, s.d.o.f. ≥ d . As such, to complete the proofof Lemma 4, we only need to prove s.d.o.f. ≤ d . In the sequel,we show that, for any given point of { V , W } , we can alwaysfind another feasible point for the problem of (30), { V ′ , W ′ } ,such that h ( V , W ) ≤ rank { H V ′ } ≤ d , thus giving the proofof s.d.o.f. ≤ d .Without lose of generality, denote V ∈ C N a × K a and W ∈ C N j × K j . With the GSVD Transform of ( W H G H , V H H H ) ,we obtain unitary matrices ˆ Ψ ∈ C K j × K j and ˆ Ψ ∈ C K a × K a ,non-negative diagonal matrices ˆ D ∈ C K j × k and ˆ D ∈ C K a × k , and a matrix ˆ X ∈ C N r × k with rank { ˆ X } = k ,such that G W ˆ Ψ = ˆ X ˆ D H (50a) H V ˆ Ψ = ˆ X ˆ D H , (50b)where k = min { K a + K j , N r } , r = k − min { K a , N r } and s = min { K a , N r } + min { K j , N r } − k .Let ˆ Ψ = ˆ Ψ (: , r + 1 : r + s ) (51a) ˆ Ψ = [ ˆ Ψ (: , r ) , ˆ Ψ (: , r + s : K j )] (51b) ˆ Ψ = ˆ Ψ (: , c in + 1 : c in + s ) (51c) ˆ Ψ = [ ˆ Ψ (: , c in ) , ˆ Ψ (: , c in + s + 1 : K a )] , (51d)in which c in = r + K a − k . Since ˆ Ψ and ˆ Ψ are invertiblematrices, ˆ Ψ ′ = [ ˆ Ψ , ˆ Ψ ] and ˆ Ψ ′ = [ ˆ Ψ , ˆ Ψ ] are alsoinvertible matrices. Applying (47) and (48), we have h ( V , W ) = h ( V ˆ Ψ ′ , W ˆ Ψ ′ ) (52a) = rank { H V ˆ Ψ } − m ( V ˆ Ψ ′ , W ˆ Ψ ′ ) (52b) ≤ rank { H V ˆ Ψ } − m ( V ˆ Ψ , W ˆ Ψ ′ ) , (52c)in which (52b) can be justified with span( G W ˆ Ψ ′ ) ∩ span( H V ˆ Ψ ′ ) = span( H V ˆ Ψ ). Besides, (52c) comes fromthe fact that m ( V ˆ Ψ ′ , W ˆ Ψ ′ ) ≥ m ( V ˆ Ψ , W ˆ Ψ ′ ) . With the
GSVD Transform of (( H W ˆ Ψ ′ ) H , ( G V ˆ Ψ ) H ) ,we obtain unitary matrices ˘ Ψ ∈ C K j × K j and ˘ Ψ ∈ C ( K a − s ) × ( K a − s ) , non-negative diagonal matrices ˘ D ∈ C K j × k and ˘ D ∈ C ( K a − s ) × k , and a matrix ˘ X ∈ C N e × k with rank { ˘ X } = k , such that H W ˆ Ψ ′ ˘ Ψ = ˘ X ˘ D H (53a) G V ˆ Ψ ˘ Ψ = ˘ X ˘ D H , (53b)where k = min { K a − s + K j , N e } , r = k − min { K a − s , N e } and s = min { K j , N e } + min { K a − s , N e } − k .Let ˘ Ψ = ˘ Ψ (: , r + s ) (54a) ˘ Ψ = ˘ Ψ (: , r + s : K j ) (54b) ˘ Ψ = ˘ Ψ (: , c in + s ) (54c) ˘ Ψ = ˘ Ψ (: , ˘ c in + s + 1 : K a − s ) , (54d)in which ˘ c in = r + ( K a − s ) − k . Since ˘ Ψ and ˘ Ψ areinvertible matrices, so ˘ Ψ ′ = [ ˘ Ψ , ˘ Ψ ] and ˘ Ψ ′ = [ ˘ Ψ , ˘ Ψ ] are also invertible matrices. Applying (47) and (48), we have rank { H V ˆ Ψ } − m ( V ˆ Ψ , W ˆ Ψ ′ )= rank { H V ˆ Ψ ˘ Ψ ′ } − m ( V ˆ Ψ ˘ Ψ ′ , W ˆ Ψ ′ ˘ Ψ ′ ) (55a) = rank { H V ˆ Ψ ˘ Ψ ′ } − rank { ˘ Ψ } (55b) ≤ rank { H V ˆ Ψ ˘ Ψ } , (55c)where due to span( G V ˆ Ψ ˘ Ψ ′ ) / span( H W ˆ Ψ ′ ˘ Ψ ′ ) =span( G V ˆ Ψ ˘ Ψ ) , (55b) holds true. Besides, (55c) holdstrue due to rank { H V ˆ Ψ ˘ Ψ ′ } ≤ rank { H V ˆ Ψ ˘ Ψ } +rank { H V ˆ Ψ ˘ Ψ } and rank { H V ˆ Ψ ˘ Ψ } ≤ rank { ˘ Ψ } .Combining (52) with (55), we arrive at h ( V , W ) ≤ rank { H V ˆ Ψ ˘ Ψ } . (56)With (53) and (54), we arrive at m ( V ˆ Ψ ˘ Ψ , W ˆ Ψ ′ ˘ Ψ ′ ) =0 , thus span( G V ˆ Ψ ˘ Ψ ) ⊂ span( H W ˆ Ψ ′ ˘ Ψ ′ ) . (57)In addition, with (50) and (51), we get n ( V ˆ Ψ , W ˆ Ψ ′ ) = 0 .So span( G W ˆ Ψ ′ ) ∩ span( H V ˆ Ψ ) = { } , which, togetherwith the facts that span( G W ˆ Ψ ′ ) = span( G W ˆ Ψ ′ ˘ Ψ ′ ) and span( H V ˆ Ψ ) ⊃ span( H V ˆ Ψ ˘ Ψ ) , gives span( G W ˆ Ψ ′ ˘ Ψ ′ ) ∩ span( H V ˆ Ψ ˘ Ψ ) = { } . (58)Combining (57) with (58), we know { V ˆ Ψ ˘ Ψ , W ˆ Ψ ′ ˘ Ψ ′ } is a feasible point for the problem of (30). By definition, rank { H V ˆ Ψ ˘ Ψ } ≤ d , which, together with (56), indicatesthat h ( V , W ) ≤ d .Because the above derivations hold true for any given pointof { V , W } , we conclude that s.d.o.f. ≤ d . This completesthe proof. A PPENDIX FP ROOF OF L EMMA d ≥ d ⋆ holds true. So if we can further prove d ≤ d ⋆ , then the proof of Lemma 5 is completed. In the following text, we give the proof of d ≤ d ⋆ by contra-diction. Assume that there exists a feasible point { ˜ V , ˜ W } of (30), where ˜ V ∈ C N a × K a , d ♯ , rank { H ˜ V } and K a = d ♯ > d ⋆ . In such case, we have rank { ˜ V } = d ♯ dueto d ♯ = rank { H ˜ V } ≤ rank { ˜ V } ≤ K a = d ♯ . Besides, bydefinition, d ♯ ≤ min { N a , N b } always holds true. In the sequel,we discuss the four cases in Table I and give contradictionsone by one.In Case I and
Case II , d ⋆ = min { N a , N b } . The assumption d ♯ > d ⋆ implies d ♯ > min { N a , N b } , which contradicts thefact d ♯ ≤ min { N a , N b } .In Case III , when d ⋆ = min { N a , N b } , the assumption d ♯ >d ⋆ = N b contradicts the fact d ♯ ≤ min { N a , N b } . As such, weonly need to focus on the case of d ⋆ = d + d + d , where d = ( N a − N e ) + , d = s and d = min { s , ⌊ N b − ( d + d )2 ⌋} .1) For the case of s ≤ ⌊ N b − ( d + d )2 ⌋ , d ⋆ = d + d + s .In addition, d ♯ > d ⋆ . So d ♯ > d + d + s . Therefore d ♯ − d > s + s , which contradicts (30b). Theexplanation is as follows. According to (32), s =min { N j , N e } +min { N a − d − d , N e }− min { N j + N a − d − d , N e } . With the GSVD Transform of ( H H , G H ) , s = dim { span( H ) ∩ span( G ) } = min { N j , N e } +min { N a , N e } − min { N j + N a , N e } . It is easy to verifythat s ≤ s + s . In addition, to satisfy (30b) we shouldhave rank { G ˜ V } ≤ s . Thus rank { G ˜ V } ≤ s + s .Moreover, rank { ˜ V } − d ≤ min { N a , N e } due to thefact rank { ˜ V } ≤ N a , so rank { G ˜ V } = rank { ˜ V } − d = d ♯ − d . Therefore, d ♯ − d ≤ s + s , which givesthe contradiction.2) For the case of s > ⌊ N b − ( d + d )2 ⌋ , d ⋆ = d + d + ⌊ N b − ( d + d )2 ⌋ , which, together with the assumption d ♯ >d ⋆ , gives d ♯ > d + d + ⌊ N b − ( d + d )2 ⌋ . (59)If N b − ( d + d ) is an even number, (59) is equivalent to d ♯ > N b + d + d . Otherwise, N b − ( d + d ) is an oddnumber, so (59) is equivalent to d ♯ > N b + d + d − .In addition, N b + d + d owns the same parity as N b − d − d , thus N b + d + d − is an even number.Therefore d ♯ > N b + d + d . To sum up, (59) indicates d ♯ > N b + d + d . Thus d ♯ − d > N b − d ♯ + d .Moreover, to satisfy (30c), we should have N b − d ♯ + d ≥ rank { ˜ W } . So d ♯ − d > rank { ˜ W } . However, rank { G ˜ V } = rank { ˜ V } − d due to rank { ˜ V } − d ≤ min { N a , N e } . Thus, rank { G ˜ V } = d ♯ − d > rank { ˜ W } ≥ rank { H ˜ W } , which contradicts (30b).In Case IV , d ⋆ = d + d , where d = ( N a − N e ) + and d = min { s , ⌊ N b − d ⌋} . Since the analysis is similar to CaseIII , so in the sequel we only give the skeleton on it.1) For the case of s ≤ ⌊ N b − d ⌋ , d ⋆ = d + s ,which, combined with the assumption d ♯ > d ⋆ , gives d ♯ > d + s . Thus d ♯ − d > s . However, with the GSVD Transform of ( H H , G H ) , s = dim { span( H ) ∩ span( G ) } = s . To satisfy (30b), we should have rank { G ˜ V } ≤ s = s , which, together with the fact rank { ˜ V } − d ≤ min { N a , N e } , indicates d ♯ − d ≤ s .
2) For the case of s > ⌊ N b − d ⌋ , d ⋆ = d + ⌊ N b − d ⌋ ,which, together with the assumption d ♯ > d ⋆ , gives d ♯ > N b + d . Thus d ♯ − d > N b − d ♯ . However,to satisfy (30c), we should have N b − d ♯ ≥ rank { ˜ W } .Therefore, rank { ˜ V }− d = d ♯ − d > rank { ˜ W } , which,together with the fact rank { G ˜ V } = rank { ˜ V } − d ,contradicts (30b).Summarizing the above four cases, for any feasible pointsfor the problem of (30), denoted by { ˜ V , ˜ W } and ˜ V ∈ C N a × K a , if K a = d ♯ , d ♯ ≤ d ⋆ . On the other hand, if K a > d ♯ , resorting to the singular value decomposition(SVD) of H ˜ V , we can always find another feasible point { ˜ V ′ , ˜ W ′ } for the problem of (30), such that ˜ V ′ ∈ C N a × K ′ a and K ′ a = rank { ˜ V ′ } = rank { H ˜ V ′ } = d ♯ . As such, theassumption d ♯ > d ⋆ also contradicts the feasibility conditionsin (30). So d ♯ ≤ d ⋆ .In conclusion, for any feasible points for the problem of(30), we should have d ♯ ≤ d ⋆ . By definition, we arrive at d ≤ d ⋆ . This completes the proof.A PPENDIX GP ROOF OF C OROLLARY s.d.o.f. = 0 canonly happen in the last case, where N a ≤ N b + N e − N j , N b < N j ≤ N e + N b − N a or N a ≤ N e , N j ≤ N b . (60)Thus, to complete the proof, we only need to focus on thecase of (60), where s.d.o.f. = min { s, ⌊ N b ⌋} in which s =min { N a , N e } + min { N j , N e } − min { N a + N j , N e } . Note thatthe case N b + N e − N j < N a ≤ N e , N b < N j ≤ N e + N b − N a never happens, so (60) is equivalent to N a ≤ N e , N j ≤ N e + N b − N a , (61)which implies s = N e + min { N j , N e } − min { N a + N j , N e } .Therefore,1) for the case of N e ≥ N a + N j , s = 0 thus s.d.o.f. = 0 ;2) for the case of N e ≤ N j , s = N a thus s.d.o.f. =min { N a , ⌊ N b ⌋} ;3) for the case of N j < N e < N a + N j , s = N a + N j − N e thus s.d.o.f. = min { N a + N j − N e , ⌊ N b ⌋} .In conclusion, when N b = 1 , s.d.o.f. = 0 if and only if N e ≥ N a + N j − ; when N b > , s.d.o.f. = 0 if and only if N e ≥ N a + N j . This completes the proof.R EFERENCES[1] A. D. Wyner, “The wire-tap channel,”
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