在推导标准化拉普拉斯矩阵的特征值范围时,用到了瑞利商,它也是LDA最大化目标函数使用的定义。
瑞利商的定义为:
R ( A , x ) = x T A x x T x R(A,x)=\frac{x^TAx}{x^Tx} R(A,x)=xTxxTAx,其中 A A A 为 n × n n\times n n×n 对称矩阵, x x x 为 n n n 维度向量。
设对称矩阵 A A A 的特征值为 λ 1 ≤ λ 2 ≤ ⋯ ≤ λ n \lambda_1\le\lambda_2\le\cdots\le\lambda_n λ1≤λ2≤⋯≤λn,对应的特征向量为 v 1 , v 2 , ⋯ , v n v_1,v_2,\cdots,v_n v1,v2,⋯,vn。有:
m a x x R ( A , x ) = λ n m i n x R ( A , x ) = λ 1 \mathop{max}\limits_{x}R(A,x)=\lambda_n \\ \mathop{min}\limits_{x}R(A,x)=\lambda_1 xmaxR(A,x)=λnxminR(A,x)=λ1
由于A为对称矩阵,可以进行相似对角化:
A = U Λ U T A=U\Lambda U^T A=UΛUT,其中特征对角阵 Λ = d i a g ( λ 1 , λ 2 , . . . , λ n ) \Lambda=diag(\lambda_1,\lambda_2,...,\lambda_n) Λ=diag(λ1,λ2,...,λn),特征向量阵 U = ( v 1 , v 2 , . . . , v n ) U=(v_1,v_2,...,v_n) U=(v1,v2,...,vn),且满足 U U T = I UU^T=I UUT=I。有:
R ( A , x ) = x T A x x T x = x T U Λ U T x x T U U T x R(A,x)=\frac{x^TAx}{x^Tx}=\frac{x^TU\Lambda U^Tx}{x^TUU^Tx} R(A,x)=xTxxTAx=xTUUTxxTUΛUTx,令 y = U T x y=U^Tx y=UTx,有:
R ( A , x ) = y T Λ y y T y \begin{aligned} R(A,x)&=\frac{y^T\Lambda y}{y^Ty}\\ \end{aligned} R(A,x)=yTyyTΛy ∵ \because ∵
y T Λ y = [ y 1 , y 2 , ⋯ , y n ] [ λ 1 λ 2 ⋱ λ n ] [ y 1 y 2 ⋮ y n ] = ∑ i n λ i y i 2 y y T = [ y 1 , y 2 , ⋯ , y n ] [ y 1 y 2 ⋮ y n ] = ∑ i n y i 2 \begin{aligned} &y^T\Lambda y=[y_1,y_2,\cdots,y_n] \begin{bmatrix} \lambda_1 & & & \\ & \lambda_2 & & \\ & & \ddots & \\ & & & \lambda_n \\ \end{bmatrix} \begin{bmatrix} y_1\\ y_2\\ \vdots\\ y_n\\ \end{bmatrix}=\sum_i^n\lambda_iy_i^2 \\ &yy^T=[y_1,y_2,\cdots,y_n] \begin{bmatrix} y_1\\ y_2\\ \vdots\\ y_n\\ \end{bmatrix}=\sum_i^ny_i^2 \\ \end{aligned} yTΛy=[y1,y2,⋯,yn] λ1λ2⋱λn y1y2⋮yn =i∑nλiyi2yyT=[y1,y2,⋯,yn] y1y2⋮yn =i∑nyi2 ∴ \therefore ∴
R ( A , x ) = ∑ i n λ i y i 2 ∑ i n y i 2 \begin{aligned} R(A,x)=\frac{\mathop{\sum}\limits_{i}^n \lambda_iy_i^2}{\mathop{\sum}\limits_{i}^n y_i^2} \end{aligned} R(A,x)=i∑nyi2i∑nλiyi2 由于特征值大小关系 λ 1 ≤ λ 2 ≤ ⋯ ≤ λ n \lambda_1\le\lambda_2\le\cdots\le\lambda_n λ1≤λ2≤⋯≤λn,显然有:
λ 1 ∑ i n y i 2 ≤ ∑ i n λ i y i 2 ≤ λ n ∑ i n y i 2 \lambda_1\mathop{\sum}\limits_{i}^n y_i^2 \leq \mathop{\sum}\limits_{i}^n\lambda_i y_i^2 \leq \lambda_n \mathop{\sum}\limits_{i}^n y_i^2 λ1i∑nyi2≤i∑nλiyi2≤λni∑nyi2,于是有:
λ 1 ∑ i n y i 2 ∑ i n y i 2 ≤ ∑ i n λ i y i 2 ∑ i n y i 2 ≤ λ n ∑ i n y I 2 ∑ i n y i 2 \frac{\lambda_1\mathop{\sum}\limits_{i}^n y_i^2}{\mathop{\sum}\limits_{i}^n y_i^2} \leq \frac{\mathop{\sum}\limits_{i}^n\lambda_i y_i^2}{\mathop{\sum}\limits_{i}^n y_i^2} \leq \frac{\lambda_n \mathop{\sum}\limits_{i}^n y_I^2}{\mathop{\sum}\limits_{i}^n y_i^2} i∑nyi2λ1i∑nyi2≤i∑nyi2i∑nλiyi2≤i∑nyi2λni∑nyI2,即:
λ 1 ≤ R ( A , x ) ≤ λ n \lambda_1 \leq R(A,x) \leq \lambda_n λ1≤R(A,x)≤λn
上一节已经证明了瑞利商的范围为 [ λ m i n , λ m a x ] [\lambda_{min}, \lambda_{max}] [λmin,λmax],要想知道上、下确界为 λ m i n , λ m a x \lambda_{min}, \lambda_{max} λmin,λmax,只需要寻找特殊值使 R ( A , x ) = λ 1 R(A,x) = \lambda_1 R(A,x)=λ1 或 R ( A , x ) = λ n R(A,x) = \lambda_n R(A,x)=λn 即可。
由于 v 1 , v 2 , . . . , v n v_1,v_2,...,v_n v1,v2,...,vn 是 n n n 维空间的一组单位正交基,因此可以设 n n n 维向量 x = ∑ i = 1 n k i v i x=\sum_{i=1}^{n} k_iv_i x=∑i=1nkivi,为简单起见,我们设 x T x = 1 x^Tx=1 xTx=1。有:
y = U T x = [ v 1 T v 2 T ⋮ v n T ] ( k 1 v + k 2 v 2 + ⋯ + k n v n ) = [ k 1 k 2 ⋮ k n ] y=U^Tx= \begin{bmatrix} v_1^T\\ v_2^T\\ \vdots\\ v_n^T\\ \end{bmatrix} (k_1v+k_2v_2+\cdots+k_nv_n)= \begin{bmatrix} k_1\\ k_2\\ \vdots\\ k_n\\ \end{bmatrix} y=UTx= v1Tv2T⋮vnT (k1v+k2v2+⋯+knvn)= k1k2⋮kn ,所以有:
R ( A , x ) = x T A x x T x = x T U Λ U T x x T x = y T Λ y = [ k 1 , k 2 , ⋯ , k n ] [ λ 1 λ 2 ⋱ λ n ] [ k 1 k 2 ⋮ k n ] = ∑ i = 1 n k i 2 λ i \begin{aligned} R(A,x)&=\frac{x^TAx}{x^Tx} \\ &={\frac {x^TU\Lambda U^T x} {x^Tx}} \\ &=y^T\Lambda y\\ &=[k_1,k_2,\cdots,k_n] \begin{bmatrix} \lambda_1 & & & \\ & \lambda_2 & & \\ & & \ddots & \\ & & & \lambda_n \\ \end{bmatrix} \begin{bmatrix} k_1\\ k_2\\ \vdots\\ k_n\\ \end{bmatrix} \\ &=\sum_{i=1}^nk_i^2\lambda_i \end{aligned} R(A,x)=xTxxTAx=xTxxTUΛUTx=yTΛy=[k1,k2,⋯,kn] λ1λ2⋱λn k1k2⋮kn =i=1∑nki2λi 显然有: λ 1 ∑ i = 1 n k i 2 ≤ ∑ i = 1 n k i 2 λ i < λ n ∑ i = 1 n k i 2 \lambda_1\sum_{i=1}^nk_i^2\le\sum_{i=1}^nk_i^2\lambda_i<\lambda_n\sum_{i=1}^nk_i^2 λ1∑i=1nki2≤∑i=1nki2λi<λn∑i=1nki2,即:
λ 1 ≤ R ( A , x ) ≤ λ n \lambda_1\le R(A,x)\le\lambda_n λ1≤R(A,x)≤λn
因为 R ( A , x ) = ∑ i = 1 n k i 2 λ i = k 1 2 λ 1 + k 2 2 λ 2 + ⋯ + k n 2 λ n R(A,x)=\sum_{i=1}^nk_i^2\lambda_i=k_1^2\lambda_1+k_2^2\lambda_2+\cdots+k_n^2\lambda_n R(A,x)=∑i=1nki2λi=k12λ1+k22λ2+⋯+kn2λn, k 1 2 + k 2 2 + ⋯ + k n 2 = 1 k_1^2+k_2^2+\cdots+k_n^2=1 k12+k22+⋯+kn2=1,所以:
瑞利商的上下确界证毕。
■ \blacksquare ■
https://sharmaeklavya2.github.io/theoremdep/nodes/linear-algebra/matrices/bounding-quadratic-form-using-eigenvalues.html
https://www.cnblogs.com/spencergogo/p/16112581.html
https://zhuanlan.zhihu.com/p/440760513