Regenerative brake

In mathematics, for a given complex Hermitian matrix $M$ and nonzero vector $x$, the Rayleigh quotient $R(M,x)$, is defined as:

$R(M,x):=\frac{x^{*}Mx}{x^{*}x}.$

For real matrices and vectors, the condition of being Hermitian reduces to that of being symmetric, and the conjugate transpose $x^{*}$ to the usual transpose $x^{\prime }$. Note that $R(M,cx)=R(M,x)$ for any real scalar $c\ne 0$. Recall that a Hermitian (or real symmetric) matrix has real eigenvalues. It can be shown that, for a given matrix, the Rayleigh quotient reaches its minimum value ${\lambda }_{\min }$ (the smallest eigenvalue of $M$) when $x$ is $v_{\min }$ (the corresponding eigenvector). Similarly, $R(M,x)\le {\lambda }_{\max }$ and $R(M,v_{\max })={\lambda }_{\max }$. The Rayleigh quotient is used in the min-max theorem to get exact values of all eigenvalues. It is also used in eigenvalue algorithms to obtain an eigenvalue approximation from an eigenvector approximation. Specifically, this is the basis for Rayleigh quotient iteration.

The range of the Rayleigh quotient is called a numerical range.

Special case of covariance matrices

An empirical covariance matrix M can be represented as the product A' A of the data matrix A pre-multiplied by its transpose A'. Being a symmetrical real matrix, M has non-negative eigenvalues, and orthogonal (or othogonalisable) eigenvectors, which can be demonstrated as follows.

Firstly, that the eigenvalues ${\lambda }_i$ are non-negative:

$M{v}_i=A^{\prime }A{v}_i={\lambda }_i{v}_i$

$\Rightarrow {v_i}^{\prime }{A}^{\prime }A{v}_i={v_i}^{\prime }{\lambda }_i{v}_i$

$\Rightarrow {\Vert A{v}_i\Vert }^2={\lambda }_i{\Vert v_i\Vert }^2$

$\Rightarrow {\lambda }_i=\frac{{\Vert A{v}_i\Vert }^2}{{\Vert v_i\Vert }^2}\ge 0.$

Secondly, that the eigenvectors $v_i$ are orthogonal to one another:

$M{v}_i={\lambda }_i{v}_i$

$\Rightarrow {v_j}^{\prime }M{v}_i={\lambda }_i{v_j}^{\prime }{v}_i$

$\Rightarrow (M{v}_j{)}^{\prime }{v}_i={\lambda }_i{v_j}^{\prime }{v}_i$

$\Rightarrow {\lambda }_j{v_j}^{\prime }{v}_i={\lambda }_i{v_j}^{\prime }{v}_i$

$\Rightarrow ({\lambda }_j-{\lambda }_i){v_j}^{\prime }{v}_i=0$

$\Rightarrow {v_j}^{\prime }{v}_i=0$ (if the eigenvalues are different – in the case of multiplicity, the basis can be orthogonalized).

To now establish that the Rayleigh quotient is maximised by the eigenvector with the largest eigenvalue, consider decomposing an arbitrary vector $x$ on the basis of the eigenvectors v_i:

$x=\sum _{i=1}^n{\alpha }_i{v}_i$, where ${\alpha }_i=\frac{x^{\prime }{v}_i}{{v_i}^{\prime }{v}_i}=\frac{⟨x,v_i⟩}{{\Vert v_i\Vert }^2}$ is the coordinate of x orthogonally projected onto $v_i$

so

$R(M,x)=\frac{x^{\prime }{A}^{\prime }Ax}{x^{\prime }x}$

can be written

$R(M,x)=\frac{(\sum _{j=1}^n{\alpha }_j{v}_j{)}^{\prime }{A}^{\prime }A(\sum _{i=1}^n{\alpha }_i{v}_i)}{(\sum _{j=1}^n{\alpha }_j{v}_j{)}^{\prime }(\sum _{i=1}^n{\alpha }_i{v}_i)}$

which, by orthogonality of the eigenvectors, becomes:

$R(M,x)=\frac{\sum _{i=1}^n{\alpha }_i^2{\lambda }_i}{\sum _{i=1}^n{\alpha }_i^2}=\sum _{i=1}^n{\lambda }_i\frac{(x^{\prime }{v}_i{)}^2}{(x^{\prime }x)({v_i}^{\prime }{v}_i)}$

The last representation establishes that the Rayleigh quotient is the sum of the squared cosines of the angles formed by the vector $x$ and each eigenvector $v_i$, weighted by corresponding eigenvalues.

If a vector $x$ maximizes $R(M,x)$, then any scalar multiple $kx$ (for $k\ne 0$) also maximizes R, so the problem can be reduced to the Lagrange problem of maximizing $\sum _{i=1}^n{\alpha }_i^2{\lambda }_i$ under the constraint that $\sum _{i=1}^n{\alpha }_i^2=1$.

Let ${\beta }_i\stackrel{\text{def}}{=}{\alpha }_i^2$. This then becomes a linear program, which always attains its maximum at one of the corners of the domain. A maximum point will have ${\alpha }_1=\pm 1$ and $\mathrm{\forall }i>1,{\alpha }_i=0$ (when the eigenvalues are ordered by decreasing magnitude).

Thus, as advertised, the Rayleigh quotient is maximised by the eigenvector with the largest eigenvalue.

Formulation using Lagrange multipliers

Alternatively, this result can be arrived at by the method of Lagrange multipliers. The problem is to find the critical points of the function

$R(M,x)=x^{T}Mx$,

subject to the constraint $\Vert x{\Vert }^2=x^{T}x=1.$ I.e. to find the critical points of

$\mathcal{ℒ}(x)=x^{T}Mx-\lambda (x^{T}x-1),$

where $\lambda$ is a Lagrange multiplier. The stationary points of $\mathcal{ℒ}(x)$ occur at

$\frac{d\mathcal{ℒ}(x)}{dx}=0$

$\therefore 2x^T{M}^T-2\lambda x^T=0$

$\therefore Mx=\lambda x$

and $R(M,x)=\frac{x^{T}Mx}{x^{T}x}=\lambda \frac{x^{T}x}{x^{T}x}=\lambda .$

Therefore, the eigenvectors $x_1\dots x_n$ of M are the critical points of the Rayleigh Quotient and their corresponding eigenvalues ${\lambda }_1\dots {\lambda }_n$ are the stationary values of R.

This property is the basis for principal components analysis and canonical correlation.

Use in Sturm–Liouville theory

Sturm–Liouville theory concerns the action of the linear operator

$L(y)=\frac{1}{w(x)}(-\frac{d}{dx}[p(x)\frac{dy}{dx}]+q(x)y)$

on the inner product space defined by

$⟨y_1,y_2⟩=\underset{a}{\overset{b}{\int }}w(x)y_1(x)y_2(x)dx$

of functions satisfying some specified boundary conditions at a and b. In this case the Rayleigh quotient is

$\frac{⟨y,Ly⟩}{⟨y,y⟩}=\frac{\underset{a}{\overset{b}{\int }}y(x)(-\frac{d}{dx}[p(x)\frac{dy}{dx}]+q(x)y(x))dx}{\underset{a}{\overset{b}{\int }}w(x)y(x{)}^{2}dx}.$

This is sometimes presented in an equivalent form, obtained by separating the integral in the numerator and using integration by parts:

$\frac{⟨y,Ly⟩}{⟨y,y⟩}=\frac{\underset{a}{\overset{b}{\int }}y(x)(-\frac{d}{dx}[p(x)y^{\prime }(x)])dx+\underset{a}{\overset{b}{\int }}q(x)y(x{)}^{2}dx}{\underset{a}{\overset{b}{\int }}w(x)y(x{)}^{2}dx}$

$=\frac{-y(x)[p(x)y^{\prime }(x)]{|}_a^b+\underset{a}{\overset{b}{\int }}{y}^{\prime }(x)[p(x)y^{\prime }(x)]dx+\underset{a}{\overset{b}{\int }}q(x)y(x{)}^{2}dx}{\underset{a}{\overset{b}{\int }}w(x)y(x{)}^{2}dx}$

$=\frac{-p(x)y(x)y^{\prime }(x){|}_a^b+\underset{a}{\overset{b}{\int }}[p(x)y^{\prime }(x{)}^2+q(x)y(x{)}^2]dx}{\underset{a}{\overset{b}{\int }}w(x)y(x{)}^{2}dx}.$

Generalization

For a given pair $(A,B)$ of matrices, and a given non-zero vector $x$, the generalized Rayleigh quotient is defined as:

$R(A,B;x):=\frac{x^{*}Ax}{x^{*}Bx}.$

The Generalized Rayleigh Quotient can be reduced to the Rayleigh Quotient $R(D,C^{*}x)$ through the transformation $D=C^{-1}A{C^{*}}^{-1}$ where $C{C}^{*}$ is the Cholesky decomposition of the Hermitian positive-definite matrix $B$.

See also

• Field of values

Further reading

• Shi Yu, Léon-Charles Tranchevent, Bart Moor, Yves Moreau, Kernel-based Data Fusion for Machine Learning: Methods and Applications in Bioinformatics and Text Mining, Ch. 2, Springer, 2011.