Reciprocal of a quadratic form I am working with an expression of the form
$$ \frac{x^TAx}{{x^TBx}}$$
and would like to simplify it. I understand that vectors do not have inverses, but viewing the bottom number as a 1 by 1 matrix,
$$ \frac{x^TAx}{{x^TBx}} = (x^TAx)(x^TBx)^{-1} "=" (x^TAx)(x^{-1}B^{-1}x^{-T}) "=" x^TAB^{-1}x^{-T}. $$
Can I do something like this, but more formally?
 A: What you are trying to is not possible as inverse is defined only for matrices and not vectors. But, if you want to simplify your expression, we can do it in the following manner. But we need the assumption that $B^{-1}$ exists. Also, wlog, we can assume that $B$ is symmetric. Then, $$B=UDU^{T}=UD^{1/2}U^{T}UD^{1/2}U^{T}=B^{1/2}B^{1/2}$$ where $D$ is the diagonal matrix with the non-zero eigenvalues and $U$ is a orthogonal matrix containing the eigenvectors of $B$. Then define $y=B^{1/2}x$ and then you have
$$\frac{x^TAx}{x^TBx}=\frac{y^T(B^{-1/2}AB^{-1/2})y}{y^Ty}$$
This becomes the so-called Rayleigh-ratio of the matrix $B^{-1/2}AB^{-1/2}$ and you have the identity that 
$$\lambda_{min}(B^{-1/2}AB^{-1/2})\leq \frac{y^T(B^{-1/2}AB^{-1/2})y}{y^Ty} \leq \lambda_{max}(B^{-1/2}AB^{-1/2})$$
A: I do not think this is possible this way. Consider the matrices:
$$A = B = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$$
and $x=(1,1)^t$.
Then $AB^{-1} = I$
Therefore $$\frac{x^t A x}{x^t B x} = 1 \neq 2 = x^t AB^{-1} x $$
The thing to notice here is that $\frac{x^t A x}{x^t B x} $ is a rational function in $x$, and in general rational functions are very unlikely to be polynomial.
