Compute an upper bound on generalized eigenvalues (by using the coefficients) Consider the generalized, symmetric eigenvalueproblem:
\begin{equation}
 A x = \lambda B x,
\end{equation}
with $A, B$ symmetric and $B$ being positive definite.  
For some computations, i was trying to figure out an upper bound for the generalized eigenvalues of this problem by using the coefficients. By taking a proper matrix norm $\| \cdot\|$, you get
\begin{equation}
 |\lambda| = \frac{\|Ax\|}{\|Bx\|} \leq \frac{\|A\| \|x\|}{\|Bx\|}.
\end{equation}
The enumerator can be estimated simple, for example by taking a normed eigenvector $x$ with $\| \cdot \|_{\infty}$ (the sup-norm, i.e. you just have to add some absolute values of the matrix entries to bound that part). But what should i do with the denominator?
Edit: as mentioned by Algebraic Pavel's or user161825's answer, you could do that by estimating the smallest eigenvalue $\mu_1$ of $B$ by some $0 < \bar{\mu} < \mu_{1}$, but how do you get this $\bar{\mu}$ just by knowing the coefficients of $B$? Of course - for computations - trying to invert the Matrix $B$ isn't feasible, are there smart ways to get this bound? 
One possible answer could be using the neat Gershgorin-circles, but this can only be applied to special cases of $B$, namely when $B$ is strict diagonal dominant. Then you could iterate over all rows $k$ and, depending on the sign of the diagonal entry $B_{k,k}$ (let's here assume it's positive), you can calculate $\displaystyle \xi_k = B_{k,k} -  \sum_{\substack{i=1, \\ i \neq k}}^{n} | B_{k,i} | $. In the end you just take the minimal $\xi_k$ over $1 \leq k \leq n$ and got your lower estimate for the denominator.
 A: The problem $Ax=\lambda Bx$ is equivalent to $B^{-1}Ax=\lambda x$, so the generalized eigenvalues are the eigenvalues of $B^{-1}A\sim B^{-1/2}AB^{-1/2}$. Hence the eigenvalues can be bounded from above by any norm of $B^{-1}A$, e.g.,
$$
|\lambda|\leq\|B^{-1}A\|_2,\quad|\lambda|\leq\|B^{-1/2}AB^{-1/2}\|_2,\quad \text{etc.}
$$
Note that the latter inequality is sharper since $\|B^{-1}A\|_2\geq\|B^{-1/2}AB^{-1/2}\|_2$ (still assuming that $B$ is SPD).
Of course, if the chosen matrix norm is submultiplicative, you can take out $B$ and get
$$
|\lambda|\leq\|A\|\|B^{-1}\|.
$$
Another simple bound can be obtained as follows:
$$Ax=\lambda Ax\quad\Rightarrow\quad x^TAx=\lambda x^TBx\quad\Rightarrow\quad\lambda=\frac{x^TAx}{x^TBx}.$$
Hence
$$
|\lambda|=\frac{|x^TAx|}{x^TBx}=\left|\frac{x^TAx}{x^Tx}\right|\frac{x^Tx}{x^TBx}
\leq\max_{y\neq 0}\left|\frac{y^TAy}{y^Ty}\right|\max_{z\neq 0}\frac{z^Tz}{z^TBz}
=\frac{\max_{y\neq 0}\left|\frac{y^TAy}{y^Ty}\right|}{\min_{z\neq 0}\frac{z^TBz}{z^Tz}}
=\frac{|\lambda_{\max}(A)|}{\lambda_{\min}(B)}.
$$
Note that since the matrices $A$ and $B$ are symmetric  and $B$ is SPD, this bound is equal again to $\|A\|_2\|B^{-1}\|_2$.
A: Let $b>0$ be the smallest eigenvalue of $B$. Then $b^2>0$ is the smallest eigenvalue of $B^2$. It follows that $\langle (B^2-b^2I)x,x\rangle\geq 0$, or equivalently $\|B x\|\geq b\|x\|$.
Alternatively, rewrite your equation as
$$
B^{-1}Ax=\lambda x,
$$
which implies $|\lambda|\leq \|B^{-1}A\|$. Of course $b=\|B^{-1}\|^{-1}$, so this result implies  $|\lambda|\leq \|A\|/b$, as above.
As for computing a lower bound for the lowest eigenvalue, this may be an interesting read. Inspired by formula 2.54 there, let us assume that $b_1\geq\ldots\geq b_n$ are the non-increasingly ordered eigenvalues of $B$, counted with multiplicity. Observe that
$$
\prod_{j=1}^{n-1} b_j\leq \frac{\sum_{j=1}^{n-1} b_j^{n-1}}{n-1}<\frac{\mbox{tr }(B^{n-1})}{n-1},
$$
so that
$$
b_n=\frac{\mbox{det }B}{\prod_{j=1}^{n-1} b_j}> \frac{(n-1)\mbox{det }B}{\mbox{tr }(B^{n-1})}>0
$$
when $n>1$.
This leaves us with the following (probably poor) estimate
$$
|\lambda|< \frac{\|A\|\mbox{tr }(B^{n-1})}{(n-1)\mbox{det }B}.
$$
