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?


closed as off-topic by YCor, Franz Lemmermeyer, Alexey Ustinov, Stefan Kohl, Myshkin Mar 27 '16 at 9:32

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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.


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