양정부호 행렬과 확장된 코시-슈바르츠 부등식 증명
📂행렬대수 양정부호 행렬과 확장된 코시-슈바르츠 부등식 증명 정리 임의의 두 벡터 b , d ∈ R p \mathbf{b}, \mathbf{d} \in \mathbf{R}^{p} b , d ∈ R p 양의 정부호 행렬 A ∈ R p × p A \in \mathbf{R}^{p \times p} A ∈ R p × p ( b T d ) 2 ≤ ( b T A b ) ( d T A − 1 d )
\left( \mathbf{b}^{T} \mathbf{d} \right)^{2} \le \left( \mathbf{b}^{T} A \mathbf{b} \right) \left( \mathbf{d}^{T} A^{-1} \mathbf{d} \right)
( b T d ) 2 ≤ ( b T A b ) ( d T A − 1 d ) 동치조건 은 어떤 상수 c ∈ R c \in \mathbb{R} c ∈ R b = c A − 1 d \mathbf{b} = c A^{-1} \mathbf{d} b = c A − 1 d d = c A b \mathbf{d} = c A \mathbf{b} d = c A b
설명 이 부등식은 코시-슈바르츠 부등식 의 일반화로써, A A A 항등행렬 I I I 이차 형식 이 등장하고, 자연스럽게 수리통계학 에서의 활용도가 높다.
증명 Part 1. 부등식
양정부호 행렬의 역행렬과 제곱근 행렬 : 양의 정부호 행렬 A A A 고유쌍 { ( λ k , e k ) } k = 1 n \left\{ \left( \lambda_{k} , e_{k} \right) \right\}_{k=1}^{n} { ( λ k , e k ) } k = 1 n λ 1 > ⋯ > λ n > 0 \lambda_{1} > \cdots > \lambda_{n} > 0 λ 1 > ⋯ > λ n > 0 직교행렬 P = [ e 1 ⋯ e n ] ∈ R n × n P = \begin{bmatrix} e_{1} & \cdots & e_{n} \end{bmatrix} \in \mathbb{R}^{n \times n} P = [ e 1 ⋯ e n ] ∈ R n × n 대각행렬 Λ = diag ( λ 1 , ⋯ , λ n ) \Lambda = \diag \left( \lambda_{1} , \cdots , \lambda_{n} \right) Λ = diag ( λ 1 , ⋯ , λ n ) A A A 역행렬 A − 1 A^{-1} A − 1 제곱근 행렬 A \sqrt{A} A A − 1 = P Λ − 1 P T = ∑ k = 1 n 1 λ k e k e k T A = P Λ P T = ∑ k = 1 n λ k e k e k T
\begin{align*}
A^{-1} =& P \Lambda^{-1} P^{T} = \sum_{k=1}^{n} {{ 1 } \over { \lambda_{k} }} e_{k} e_{k}^{T}
\\ \sqrt{A} =& P \sqrt{\Lambda} P^{T} = \sum_{k=1}^{n} \sqrt{\lambda_{k}} e_{k} e_{k}^{T}
\end{align*}
A − 1 = A = P Λ − 1 P T = k = 1 ∑ n λ k 1 e k e k T P Λ P T = k = 1 ∑ n λ k e k e k T
A A A A A A 제곱근 행렬 은
A = P Λ P T = ∑ k = 1 n λ k e k e k T
\sqrt{A} = P \sqrt{\Lambda} P^{T} = \sum_{k=1}^{n} \sqrt{\lambda_{k}} e_{k} e_{k}^{T}
A = P Λ P T = k = 1 ∑ n λ k e k e k T 전치행렬 이어서 A 1 / 2 = ( A 1 / 2 ) T A^{1/2} = \left( A^{1/2} \right)^{T} A 1/2 = ( A 1/2 ) T A − 1 A^{-1} A − 1
x : = A 1 / 2 b \mathbf{x} := A^{1/2} \mathbf{b} x := A 1/2 b y : = A − 1 / 2 d \mathbf{y} := A^{-1/2} \mathbf{d} y := A − 1/2 d ( x T y ) ≤ ( x T x ) ( y T y ) \left( \mathbf{x}^{T} \mathbf{y} \right) \le \left( \mathbf{x}^{T} \mathbf{x} \right) \left( \mathbf{y}^{T} \mathbf{y} \right) ( x T y ) ≤ ( x T x ) ( y T y ) ( b T d ) 2 = ( b T A 1 / 2 A − 1 / 2 d ) 2 = ( b T ( A 1 / 2 ) T A − 1 / 2 d ) 2 = ( ( A 1 / 2 b ) T A − 1 / 2 d ) 2 = ( ( A 1 / 2 b ) T A − 1 / 2 d ) 2 = ( x T y ) 2 ≤ ( x T x ) ( y T y ) ≤ ( ( A 1 / 2 b ) T ( A 1 / 2 b ) ) ( ( A − 1 / 2 d ) T ( A − 1 / 2 d ) ) = ( b T A b ) ( d T A − 1 d )
\begin{align*}
& \left( \mathbf{b}^{T} \mathbf{d} \right)^{2}
\\ =& \left( \mathbf{b}^{T} A^{1/2} A^{-1/2} \mathbf{d} \right)^{2}
\\ =& \left( \mathbf{b}^{T} \left( A^{1/2} \right)^{T} A^{-1/2} \mathbf{d} \right)^{2}
\\ =& \left( \left( A^{1/2} \mathbf{b} \right)^{T} A^{-1/2} \mathbf{d} \right)^{2}
\\ =& \left( \left( A^{1/2} \mathbf{b} \right)^{T} A^{-1/2} \mathbf{d} \right)^{2}
\\ =& \left( \mathbf{x}^{T} \mathbf{y} \right)^{2}
\\ \le & \left( \mathbf{x}^{T} \mathbf{x} \right) \left( \mathbf{y}^{T} \mathbf{y} \right)
\\ \le & \left( \left( A^{1/2} \mathbf{b} \right)^{T} \left( A^{1/2} \mathbf{b} \right) \right) \left( \left( A^{-1/2} \mathbf{d} \right)^{T} \left( A^{-1/2} \mathbf{d} \right) \right)
\\ =& \left( \mathbf{b}^{T} A \mathbf{b} \right) \left( \mathbf{d}^{T} A^{-1} \mathbf{d} \right)
\end{align*}
= = = = = ≤ ≤ = ( b T d ) 2 ( b T A 1/2 A − 1/2 d ) 2 ( b T ( A 1/2 ) T A − 1/2 d ) 2 ( ( A 1/2 b ) T A − 1/2 d ) 2 ( ( A 1/2 b ) T A − 1/2 d ) 2 ( x T y ) 2 ( x T x ) ( y T y ) ( ( A 1/2 b ) T ( A 1/2 b ) ) ( ( A − 1/2 d ) T ( A − 1/2 d ) ) ( b T A b ) ( d T A − 1 d )
Part 2. 등식
어떤 상수가 c = 0 c = 0 c = 0 b = 0 \mathbf{b} = \mathbf{0} b = 0 d = 0 \mathbf{d} = \mathbf{0} d = 0 영벡터 가 아니고, 일반성을 잃지 않고 d = c A b \mathbf{d} = c A \mathbf{b} d = c A b b T d = b T c A b = c b T A b
\begin{align*}
\mathbf{b}^{T} \mathbf{d} =& \mathbf{b}^{T} c A \mathbf{b}
\\ =& c \mathbf{b}^{T} A \mathbf{b}
\end{align*}
b T d = = b T c A b c b T A b b T d = ( 1 c A − 1 d ) T d = 1 c d T A − 1 d
\begin{align*}
\mathbf{b}^{T} \mathbf{d} =& \left( {{ 1 } \over { c }} A^{-1} \mathbf{d} \right)^{T} \mathbf{d}
\\ =& {{ 1 } \over { c }} \mathbf{d}^{T} A^{-1} \mathbf{d}
\end{align*}
b T d = = ( c 1 A − 1 d ) T d c 1 d T A − 1 d ( b T d ) 2 = ( b T A b ) ( d T A − 1 d )
\left( \mathbf{b}^{T} \mathbf{d} \right)^{2} = \left( \mathbf{b}^{T} A \mathbf{b} \right) \left( \mathbf{d}^{T} A^{-1} \mathbf{d} \right)
( b T d ) 2 = ( b T A b ) ( d T A − 1 d )
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