I have a question about the effect of applying a linear transformation $M$ in $\mathbb{R}^{n \times n}$ to a vector $v \in \mathbb{R^n}$. 


I know that if $M$ has p-norm $\|M\|\_p = \lambda$, then by definition I can guarantee that for every vector $v \in \mathbb{R^n}$
$$\|Mv\|\_p \leq \lambda \|v\|\_p.$$


Is there a corresponding lower bound? In general the answer is no, (for example, choose $v \in Nullspace(M)$ if the matrix is not full rank. )But I am interested in a corresponding lower bound where


1. $v$ is chosen ``generically'', so that the probability of it being in a (fixed in advance) subspace is 0.
2. $\|v\|\_p$ is sufficiently large.


In this case, is it true that
$$\|Mv\|\_p \geq C \|v\|\_p$$
where $C$ is some positive constant greater than zero?



The effect of linear transformation on generic vectors
Note that your constraint (2) is not relevant, because an inequality
$\|Mv\|\_p\ge C\|v\|\_p$ is preserved by scalar multiplication. So we
might as well assume $\|v\|\_p=1$. This is a compact set, so any
continuous real-valued function on it will have a maximum value. If
$M$ is nonsingular, then $\frac{\|v\|\_p}{\|Mv\|\_p}$ is such a
function, so taking $C$ to be its maximum gives the inequality you
want, and it holds for all $v\ne0$. In fact, $C$ is just the $p$-norm
of $M^{-1}$.


If $M$ is singular, the answer really depends on what you mean by
"chosen generically". If you want an inequality that holds for all
$v\notin\ker(M)$, the answer is no: There is no upper bound on
$\frac{\|v\|\_p}{\|Mv\|\_p}$, since $v$ can be arbitrarily close to the
kernel.


The problem is that the set $S-\ker(M)$ is not compact, where
$S=\{v:\|v\|\_p=1\}$. But if we choose an open neighborhood $U$ of
$\ker(M)$, then $S-U$ is compact, so there will be an inequality
$\|Mv\|\_p\ge{}C\|v\|\_p$ that holds on $S-U$, but with the constant
$C=C\_U$ depending on $U$. In fact if we take $U$ to be an open "cone"
containing $\ker(M)-\{0\}$ — *i.e.* closed under multiplication
by positive scalars — then the inequality holds for all $v\notin{}U$.


For example, if $n=p=2$ and $M(x,y)=(0,y)$, we could take
$U=\{(x,y):0<|y|<c\,|x|\}$, and then
$\|M(x,y)\|\ge{}C\|(x,y)\|$ with $C=\sqrt{1+c^{-2}}$, for all
$(x,y)\notin{}U\cup\{0\}$. Of course, $C$ can be made arbitrarily
large by taking $c$ sufficiently small.


