Re-norming a contractive map into a self-similarity

Question

Let $V = \mathbb{R}^{d}$. Then we say a linear map $T: V \to V$ is Lipschitz if there exists a constant $K < \infty$ such that $\| T(v) \| / \| v \| \leq K$ for $\| v \| \neq 0$. Further, if the unit sphere is compact (as it is in finite dimensions), then $K = \max_{\| v \| = 1} \| T(v) \|$. Moreover, we say $T$ is a self-similarity if we replace the inequality with a strict equality, and examples of $T$ which are one and not the other are plentiful (e.g. if $T$ is not injective and not identically zero).

Now suppose $T$ is a linear, injective, contractive, non-self-similar map in the norm $\| \cdot \|$. What interests me is, might there exist another norm $\| \cdot \|_{1}$ in which $T$ is a self-similarity, i.e. such that there exists $K_{1}$ for which $\| T(v) \|_{1} / \| v \|_{1} = K_{1}$ identically for non-zero $v \in V$? What can we say about our possible choices of $K_{1}$? Do we know if $\| \cdot \|_{1}$ can be induced by an inner product? How well can we express $\| \cdot \|_{1}$?

As for work I've done, I tried it on $\mathbb{R}^{2}$, and tried assuming it could be induced by inner product (as otherwise I'm not sure how I'd write it out), but I ended up with several-variable polynomials I had no clue how to start on.

Thanks!

Answer 1

I find your question hard to understand.

• What's the purpose of mentioning this "Lipschitz" stuff? According to your definition, every linear map on $\mathbb R^d$ is Lipschitz.
• What's the purpose of mentioning a contraction? If $\frac{\|Tv\|_1}{\|v\|_1}$ is a constant for all nonzero $v$, so is $\frac{\|cTv\|_1}{\|v\|_1}$ for every real number $c$.
• Who coined the term "self-similarity"? What does "similar" refer to?

Anyway, I suppose that you actually intended to ask the following question:

Let $V=\mathbb R^d$. For every linear map $T$ on $V$ and every vector norm $\|\cdot\|$ on $V$, denote the induced matrix norm $\max_{\|v\|=1}\|Tv\|$ (often known as the operator norm) by $\|T\|$. We say that $T$ is a self-similarity with respect to the vector norm $\|\cdot\|$ if $\frac{\|Tv\|}{\|v\|}=\|T\|$ for every nonzero $v\in V$. Now assume that $T$ is injective (hence nonsingular). Does there always exist a vector norm $\|\cdot\|$ on $V$ such that $T$ is a self-similarity with respect to $\|\cdot\|$?

If this is indeed your question, the answer is no. It can be shown that in general, $T$ is self-similar with respect to some norm if and only if the matrix of $T$ is $cP^{-1}QP$ for some real scalar $c$, real invertible matrix $P$ and real orthogonal matrix $Q$. But to answer your question, it suffices to give an example of $T$ that is never self-similar, regardless of the norm.

Consider the linear map $T$ whose matrix with respect to the standard basis of $\mathbb R^n$ is $\pmatrix{1&1\\ 0&1}$. If $T$ is "self-similar" with respect to some norm $\|\cdot\|$, then with $e_1=(1,0)^\top$, we have $Te_1=e_1$ and hence $\|T\|=\frac{\|Te_1\|}{\|e_1\|}=1$. But then we would have $\|Tv\|=\|T\|\|v\|=\|v\|$ and in turn $\|T^kv\|=\|v\|$ for every vector $v$ and every positive integer $k$. In particular, when $v=e_2=(0,1)^\top$, we get $\|(k,1)^\top\|=\|T^ke_2\|=\|e_2\|$. Hence $S=\{(k,1)^\top:\ k\in\mathbb N\}$ is bounded with respect to $\|\cdot\|$. Yet this is impossible, as all norms on $\mathbb R^n$ are equivalent and $S$ is unbounded with respect to the Euclidean norm.