半正定Toeplitz矩阵的范德蒙德分解

半正定Toeplitz矩阵的范德蒙德分解

Toeplitz矩阵定义:Matrices whose entries are constant along each diagonal are called Toeplitz matrices.
形如
T = [ r 0 r 1 r 2 r 3 r − 1 r 0 r 1 r 2 r − 2 r − 1 r 0 r 1 r − 3 r − 2 r − 1 r 0 ] (1) \boldsymbol{T}=\left[ \begin{matrix} r_0& r_1& r_2& r_3\\ r_{-1}& r_0& r_1& r_2\\ r_{-2}& r_{-1}& r_0& r_1\\ r_{-3}& r_{-2}& r_{-1}& r_0\\ \end{matrix} \right] \tag{1} T=r0r1r2r3r1r0r1r2r2r1r0r1r3r2r1r0(1)

半正定的Toeplitz矩阵:Positive-Semi-Definite Toeplitz, PSD
形如
T = [ r 0 r 1 r 2 r 3 r 1 ∗ r 0 r 1 r 2 r 2 ∗ r 1 ∗ r 0 r 1 r 3 ∗ r 2 ∗ r 1 ∗ r 0 ] ,     T ≽ 0 (2) \boldsymbol{T}=\left[ \begin{matrix} r_0& r_1& r_2& r_3\\ r_{1}^{*}& r_0& r_1& r_2\\ r_{2}^{*}& r_{1}^{*}& r_0& r_1\\ r_{3}^{*}& r_{2}^{*}& r_{1}^{*}& r_0\\ \end{matrix} \right] , \ \ \ \boldsymbol{T} \succcurlyeq \boldsymbol 0 \tag{2} T=r0r1r2r3r1r0r1r2r2r1r0r1r3r2r1r0,   T0(2)
其中 T ≽ 0 \boldsymbol{T} \succcurlyeq \boldsymbol 0 T0表示 T \boldsymbol{T} T是半正定矩阵。

定理半正定Toeplitz矩阵的范德蒙德分解
Any PSD Toeplitz matrix T ( u ) ∈ C N × N \boldsymbol T(\boldsymbol u) \in \mathbb C^{N \times N} T(u)CN×N of rank r ≤ N r \leq N rN admits the following r-atomic Vandermonde decomposition:
T = ∑ k = 1 r p k a ( f k ) a H ( f k ) = A ( f ) d i a g ( p ) A H ( f ) (3) \boldsymbol T = \sum_{k=1}^r p_k \boldsymbol a (f_k) \boldsymbol a^H(f_k) = \boldsymbol A( \boldsymbol f ) diag(\boldsymbol p) \boldsymbol A^H( \boldsymbol f ) \tag{3} T=k=1rpka(fk)aH(fk)=A(f)diag(p)AH(f)(3)
where p k > 0 p_k >0 pk>0, and f k ∈ T f_k \in \mathbb T fkT, T = ( − 1 2 , 1 2 ] \mathbb T=(-\frac{1}{2}, \frac{1}{2}] T=(21,21] k = 1 , 2 , ⋯   , r k=1,2,\cdots,r k=1,2,,r are distinct. Moreover, the decompostion above is unique if r < N r < N r<N.
其中
a ( f ) = [ 1 , e i 2 π f , ⋯   , e i 2 π ( N − 1 ) f ] T ∈ C N × 1 \boldsymbol a(f) = [1, e^{i 2 \pi f}, \cdots, e^{i 2 \pi (N-1) f} ]^T \in \mathbb C^{N \times 1} a(f)=[1,ei2πf,,ei2π(N1)f]TCN×1

证明
(1)首先考虑 r = rank ( T ) ≤ N − 1 r=\text{rank}(\boldsymbol T) \leq N - 1 r=rank(T)N1的情况
因为 T ≽ 0 \boldsymbol T \succcurlyeq 0 T0,因此存在 V ∈ C N × r \boldsymbol V \in \mathbb C^{N \times r} VCN×r满足: T = V V H \boldsymbol T= \boldsymbol {VV}^H T=VVH
V − N \boldsymbol V_{-N} VN V \boldsymbol V V去掉第 N N N行(最后一行)的矩阵: V − N ∈ C ( N − 1 ) × r \boldsymbol V_{-N} \in \mathbb{C}^{(N-1) \times r} VNC(N1)×r
V − 1 \boldsymbol V_{-1} V1 V \boldsymbol V V去掉第 1 1 1行(第一行)的矩阵: V − 1 ∈ C ( N − 1 ) × r \boldsymbol V_{-1} \in \mathbb{C}^{(N-1) \times r} V1C(N1)×r
因为半正定Toeplitz矩阵的特殊结构,必然有: V − N V − N H = V − 1 V − 1 H \boldsymbol V_{-N} \boldsymbol V^H_{-N}=\boldsymbol V_{-1} \boldsymbol V^H_{-1} VNVNH=V1V1H (下图给了一个直观解释, V − N V − N H \boldsymbol V_{-N} \boldsymbol V^H_{-N} VNVNH对应红色方框中的矩阵, V − 1 V − 1 H \boldsymbol V_{-1} \boldsymbol V^H_{-1} V1V1H对应绿色方框中的矩阵,两者是一样的)

半正定Toeplitz矩阵的范德蒙德分解_第1张图片

因此,一定存在某个酉阵 Q ∈ C r × r \boldsymbol Q \in \mathbb C^{r \times r} QCr×r,使得 V − 1 = V − N Q \boldsymbol V_{-1} = \boldsymbol V_{-N} \boldsymbol Q V1=VNQ,据此我们可以进一步得到(=> it follows that) V j , : = V 1 , : Q j − 1 , j = 2 , ⋯   , N \boldsymbol V_{j,:}=\boldsymbol V_{1,:} \boldsymbol Q^{j-1},j=2,\cdots,N Vj,:=V1,:Qj1,j=2,,N,因此
u j = V 1 , : ( V j , : ) T = V 1 , : ( Q j − 1 ) H ( V 1 , : ) H = V 1 , : Q 1 − j ( V 1 , : ) H (4) \begin{aligned} u_j &= \boldsymbol V_{1,:} (\boldsymbol V_{j,:}) ^T \\ &= \boldsymbol V_{1,:} (\boldsymbol Q^{j-1})^H (\boldsymbol V_{1,:})^H \\ &= \boldsymbol V_{1,:} \boldsymbol Q^{1-j} (\boldsymbol V_{1,:})^H \end{aligned} \tag{4} uj=V1,:(Vj,:)T=V1,:(Qj1)H(V1,:)H=V1,:Q1j(V1,:)H(4)

我们可以将 Q ∈ C r × r \boldsymbol Q \in \mathbb C^{r \times r} QCr×r特征分解为(注意 Q \boldsymbol Q Q是酉阵,特征分解必然存在):
Q = Q ~ d i a g ( z 1 , z 2 , ⋯   , z r ) Q ~ H (5) \boldsymbol Q = \tilde{\boldsymbol Q} diag(z_1, z_2, \cdots, z_r) \tilde{\boldsymbol Q}^H \tag{5} Q=Q~diag(z1,z2,,zr)Q~H(5)

其中 Q ~ ∈ C r × r \tilde{\boldsymbol Q} \in \mathbb C^{r \times r} Q~Cr×r是酉阵。因为酉阵的特征值的模都等于1,因此我们可以找到 f k ∈ T , k = 1 , 2 , ⋯   , r f_k \in \mathbb T, k= 1,2,\cdots,r fkT,k=1,2,,r满足 z k = e i 2 π f k , k = 1 , 2 , ⋯   , r z_k=e^{i 2 \pi f_k}, k=1,2,\cdots,r zk=ei2πfk,k=1,2,,r。令 p k = ∣ V 1 , : Q ~ : , k ∣ 2 > 0 , k = 1 , ⋯   , r p_k = \vert \boldsymbol V_{1,:} \tilde{\boldsymbol Q}_{:,k} \vert^2 > 0, k =1,\cdots,r pk=V1,:Q~:,k2>0,k=1,,r,将式(5)代入式(4),我们得到
u j = V 1 , : Q ~ d i a g ( z 1 , z 2 , ⋯   , z r ) 1 − j Q ~ H ( V 1 , : ) H = ∑ k = 1 r p k z k 1 − j = ∑ k = 1 r p k e − i 2 π ( j − 1 ) f k (6) \begin{aligned} u_j &= \boldsymbol V_{1,:} \tilde{\boldsymbol Q} diag(z_1, z_2, \cdots, z_r)^{1-j} \tilde{\boldsymbol Q}^H (\boldsymbol V_{1,:})^H \\ &= \sum_{k=1}^r p_k z_k^{1-j} \\ &= \sum_{k=1}^r p_k e^{-i 2 \pi (j-1) f_k} \end{aligned} \tag{6} uj=V1,:Q~diag(z1,z2,,zr)1jQ~H(V1,:)H=k=1rpkzk1j=k=1rpkei2π(j1)fk(6)

由此可以得出式(3)是成立的。另外, f k 1 ≠ f k 2 , k 1 ≠ k 2 f_{k_1} \neq f_{k_2}, k_1 \neq k_2 fk1=fk2,k1=k2,否则 rank ( T ) < r \text{rank}(\boldsymbol T) < r rank(T)<r(与假设矛盾)。

(2)然后考虑 r = rank ( T ) = N r=\text{rank}(\boldsymbol T) = N r=rank(T)=N的情况
这时, T ≻ 0 \boldsymbol T \succ 0 T0。我们随机地选 f N ∈ T f_N \in \mathbb T fNT,并且令 p N = ( a H ( f N ) T − 1 a ( f N ) ) p_N= { \left ( \boldsymbol a^H(f_N) \boldsymbol T^{-1} \boldsymbol a (f_N) \right ) } pN=(aH(fN)T1a(fN))。另外,我们定义一个新的向量 u ′ ∈ C N × 1 \boldsymbol u^{\prime} \in \mathbb C^{N \times 1} uCN×1
u j ′ = u j − p N e − i 2 π ( j − 1 ) f N u^{\prime}_j = u_j - p_N e^{-i 2 \pi (j-1) f_N} uj=ujpNei2π(j1)fN

可以被证明:
T ( u ′ ) = T ( u ) − p N a ( f N ) a H ( f N ) T ( u ′ ) ≽ 0 rank ( T ( u ′ ) ) = N − 1 \begin{aligned} \boldsymbol T(\boldsymbol u^{\prime}) &= \boldsymbol T(\boldsymbol u) - p_N \boldsymbol a (f_N) \boldsymbol a^H(f_N) \\ \boldsymbol T(\boldsymbol u^{\prime}) & \succcurlyeq \boldsymbol 0 \\ \text{rank} \left ( \boldsymbol T(\boldsymbol u^{\prime}) \right ) &= N-1 \end{aligned} T(u)T(u)rank(T(u))=T(u)pNa(fN)aH(fN)0=N1

因此, T ( u ′ ) \boldsymbol T(\boldsymbol u^{\prime}) T(u)满足第一种 r ≤ N − 1 r \leq N-1 rN1的情况。因此,当 r = N r=N r=N时,分解并不唯一。

最后我们来证明 r ≤ N − 1 r \leq N-1 rN1时分解的唯一性,如果假设存在另一种分解形式: T = A ( f ′ ) P ′ A H ( f ′ ) ,   p j ′ > 0 \boldsymbol T = \boldsymbol A(f^{\prime}) \boldsymbol P^{\prime} \boldsymbol A^H(f^{\prime}), \ p^{\prime}_j > 0 T=A(f)PAH(f), pj>0,且 f j ′ ∈ T f_j^{\prime} \in \mathbb T fjT各不相同,这时,我们有
A ( f ′ ) P ′ A H ( f ′ ) = A ( f ) P A H ( f ) \boldsymbol A(f^{\prime}) \boldsymbol P^{\prime} \boldsymbol A^H(f^{\prime}) = \boldsymbol A( \boldsymbol f ) \boldsymbol P \boldsymbol A^H( \boldsymbol f ) A(f)PAH(f)=A(f)PAH(f)

那么,存在一个酉阵 Q ′ ∈ C r × r \boldsymbol Q^{\prime} \in \mathbb C^{r \times r} QCr×r 使得 A ( f ′ ) P ′ 1 2 = A ( f ) P 1 2 Q ′ \boldsymbol A(f^{\prime}) \boldsymbol P^{\prime \frac{1}{2}}=\boldsymbol A( \boldsymbol f ) \boldsymbol P^{\frac{1}{2}} \boldsymbol Q^{\prime} A(f)P21=A(f)P21Q,因此
A ( f ′ ) = A ( f ) P 1 2 Q ′ P ′ − 1 2 \boldsymbol A(f^{\prime}) = \boldsymbol A( \boldsymbol f ) \boldsymbol P^{\frac{1}{2}} \boldsymbol Q^{\prime} \boldsymbol P^{\prime -\frac{1}{2}} A(f)=A(f)P21QP21

上式意味着,对于 ∀ j ∈ { 1 , 2 , ⋯   , r } \forall j \in \{1,2,\cdots, r\} j{1,2,,r} a ( f j ′ ) ∈ span { a ( f 1 ) , ⋯   , a ( f r ) } \boldsymbol a(f^{\prime}_j) \in \text{span} \left \{ \boldsymbol a(f_1), \cdots, \boldsymbol a(f_r) \right \} a(fj)span{a(f1),,a(fr)}。又因为在 r ≤ N − 1 r \leq N-1 rN1时,任意两个分量 a ( f i ) \boldsymbol a(f_i) a(fi) a ( f j ) , i ≠ j \boldsymbol a(f_j), i\neq j a(fj),i=j都是线性独立的,因此必然有, { f j ′ } j = 1 r \{f^{\prime}_j\}_{j=1}^r {fj}j=1r { f j } j = 1 r \{f^{}_j\}_{j=1}^r {fj}j=1r相等。由此可以得出,当 r ≤ N − 1 r \leq N-1 rN1时,分解具有唯一性。

总结

  • r ≤ N − 1 r \leq N-1 rN1时,半正定Toeplitz矩阵的范德蒙德分解是唯一地;
  • r = N r = N r=N时,半正定Toeplitz矩阵的范德蒙德分解不唯一。

推论:任意PSD Toeplitz矩阵 T ( u ) ∈ C N × N \boldsymbol T(\boldsymbol u) \in \mathbb C^{ N \times N} T(u)CN×N可以被唯一地分解为:
T = ∑ k = 1 r p k a ( f k ) a H ( f k ) + σ I = A ( f ) d i a g ( p ) A H ( f ) + σ I \boldsymbol T = \sum_{k=1}^r p_k \boldsymbol a(f_k) \boldsymbol a^H(f_k) + \sigma \boldsymbol I = \boldsymbol A( \boldsymbol f ) diag(\boldsymbol p) \boldsymbol A^H( \boldsymbol f ) + \sigma \boldsymbol I T=k=1rpka(fk)aH(fk)+σI=A(f)diag(p)AH(f)+σI
其中 σ = λ m i n ( T ) \sigma = \lambda_{min}(\boldsymbol T) σ=λmin(T) r = rank ( T − σ I ) < N r = \text{rank}(\boldsymbol T - \sigma \boldsymbol I) < N r=rank(TσI)<N p k > 0 p_k >0 pk>0 f k ∈ T , k = 1 , ⋯   , r f_k \in \mathbb T, k=1,\cdots,r fkT,k=1,,r are disjoint.
Remark: Note that the uniqueness of the decomposition above is guranteed by the condition that σ = λ m i n ( T ) \sigma=\lambda_{min}(\boldsymbol T) σ=λmin(T). If the condition is violated by letting 0 ≤ σ < λ m i n ( T ) 0 \leq \sigma < \lambda_{min}(\boldsymbol T) 0σ<λmin(T) (in such a case T \boldsymbol T T has full rank and r ≥ N r \geq N rN), then the deomposition cannot be unique.

参考

[1] Yang, Z., Li, J., Stoica, P., & Xie, L. (2016). Sparse Methods for Direction-of-Arrival Estimation. ArXiv, abs/1609.09596.

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