I understand that the Rayleigh quotient is defined and used primarily for symmetric matrices. How can it be extended and applied to asymmetric ones? What does it mean when x^T A x = 0 for nonzero x? How can we tell if there are or aren't any such x for a given A?
Maybe, refer to http://www.math.utah.edu/~zhu/6610.d/homework_12.pdf for a discussion.
Xian-Ming Gu's answer helped me understand my issue. It's not about asymmetry per se, rather than A being non-SPD.
Consider A = [0, 1; 1, 0], mirroring in the line x_1 = x_2, with \lambda_1=1, v_1=[1;1] and \lambda_2=-1, v_2=[1;-1]. The Rayleigh quotient takes values in the interval [min(\lambda),\max(\lambda)] = [-1, 1], including zero. One vector that produces zero is x=v_1+v_2=[2;0].
Consider B = [0, -1; 1, 0], rotating vectors by 90 degrees, with \lambda_1=i, v_1=[i;1], \lambda_2=-i, v_2=[-i;1]. This is skew-symmetric. With symmetric part (B+B^T)/2 equal to zero, the Rayleigh quotient is zero for all vectors x.
Consider a non-singular 2x2 matrix C with eigenvalues \lambda_1, \lambda_2 \neq 0. Consider a vector x = \alpha v_1 + \beta v_2, for which x^T A x = \lambda_1 \alpha^2 v_1^T v_1 + \lambda_2 \beta^2 v_2^T v_2. If \lambda_1 and \lambda_2 have opposite signs, then we can choose \alpha, \beta to get x^T A x = 0.
One possibility is to work with singular values instead of eigenvalues, singular values are all real and non-negative and thus can be ordered just as eigenvalues of hermitian, PD or PSD.
I suggest that in the Rayleigh Quotient that you can go with A*AH or AH*A, AH denotes the hermitian conjugate of matrix A, instead of A itself.
If A is non symmetric you can do A´=(A^{T} + A)/2 then A´ is symmetric moreover q_{A} : C^{n} --->C and q_{A'} : C^{n} ---> C , are equal where q_{A} = x^{T}Ax is a quadratic form. If x^{T}Ax = 0 then x is a zero of the polynomial q_{A} with grade =< 2 .
If A is diagonalizable and x^{T}Ax : R^{n} ----> R you have,
1.x^{T}Ax >0 (positive definite) \lamba >0 for all eigenvalaues.
2.x^{T}Ax >= 0 (positive semidefinite) exists \lamba = 0 and the others eigenvalues are positive.
3.x^{T}Ax
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