冯诺依曼迹不等式的定义
假设
A
∈
S
+
+
n
\mathbf{A} \in {\mathbb{S}_{++}^n}
A∈S++n，
B
∈
S
+
+
n
\mathbf{B} \in {\mathbb{S}_{++}^n}
B∈S++n，其特征值分别为：
λ
1
(
A
)
≥
λ
2
(
A
)
≥
⋯
≥
λ
n
(
A
)
>
0
,
λ
1
(
B
)
≥
λ
2
(
B
)
≥
⋯
≥
λ
n
(
B
)
>
0
\lambda_1(\mathbf{A}) \geq \lambda_2(\mathbf{A}) \geq \cdots \geq \lambda_n(\mathbf{A}) >0,\lambda_1(\mathbf{B}) \geq \lambda_2(\mathbf{B}) \geq \cdots \geq \lambda_n(\mathbf{B})>0
λ1(A)≥λ2(A)≥⋯≥λn(A)>0,λ1(B)≥λ2(B)≥⋯≥λn(B)>0
有如不等式成立：
∑
i
=
1
n
λ
i
(
A
)
λ
n
−
i
+
1
(
B
)
≤
t
r
(
A
B
)
=
∑
i
=
1
n
λ
i
(
A
B
)
≤
∑
i
=
1
n
λ
i
(
A
)
λ
i
(
B
)
\sum_{i=1}^n \lambda_i(\mathbf{A})\lambda_{n-i+1}(\mathbf{B})\leq\mathrm{tr}(\mathbf{AB})=\sum_{i=1}^n \lambda_i(\mathbf{A B})\leq \sum_{i=1}^n \lambda_i(\mathbf{A})\lambda_{i}(\mathbf{B})
i=1∑nλi(A)λn−i+1(B)≤tr(AB)=i=1∑nλi(AB)≤i=1∑nλi(A)λi(B)
该不等式称为：冯诺依曼不等式。
本文主要证明冯诺依曼迹不等式的左边部分，右边不等式较为容易理解。
冯诺依曼迹不等式的证明
柯西中值定理
在证明之前需要引入一个重要的柯西中值定理（Cauchy Interlacing Theorem）：
λ
i
+
1
(
C
)
≤
λ
i
(
C
~
)
≤
λ
i
(
C
)
\lambda_{i+1}(\mathbf{C}) \leq \lambda_i(\widetilde{\mathbf{C}}) \leq \lambda_i(\mathbf{C})
λi+1(C)≤λi(C
)≤λi(C)
其中
C
~
∈
S
+
+
n
−
1
\widetilde{\mathbf{C}} \in {\mathbb{S}_{++}^{n-1}}
C
∈S++n−1是
C
∈
S
+
+
n
\mathbf{C} \in {\mathbb{S}_{++}^n}
C∈S++n
n
−
1
n-1
n−1维度的主子矩阵(Principal submatrix)。
证：
存在一个正交矩阵
P
∈
R
(
n
)
×
(
n
−
1
)
\mathbf{P} \in \mathbb{R}^{(n) \times(n-1)}
P∈R(n)×(n−1)使得
P
T
C
P
=
C
~
\mathbf{P}^{\mathrm{T}} \mathbf{C P}=\widetilde{\mathbf{C}}
PTCP=C
，对于
i
≤
n
−
1
i\leq n-1
i≤n−1基于Courant-Fischer theorem，有：
λ
i
(
C
~
)
=
max
S
i
⊆
R
n
−
1
min
x
∈
S
i
,
∥
x
∥
2
=
1
x
T
C
~
x
=
max
S
i
⊆
R
n
−
1
min
x
∈
S
i
,
∥
x
∥
2
=
1
(
P
x
)
T
C
(
P
x
)
≤
max
P
i
∈
R
n
min
y
∈
P
i
,
∥
y
∥
2
=
1
y
T
C
y
=
λ
i
(
C
)
\lambda_i(\widetilde{\mathbf{C}})=\max _{\mathcal{S}_i \subseteq \mathbb{R}^{n-1}} \min _{\mathbf{x} \in \mathcal{S}_i,\|\mathbf{x}\|_2=1} \mathbf{x}^{\mathrm{T}} \widetilde{\mathbf{C}} \mathbf{x}=\max _{\mathcal{S}_i \subseteq \mathbb{R}^{n-1}} \min _{\mathbf{x} \in \mathcal{S}_i,\|\mathbf{x}\|_2=1}(\mathbf{P x})^{\mathrm{T}} \mathbf{C}(\mathbf{P x})\leq \max _{\mathcal{P}_i \in \mathbb{R}^n} \min _{\mathbf{y} \in \mathcal{P}_i,\|\boldsymbol{y}\|_2=1} \mathbf{y}^{\mathrm{T}} \mathbf{C y}=\lambda_i(\mathbf{C})
λi(C
)=Si⊆Rn−1maxx∈Si,∥x∥2=1minxTC
x=Si⊆Rn−1maxx∈Si,∥x∥2=1min(Px)TC(Px)≤Pi∈Rnmaxy∈Pi,∥y∥2=1minyTCy=λi(C)
同时有：
λ
i
(
C
~
)
=
min
S
n
−
i
∈
R
n
−
1
max
x
∈
S
n
−
i
,
∥
x
∥
2
=
1
x
T
C
~
x
=
min
S
n
−
i
⊆
R
n
−
1
max
x
∈
S
n
−
i
,
∥
x
∥
2
=
1
(
P
x
)
T
C
(
P
x
)
≥
min
P
n
−
i
∈
R
n
max
y
∈
P
n
−
i
,
∥
y
∥
2
=
1
y
T
C
y
=
λ
i
+
1
(
C
)
\lambda_i(\widetilde{\mathbf{C}})=\min _{\mathcal{S}_{n-i} \in \mathbb{R}^{n-1}} \max _{\mathbf{x} \in \mathcal{S}_{n-i},\|\mathbf{x}\|_2=1} \mathbf{x}^{\mathrm{T}} \widetilde{\mathbf{C}} \mathbf{x}=\min _{\mathcal{S}_{n-i} \subseteq \mathbb{R}^{n-1}} \max _{\mathbf{x} \in \mathcal{S}_{n-i},\|\mathbf{x}\|_2=1}(\mathbf{P x})^{\mathrm{T}} \mathbf{C}(\mathbf{P x})\geq\min _{\mathcal{P}_{n-i} \in \mathbb{R}^n} \max _{\mathbf{y} \in \mathcal{P}_{n-i},\|y\|_2=1} \mathbf{y}^{\mathrm{T}} \mathbf{C y}=\lambda_{i+1}(\mathbf{C})
λi(C
)=Sn−i∈Rn−1minx∈Sn−i,∥x∥2=1maxxTC
x=Sn−i⊆Rn−1minx∈Sn−i,∥x∥2=1max(Px)TC(Px)≥Pn−i∈Rnminy∈Pn−i,∥y∥2=1maxyTCy=λi+1(C)
综上：
λ
i
+
1
(
C
)
≤
λ
i
(
C
~
)
≤
λ
i
(
C
)
\lambda_{i+1}(\mathbf{C}) \leq \lambda_i(\widetilde{\mathbf{C}}) \leq \lambda_i(\mathbf{C})
λi+1(C)≤λi(C
)≤λi(C)
证毕。
数学归纳法
基于上面的柯西中值定理，然后利用数学归纳法证明：
∑ i = 1 n − 1 ( λ i ( Λ A ) − λ n ( Λ A ) ) C i , i ≥ ∑ i = 1 n − 1 ( λ i ( Λ A ) − λ n ( Λ A ) ) λ n − i ( C ~ ) \sum_{i=1}^{n-1}( \lambda_i(\boldsymbol{\Lambda}_{\mathbf{A}})-\lambda_n(\boldsymbol{\Lambda}_{\mathbf{A}}))\mathbf{C}_{i, i}\geq\sum_{i=1}^{n-1}(\lambda_i(\boldsymbol{\Lambda}_{\mathbf{A}})-\lambda_n(\boldsymbol{\Lambda}_{\mathbf{A}})) \lambda_{n-i}(\widetilde{\mathbf{C}}) i=1∑n−1(λi(ΛA)−λn(ΛA))Ci,i≥i=1∑n−1(λi(ΛA)−λn(ΛA))λn−i(C )
证明：
定义对角矩阵 Λ A ′ ∈ S + + k + 1 \boldsymbol{\Lambda}^{'}_{\mathbf{A}} \in {\mathbb{S}_{++}^{k+1}} ΛA′∈S++k+1和 C ′ ∈ S + + k + 1 \mathbf{C}^{'} \in {\mathbb{S}_{++}^{k+1}} C′∈S++k+1，其中对角阵 Λ A ∈ S + + k \boldsymbol{\Lambda}_{\mathbf{A}}\in \mathbb{S}_{++}^{k} ΛA∈S++k和 C ∈ S + + k \mathbf{C}\in \mathbb{S}_{++}^{k} C∈S++k分别是 Λ A ′ \boldsymbol{\Lambda}^{'}_{\mathbf{A}} ΛA′和 C ′ \mathbf{C}^{'} C′的 k k k维主子矩阵，因此对于任意的 1 ≤ i ≤ k 1 \leq i \leq k 1≤i≤k，有 λ i ( Λ A ′ ) = λ i ( Λ A ) \lambda_i(\boldsymbol{\Lambda}^{'}_{\mathbf{A}})=\lambda_i(\boldsymbol{\Lambda}_{\mathbf{A}}) λi(ΛA′)=λi(ΛA)和 C i , i ′ = C i , i \mathbf{C}_{i, i}^{\prime}=\mathbf{C}_{i, i} Ci,i′=Ci,i成立。
当 n = 2 n=2 n=2时，显然成立。
假定当
n
=
k
n=k
n=k时成立，有：
∑
i
=
1
k
−
1
(
λ
i
(
Λ
A
)
−
λ
k
(
Λ
A
)
)
C
i
,
i
≥
∑
i
=
1
k
−
1
(
λ
i
(
Λ
A
)
−
λ
k
(
Λ
A
)
)
λ
k
−
i
(
C
~
)
\sum_{i=1}^{k-1}\left(\lambda_i(\boldsymbol{\Lambda}_{\mathbf{A}})-\lambda_k(\boldsymbol{\Lambda}_{\mathbf{A}})\right) \mathbf{C}_{i, i} \geq \sum_{i=1}^{k-1}\left(\lambda_i(\boldsymbol{\Lambda}_{\mathbf{A}})-\lambda_k(\boldsymbol{\Lambda}_{\mathbf{A}})\right) \lambda_{k-i}(\widetilde{\mathbf{C}})
i=1∑k−1(λi(ΛA)−λk(ΛA))Ci,i≥i=1∑k−1(λi(ΛA)−λk(ΛA))λk−i(C
)
则当
n
=
k
+
1
n=k+1
n=k+1时有：
∑
i
=
1
k
(
λ
i
(
Λ
A
′
)
−
λ
k
+
1
(
Λ
A
′
)
)
C
i
,
i
′
=
∑
i
=
1
k
(
λ
i
(
Λ
A
′
)
−
λ
k
(
Λ
A
′
)
+
λ
k
(
Λ
A
′
)
−
λ
k
+
1
(
A
′
)
)
C
i
,
i
′
=
∑
i
=
1
k
−
1
(
λ
i
(
Λ
A
′
)
−
λ
k
(
Λ
A
′
)
)
C
i
,
i
′
+
(
λ
k
(
Λ
A
′
)
−
λ
k
+
1
(
Λ
A
′
)
)
∑
i
=
1
k
C
i
,
i
′
=
∑
i
=
1
k
−
1
(
λ
i
(
Λ
A
)
−
λ
k
(
Λ
A
)
)
C
i
,
i
+
(
λ
k
(
Λ
A
)
−
λ
k
+
1
(
Λ
A
′
)
)
∑
i
=
1
k
C
i
,
i
′
≥
∑
i
=
1
k
−
1
(
λ
i
(
Λ
A
)
−
λ
k
(
Λ
A
)
)
λ
k
−
i
(
C
~
)
+
(
λ
k
(
Λ
A
)
−
λ
k
+
1
(
Λ
A
′
)
)
∑
i
=
1
k
C
i
,
i
′
≥
∑
i
=
1
k
−
1
(
λ
i
(
Λ
A
′
)
−
λ
k
(
Λ
A
′
)
)
λ
k
−
i
+
1
(
C
)
+
(
λ
k
(
Λ
A
′
)
−
λ
k
+
1
(
Λ
A
′
)
)
∑
i
=
1
k
λ
k
−
i
+
1
(
C
)
=
∑
i
=
1
k
(
λ
i
(
Λ
A
′
)
−
λ
k
+
1
(
Λ
A
′
)
)
λ
k
−
i
+
1
(
C
)
\begin{aligned} & \sum_{i=1}^k(\lambda_i(\boldsymbol{\Lambda}^{'}_{\mathbf{A}})-\lambda_{k+1}(\boldsymbol{\Lambda}^{'}_{\mathbf{A}} )) \mathbf{C}_{i, i}^{\prime} \\ =& \sum_{i=1}^k(\lambda_i(\boldsymbol{\Lambda}^{'}_{\mathbf{A}})-\lambda_{k}(\boldsymbol{\Lambda}^{'}_{\mathbf{A}} )+\lambda_{k}(\boldsymbol{\Lambda}^{'}_{\mathbf{A}} )-\lambda_{k+1}(\mathbf{A}^{\prime})) \mathbf{C}_{i, i}^{\prime} \\ =& \sum_{i=1}^{k-1}(\lambda_i(\boldsymbol{\Lambda}^{'}_{\mathbf{A}})-\lambda_{k}(\boldsymbol{\Lambda}^{'}_{\mathbf{A}} )) \mathbf{C}_{i, i}^{\prime}+(\lambda_k(\boldsymbol{\Lambda}^{'}_{\mathbf{A}})-\lambda_{k+1}(\boldsymbol{\Lambda}^{'}_{\mathbf{A}} )) \sum_{i=1}^k \mathbf{C}_{i, i}^{\prime}\\=& \sum_{i=1}^{k-1}(\lambda_i(\boldsymbol{\Lambda}_{\mathbf{A}})-\lambda_{k}(\boldsymbol{\Lambda}_{\mathbf{A}} )) \mathbf{C}_{i, i}+(\lambda_k(\boldsymbol{\Lambda}_{\mathbf{A}})-\lambda_{k+1}(\boldsymbol{\Lambda}^{'}_{\mathbf{A}} )) \sum_{i=1}^k \mathbf{C}_{i, i}^{\prime}\\\geq&\sum_{i=1}^{k-1}(\lambda_i(\boldsymbol{\Lambda}_{\mathbf{A}})-\lambda_k(\boldsymbol{\Lambda}_{\mathbf{A}})) \lambda_{k-i}(\widetilde{\mathbf{C}})+(\lambda_k(\boldsymbol{\Lambda}_{\mathbf{A}})-\lambda_{k+1}(\boldsymbol{\Lambda}^{'}_{\mathbf{A}} )) \sum_{i=1}^k \mathbf{C}_{i, i}^{\prime}\\\geq&\sum_{i=1}^{k-1}(\lambda_i(\boldsymbol{\Lambda}^{'}_{\mathbf{A}})-\lambda_{k}(\boldsymbol{\Lambda}^{'}_{\mathbf{A}} )) \lambda_{k-i+1}({\mathbf{C}})+(\lambda_k(\boldsymbol{\Lambda}_{\mathbf{A}}^{\prime})-\lambda_{k+1}(\boldsymbol{\Lambda}^{'}_{\mathbf{A}} )) \sum_{i=1}^k \lambda_{k-i+1}(\mathbf{C})\\=&\sum_{i=1}^{k}(\lambda_i(\boldsymbol{\Lambda}^{'}_{\mathbf{A}})-\lambda_{k+1}(\boldsymbol{\Lambda}^{'}_{\mathbf{A}} )) \lambda_{k-i+1}({\mathbf{C}}) \end{aligned}
===≥≥=i=1∑k(λi(ΛA′)−λk+1(ΛA′))Ci,i′i=1∑k(λi(ΛA′)−λk(ΛA′)+λk(ΛA′)−λk+1(A′))Ci,i′i=1∑k−1(λi(ΛA′)−λk(ΛA′))Ci,i′+(λk(ΛA′)−λk+1(ΛA′))i=1∑kCi,i′i=1∑k−1(λi(ΛA)−λk(ΛA))Ci,i+(λk(ΛA)−λk+1(ΛA′))i=1∑kCi,i′i=1∑k−1(λi(ΛA)−λk(ΛA))λk−i(C
)+(λk(ΛA)−λk+1(ΛA′))i=1∑kCi,i′i=1∑k−1(λi(ΛA′)−λk(ΛA′))λk−i+1(C)+(λk(ΛA′)−λk+1(ΛA′))i=1∑kλk−i+1(C)i=1∑k(λi(ΛA′)−λk+1(ΛA′))λk−i+1(C)
-
注：第一个不等式利用了
n
=
k
n=k
n=k时成立的不等式，第二个不等式利用了前面的柯西中值定理。
即：
∑ i = 1 k ( λ i ( Λ A ′ ) − λ k + 1 ( Λ A ′ ) ) C i , i ′ ≥ ∑ i = 1 k ( λ i ( Λ A ′ ) − λ k + 1 ( Λ A ′ ) ) λ k − i + 1 ( C ) \sum_{i=1}^k(\lambda_i(\boldsymbol{\Lambda}^{'}_{\mathbf{A}})-\lambda_{k+1}(\boldsymbol{\Lambda}^{'}_{\mathbf{A}} )) \mathbf{C}_{i, i}^{\prime}\geq\sum_{i=1}^{k}(\lambda_i(\boldsymbol{\Lambda}^{'}_{\mathbf{A}})-\lambda_{k+1}(\boldsymbol{\Lambda}^{'}_{\mathbf{A}} )) \lambda_{k-i+1}({\mathbf{C}}) i=1∑k(λi(ΛA′)−λk+1(ΛA′))Ci,i′≥i=1∑k(λi(ΛA′)−λk+1(ΛA′))λk−i+1(C)
证毕。
冯诺依曼迹不等式的证明
对
A
\mathbf{A}
A和
A
\mathbf{A}
A做特征值分解分别有：
A
=
U
Λ
A
U
T
\mathbf{A}=\mathbf{U} \boldsymbol{\Lambda}_{\mathbf{A}} \mathbf{U}^{\mathrm{T}}
A=UΛAUT和
B
=
V
Λ
B
V
T
\mathbf{B}=\mathbf{V} \boldsymbol{\Lambda}_{\mathbf{B}} \mathbf{V}^{\mathrm{T}}
B=VΛBVT，则：
tr
(
A
B
)
=
tr
(
U
Λ
A
U
T
V
Λ
B
V
T
)
=
tr
(
Λ
A
U
T
V
⏟
Q
Λ
B
V
T
U
⏟
Q
T
)
=
tr
(
Λ
A
Q
Λ
B
Q
T
⏟
C
)
=
tr
(
Λ
A
C
)
=
∑
i
=
1
n
λ
i
(
Λ
A
)
C
i
,
i
=
∑
i
=
1
n
λ
i
(
A
)
C
i
,
i
\operatorname{tr}(\mathbf{A B})=\operatorname{tr}\left(\mathbf{U} \boldsymbol{\Lambda}_{\mathbf{A}} \mathbf{U}^{\mathrm{T}} \mathbf{V} \boldsymbol{\Lambda}_{\mathbf{B}} \mathbf{V}^{\mathrm{T}}\right)=\operatorname{tr}(\boldsymbol{\Lambda}_{\mathbf{A}} \underbrace{\mathbf{U}^{\mathrm{T}} \mathbf{V}}_{\mathbf{Q}} \boldsymbol{\Lambda}_{\mathbf{B}} \underbrace{\mathbf{V}^{\mathrm{T}} \mathbf{U}}_{\mathbf{Q}^{\mathrm{T}}})=\operatorname{tr}(\underbrace{\boldsymbol{\Lambda}_{\mathrm{A}} \mathbf{Q} \boldsymbol{\Lambda}_{\mathbf{B}} \mathbf{Q}^{\mathrm{T}}}_{\mathrm{C}})=\operatorname{tr}\left(\boldsymbol{\Lambda}_{\mathbf{A}} \mathbf{C}\right)=\sum_{i=1}^n \lambda_i(\boldsymbol{\Lambda}_{\mathbf{A}}) \mathbf{C}_{i, i}=\sum_{i=1}^n \lambda_i(\mathbf{A}) \mathbf{C}_{i, i}
tr(AB)=tr(UΛAUTVΛBVT)=tr(ΛAQ
UTVΛBQT
VTU)=tr(C
ΛAQΛBQT)=tr(ΛAC)=i=1∑nλi(ΛA)Ci,i=i=1∑nλi(A)Ci,i
不难发现， Λ A \boldsymbol{\Lambda}_{\mathbf{A}} ΛA和 A \mathbf{A} A有相同的特征值， C \mathbf{C} C和 B \mathbf{B} B有相同的特征值。
∑ i = 1 n λ i ( Λ A ) C i , i = ∑ i = 1 n − 1 λ i ( Λ A ) C i , i + λ n ( Λ A ) ( ∑ i = 1 n C i , i − ∑ i = 1 n − 1 C i , i ) = ∑ i = 1 n − 1 ( λ i ( Λ A ) − λ n ( Λ A ) ) C i , i + λ n ( Λ A ) ∑ i = 1 n C i , i ≥ ∑ i = 1 n − 1 ( λ i ( Λ A ) − λ n ( Λ A ) ) λ n − i ( C ~ ) + λ n ( Λ A ) ∑ i = 1 n C i , i ≥ ∑ i = 1 n − 1 ( λ i ( Λ A ) − λ n ( Λ A ) ) λ n − i + 1 ( C ) + λ n ( Λ A ) ∑ i = 1 n λ n − i + 1 ( C ) = ∑ i = 1 n λ i ( Λ A ) λ n − i + 1 ( C ) = ∑ i = 1 n λ i ( A ) λ n − i + 1 ( B ) \begin{aligned} \sum_{i=1}^n \lambda_i(\boldsymbol{\Lambda}_{\mathbf{A}}) \mathbf{C}_{i, i}&=\sum_{i=1}^{n-1}\lambda_i(\boldsymbol{\Lambda}_{\mathbf{A}}) \mathbf{C}_{i, i}+\lambda_n(\boldsymbol{\Lambda}_{\mathbf{A}}) (\sum_{i=1}^n\mathbf{C}_{i, i}-\sum_{i=1}^{n-1}\mathbf{C}_{i, i})\\&=\sum_{i=1}^{n-1}( \lambda_i(\boldsymbol{\Lambda}_{\mathbf{A}})-\lambda_n(\boldsymbol{\Lambda}_{\mathbf{A}}))\mathbf{C}_{i, i}+\lambda_n(\boldsymbol{\Lambda}_{\mathbf{A}})\sum_{i=1}^n\mathbf{C}_{i, i}\\&\geq \sum_{i=1}^{n-1}(\lambda_i(\boldsymbol{\Lambda}_{\mathbf{A}})-\lambda_n(\boldsymbol{\Lambda}_{\mathbf{A}})) \lambda_{n-i}(\widetilde{\mathbf{C}})+\lambda_n(\boldsymbol{\Lambda}_{\mathbf{A}})\sum_{i=1}^n\mathbf{C}_{i, i}\\&\geq\sum_{i=1}^{n-1}(\lambda_i(\boldsymbol{\Lambda}_{\mathbf{A}})-\lambda_n(\boldsymbol{\Lambda}_{\mathbf{A}})) \lambda_{n-i+1}({\mathbf{C}})+\lambda_n(\boldsymbol{\Lambda}_{\mathbf{A}})\sum_{i=1}^n\lambda_{n-i+1}(\mathbf{C})\\&=\sum_{i=1}^{n}\lambda_i(\boldsymbol{\Lambda}_{\mathbf{A}}) \lambda_{n-i+1}({\mathbf{C}})=\sum_{i=1}^n \lambda_i(\mathbf{A})\lambda_{n-i+1}(\mathbf{B}) \end{aligned} i=1∑nλi(ΛA)Ci,i=i=1∑n−1λi(ΛA)Ci,i+λn(ΛA)(i=1∑nCi,i−i=1∑n−1Ci,i)=i=1∑n−1(λi(ΛA)−λn(ΛA))Ci,i+λn(ΛA)i=1∑nCi,i≥i=1∑n−1(λi(ΛA)−λn(ΛA))λn−i(C )+λn(ΛA)i=1∑nCi,i≥i=1∑n−1(λi(ΛA)−λn(ΛA))λn−i+1(C)+λn(ΛA)i=1∑nλn−i+1(C)=i=1∑nλi(ΛA)λn−i+1(C)=i=1∑nλi(A)λn−i+1(B)文章来源：https://www.toymoban.com/news/detail-628155.html
即： tr ( A B ) ≥ ∑ i = 1 n λ i ( A ) λ n − i + 1 ( B ) \operatorname{tr}(\mathbf{A B})\geq\sum_{i=1}^n \lambda_i(\mathbf{A})\lambda_{n-i+1}(\mathbf{B}) tr(AB)≥∑i=1nλi(A)λn−i+1(B)。文章来源地址https://www.toymoban.com/news/detail-628155.html
- 注：第一个不等式利用了数学归纳法证明的不等式，第二个不等式利用了前面的柯西中值定理。
到了这里，关于冯诺依曼(Von Neumann)迹不等式证明的文章就介绍完了。如果您还想了解更多内容，请在右上角搜索TOY模板网以前的文章或继续浏览下面的相关文章，希望大家以后多多支持TOY模板网！
