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Consider two symmetric positive semi-definite matrices $A, B \in \mathbb{R}^{n\times n}$. Suppose that $A$ and $B$ have the same null space $\mathcal{N}\subset \mathbb{R}^n$. Now consider the objective function $J=\frac{x^TAx}{x^TBx}$ where $x\in\mathbb{R}^n$ is an arbitrary unit-norm vector with $x^Tx=1$ and $x\notin \mathcal{N}$. Although $x^TAx\ne 0$ since $x\notin \mathcal{N}$, it is possible $x^TAx\rightarrow 0$ when $x$ is very close to $\mathcal{N}$. So neither $x^TAx$ nor $x^TBx$ has lower bound. But under what condition does the objective function $J$ have a positive lower bound? I mean what special properties (e.g., eigenvalues) should $A$ and $B$ have to make $J$ have a positive lower bound?

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    I wish I were:)2012-08-13

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I am not exactly sure what you are looking for, as $J(x) \geq 0$ for all $x$ such that $J$ is defined. In fact, both $x^TAx \geq 0$ and $x^TBx \geq 0$, so both of these have lower bounds as well.

However, if $x \notin \mathcal{N}$ you can get a better lower bound:

We have $\mathcal{N} = \ker A = \ker B$. Let $\underline{\lambda}_A$ be the smallest non-zero eigenvalue of $A$. Then if $x = x_1 + x_2$, where $x_1 \in \mathcal{N}$, and $x_2 \in \mathcal{N}^\bot$, then we have $x^TAx = x_2^T A x_2 \geq \underline{\lambda}_A \| x_2 \|^2, \ \ \ x^TBx = x_2^TBx_2 \leq \|B\| \|x_2\|^2.$ So, if $x \notin \mathcal{N}$, then $x_2 \neq 0$, which gives the estimate $J(x) = \frac{x_2^T A x_2}{x_2^TBx_2} \geq \frac{\underline{\lambda}_A}{\|B\|} > 0.$

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    I thought $\|B\|$ is the Frobenius norm of $B$. Thanks for the answer and the suggestion!2012-08-13