$|Av||A^{-1}v|$, $A$ non-singular, $|v|=1$. Let $A$ a non-singular $n \times n$ matrix, $v \in \mathbb{R}^{n}$ a variable vector.
The operator norm of $A$ is defined to be $|| A ||=\max_{|v|=1} |Av|$ where $|Av|$ is the standard Euclidean norm of the vector $|Av|$. It is known that $||A||$ is equal to the largest singular value of $A$, from which it follows that $\min_{|v|=1} |Av|=1/|| A^{-1}||$.
Suppose we fix a unit vector $v$. Are there any nice lower bounds on $|Av||A^{-1}v|$ besides the obvious one $1/||A||||A^{-1}||$? What if we make additional assumptions on $A$? This seems to be partly a question about how the singular vectors of $A$ are related to those of $A^{-1}$.
 A: Without some assumptions about $A$, the trivial bound 
$$|Av|\,|A^{-1}v|\ge \|A\|^{-1}\|A^{-1}\|^{-1} \tag{1}$$
cannot be improved. Indeed, let $A$ be the transformation $(x,y)\mapsto (My,x/M)$. Then for the vector $v=(1,0)$ we have $Av=(0,1/M)=A^{-1}v$, since $A$ is its own inverse. Hence, equality holds in (1). 
Note that the singular vectors of $A$ and $A^{-1}$ in this case are the same; it just so happens that the corresponding singular values are not in the relation you'd expect. In general, you can multiply $A$ with an orthogonal matrix to make singular vectors go wherever you want. 
But if you assume that $A$ is symmetric, then there is a much nicer bound: 
$$|Av|\,|A^{-1}v|\ge 1 \tag{2}$$
Indeed, $A$ can be diagonalized; let's say that $\lambda_1,\dots,\lambda_n$ are the diagonal entries. By the Cauchy-Schwarz inequality, for every unit vector $v$ we have
$$
1 =  \sum_{i}(\lambda_i v_i) (\lambda_i^{-1} v_i)
\le \sqrt{\sum_i \lambda_i^2 v_i^2}\, \sqrt{\sum_i \lambda_i^{-2} v_i^2} 
=|Av|\,|A^{-1}v|
$$
