Eigenvectors from eigenvalues的证明

Eigenvectors from eigenvalues的证明

论文: Eigenvectors from eigenvalues 提出并简要证明了一种根据矩阵和子矩阵的特征值判断Hermitian矩阵的normed特征向量的取值(约束条件见下文中的“方法适用约束条件”部分)的方法,即下文中的Lemma 2。本文给出部分命题的详细证明过程。

方法适用约束条件

  1. 矩阵只能是Hermitian矩阵。矩阵各元素为复数。
  2. 只能计算Hermitian矩阵的normed特征向量中各元素的绝对值,不能判断符号。
  3. 假如存在 λ i = λ j , i ≠ j \lambda _{i} = \lambda _{j}, i \neq j λi=λj,i=j,则不能计算 v i , v j v_{i}, v_{j} vivj 中各元素值。

假设

对于任意一个 n × n n \times n n×n 的Hermitian矩阵 A A A, 假设特征值为 λ i ( A ) \lambda _ {i} {(A)} λi(A), 对应的线性无关的 normed特征向量为 v i v_{i} vi, i ∈ { 1 , … , n } i \in \{1, \dots, n \} i{1,,n}。假设 v i v_{i} vi的各元素为 v i , j v_{i, j} vi,j, j ∈ { 1 , … , n } j \in \{1, \dots, n \} j{1,,n}
假如 A A A有一个特征值为 0 0 0,不失一般性,假如 λ n ( A ) = 0 \lambda _ {n} {(A)} = 0 λn(A)=0

Lemma 1

命题

∏ i = 1 n − 1 λ i ( A ) ∣ det ⁡ ( B , v n ) ∣ 2 = det ⁡ ( B ∗ A B ) (1) \prod \limits_{i = 1} ^{n - 1} \lambda _ {i} {(A)} \left \vert \det \left ( \begin{matrix} B, v_{n} \end{matrix} \right ) \right \vert ^{2}=\det (B^{*}AB) \tag {1} i=1n1λi(A)det(B,vn)2=det(BAB)(1)
其中 B B B 是任意一个 n × ( n − 1 ) n \times (n - 1) n×(n1) 的矩阵。 ( B , v n ) \begin{pmatrix} B, v_{n} \end{pmatrix} (B,vn) 是由 B B B v n v_{n} vn 按列拼接起来的 n × n n \times n n×n 矩阵。

证明

由于 A A A 是 Hermitian矩阵,令矩阵 V = ( v 1 , ⋯   , v n ) V = \begin{pmatrix} v_{1}, \cdots , v_{n} \end{pmatrix} V=(v1,,vn) D = ( λ 1 ( A ) ⋱ λ n − 1 ( A ) 0 ) = ( Λ 0 ) D = \begin{pmatrix} \lambda _ {1} {(A)} & & & \\ & \ddots & \\ & & \lambda _ {n - 1} {(A)} & \\ & & & 0 \end{pmatrix} = \begin{pmatrix} \Lambda & \\ & 0 \end{pmatrix} D=λ1(A)λn1(A)0=(Λ0),则 A = V D V ∗ A = VDV ^{*} A=VDV,且 V V V是酉矩阵。
V ∗ B = B 1 = ( B ′ X ) V^{*} B = B_{1} = \begin{pmatrix} B' \\ X \end{pmatrix} VB=B1=(BX),其中 B ′ B' B是一个 ( n − 1 ) × ( n − 1 ) (n - 1) \times (n - 1) (n1)×(n1) 的矩阵, X X X是一个 1 × ( n − 1 ) 1 \times (n - 1) 1×(n1) 的矩阵。
易知 V e n = v n Ve_{n}=v _{n} Ven=vn,则 e n = V ∗ v n e_{n}=V^{*} v _{n} en=Vvn
V ∗ ( B , v n ) = ( V ∗ B , V ∗ v n ) = ( B 1 , e n ) = ( B ′ 0 1 × ( n − 1 ) X 1 ) V^{*} \begin{pmatrix} B, v_{n} \end{pmatrix} = \begin{pmatrix} V^{*} B, V^{*} v_{n} \end{pmatrix} = \begin{pmatrix} B_{1}, e_{n} \end{pmatrix} = \begin{pmatrix} B' & 0_{1 \times (n - 1)} \\ X & 1 \end{pmatrix} V(B,vn)=(VB,Vvn)=(B1,en)=(BX01×(n1)1)

因此 det ⁡ ( B , v n ) = det ⁡ ( V ( B ′ 0 1 × ( n − 1 ) X 1 ) ) = det ⁡ ( V ) det ⁡ ( B ′ ) = ± 1 det ⁡ ( B ′ ) \det \left ( \begin{matrix} B, v_{n} \end{matrix} \right ) = \det \left ( V \begin{pmatrix} B' & 0_{1 \times (n - 1)} \\ X & 1 \end{pmatrix} \right ) = \det (V) \det (B') = \pm1 \det (B') det(B,vn)=det(V(BX01×(n1)1))=det(V)det(B)=±1det(B)
于是 ∏ i = 1 n − 1 λ i ( A ) ∣ det ⁡ ( B , v n ) ∣ 2 = ∏ i = 1 n − 1 λ i ( A ) ∣ det ⁡ ( B ′ ) ∣ 2 \prod \limits_{i = 1} ^{n - 1} \lambda _ {i} {(A)} \left \vert \det \left ( \begin{matrix} B, v_{n} \end{matrix} \right ) \right \vert ^{2}=\prod \limits_{i = 1} ^{n - 1} \lambda _ {i} {(A)} \left \vert \det \left ( \begin{matrix} B' \end{matrix} \right ) \right \vert ^{2} i=1n1λi(A)det(B,vn)2=i=1n1λi(A)det(B)2

B ∗ A B = B ∗ V D V ∗ B = B 1 ∗ D B 1 = ( ( B ′ ) ∗ X ∗ ) ( Λ 0 ) ( B ′ X ) = ( B ′ ) ∗ Λ B ′ B^{*}AB =B^{*}VDV^{*}B = { B_{1} } ^{*} D B_{1} = \begin{pmatrix} {(B')} ^{*} & X ^{*} \end{pmatrix} \begin{pmatrix} \Lambda & \\ & 0 \end{pmatrix} \begin{pmatrix} B' \\ X \end{pmatrix} = {(B')} ^{*} \Lambda B' BAB=BVDVB=B1DB1=((B)X)(Λ0)(BX)=(B)ΛB
因此 det ⁡ ( B ∗ A B ) = ∏ i = 1 n − 1 λ i ( A ) ∣ det ⁡ ( B ′ ) ∣ 2 \det (B^{*}AB) =\prod \limits_{i = 1} ^{n - 1} \lambda _ {i} {(A)} \left \vert \det \left ( \begin{matrix} B' \end{matrix} \right ) \right \vert ^{2} det(BAB)=i=1n1λi(A)det(B)2

因此命题成立。

Lemma 2

命题

特征向量的每个元素的绝对值的平方与矩阵和子矩阵的特征值的关系:
∣ v i , j ∣ 2 ∏ k = 1 , k ≠ i n [ λ i ( A ) − λ k ( A ) ] = ∏ k = 1 n − 1 [ λ i ( A ) − λ k ( M j ) ] (2) \left \vert v_{i, j} \right \vert ^{2} \prod \limits_{k = 1, k \neq i} ^{n} \left [ \lambda _ {i} {(A)} - \lambda _{k} {(A)} \right ] = \prod \limits_{k = 1} ^{n - 1} \left [ \lambda _{i} {(A)} - \lambda _{k} {(M_{j})} \right ] \tag {2} vi,j2k=1,k=in[λi(A)λk(A)]=k=1n1[λi(A)λk(Mj)](2)

证明

不失一般性,只证明 i = n , j = 1 i = n, j = 1 i=n,j=1的情况。
A 1 = ( a 11 ⋯ ⋮ M 1 ′ ) = A − λ n I A_1 = \begin{pmatrix} a_{11} & \cdots \\ \vdots & M_{1}' \end{pmatrix} = A - \lambda _{n} I A1=(a11M1)=AλnI,则 λ n ( A 1 ) = 0 \lambda _{n} {(A_{1})} = 0 λn(A1)=0
等式 (2)可化为:
∣ v n , 1 ∣ 2 ∏ k = 1 n − 1 λ k ( A 1 ) = ∏ k = 1 n − 1 λ k ( M 1 ′ ) (3) \left \vert v_{n, 1} \right \vert ^{2} \prod \limits_{k = 1} ^{n - 1} \lambda _{k} {(A_{1})} = \prod \limits_{k = 1} ^{n - 1} \lambda _{k} {(M_{1}')} \tag {3} vn,12k=1n1λk(A1)=k=1n1λk(M1)(3)

在 Lemma 1中令 B = ( 0 1 , n − 1 I n − 1 ) B = \begin{pmatrix} 0_{1, n-1} \\ I_{n - 1} \end{pmatrix} B=(01,n1In1),则
det ⁡ ( B , v n ) = v n , 1 \det \left ( \begin{matrix} B, v_{n} \end{matrix} \right ) = v_{n, 1} det(B,vn)=vn,1
B ∗ A 1 B = ( 0 n − 1 , 1 I n − 1 ) ( a 11 ⋯ ⋮ M 1 ′ ) ( 0 1 , n − 1 I n − 1 ) = M 1 ′ B^{*}A _{1} B = \begin{pmatrix} 0_{n-1, 1} & I_{n - 1} \end{pmatrix} \begin{pmatrix} a_{11} & \cdots \\ \vdots & M_{1}' \end{pmatrix} \begin{pmatrix} 0_{1, n-1} \\ I_{n - 1} \end{pmatrix} = M_{1}' BA1B=(0n1,1In1)(a11M1)(01,n1In1)=M1
于是
∣ v n , 1 ∣ 2 ∏ k = 1 n − 1 λ k ( A 1 ) = ∏ i = 1 n − 1 λ i ( A 1 ) ∣ v n , 1 ∣ 2 = ∏ i = 1 n − 1 λ i ( A 1 ) ∣ det ⁡ ( B , v n ) ∣ 2 = det ⁡ ( B ∗ A 1 B ) = det ⁡ ( M 1 ′ ) = ∏ k = 1 n − 1 λ k ( M 1 ′ ) \left \vert v_{n, 1} \right \vert ^{2} \prod \limits_{k = 1} ^{n - 1} \lambda _{k} {(A_{1})} = \prod \limits_{i = 1} ^{n - 1} \lambda _{i} {(A_{1})} \left \vert v_{n, 1} \right \vert ^{2} = \prod \limits_{i = 1} ^{n - 1} \lambda _ {i} {(A_{1})} \left \vert \det \left ( \begin{matrix} B, v_{n} \end{matrix} \right ) \right \vert ^{2} = \det (B^{*}A_{1} B) = \det(M_{1}') = \prod \limits_{k = 1} ^{n - 1} \lambda _{k} {(M_{1}')} vn,12k=1n1λk(A1)=i=1n1λi(A1)vn,12=i=1n1λi(A1)det(B,vn)2=det(BA1B)=det(M1)=k=1n1λk(M1)

因此命题成立。

References

https://terrytao.wordpress.com/2019/08/13/eigenvectors-from-eigenvalues/
https://arxiv.org/abs/1908.03795

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