2

Suppose $M \colon \mathbb R^n \to \mathbb R^n$ is a linear transformation. We know the operator norm induced by Euclidean 2-norm is lower bounded by $\rho(M)$, i.e., $\|M\|_2 \ge \rho(M)$. I find here that the characterization of $M$ such that $\|M\|_2 = \rho(M)$.

I am wondering if there exists characterization of the set of vectors $U$ on which we have $\|Mu\|_2 > \rho(M) \|u\|_2$ for $u \in U$ when $\|M\|_2 > \rho(M)$. For instance, we know $\|M\|_2^2 = \rho(M^*M)$. If $\|M\|_2 > \rho(M)$, it much be the vector $v$ corresponding to the largest singular value has the property. But in general, do we know how many vectors out there?

Martin Argerami
  • 217,281
user1101010
  • 3,638
  • 1
  • 17
  • 40

1 Answers1

1

I don't think you'll get an explicit characterization in general: it depends too much on the structure of $M$.

On one extreme, if $M$ is nilpotent, then $U=\mathbb R^n\setminus\{0\}$.

If you think about the complement of $U$, those would be the vectors with $\|Mu\|\leq\rho(M)\,\|u\|$. This clearly includes all eigenvectors, but it is not necessarily a subspace: if $$ M=\begin{bmatrix}1&2\\0&-1 \end{bmatrix},\ \ u=\begin{bmatrix} 1\\0\end{bmatrix},\ \ v=\begin{bmatrix} -1\\1/2\end{bmatrix}, $$ then $$ Mu=u,\ \ \ \ \ Mv=-v. $$ Also, $\rho(M)=1$, and $$ \|Mu\|=\|u\|,\ \ \ \|Mv\|=\|v\|,\ \ \ \|M(u+v)\|=\left\|\begin{bmatrix}1\\-1/2\end{bmatrix}\right\|=\frac{\sqrt5}2>\frac12=\|u+v\|. $$ So neither $U$ nor its complement are subspaces in general.

Martin Argerami
  • 217,281