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Can the "inducing" vector norm be deduced or "recovered" from an induced (operator) norm?

This question occurred to me after seeing this question. I'm hoping that perhaps there exists something like the polarization identity, i.e. some identity one may use to "recover" the vector norm the same way the polarization identity "recovers" the inner product.

I am aware that any induced norm satisfies the inequality $$ \left|\|A^r\|\right|^{1/r} \geq \rho(A) $$ Also, there exists some invertible matrix $S$ such that $$ \left|\|SAS^{-1}\|\right| = \rho(A) $$ I'm wondering if one or both of the above might also be a sufficient condition for $\left|\|\cdot \|\right|$ to be a derived norm, and that they might somehow be used to derive some sort of identity producing the necessary vector norm.

Any input is appreciated!


Runaway train of thought below:

Following HHO's advice (see comment below), here's a neat way to recover the vector norm (assuming that a suitable vector norm does exist):

Let $\|\cdot \|_O$ denote the operator norm. I will define a vector norm $\|\cdot \|$ as follows: arbitrarily, I set $\|e_1\| = 1$ (because $\alpha \|\cdot\|$ is a vector norm for any $\alpha>0$ and any vector norm $\|\cdot\|$ and since both of these result in the same induced norm, we may set $\|e_1\| = \alpha$ for any $\alpha>0$). From there, we may define $$ \|u\| = \left\| u e_1^* \right\|_O = \left\| \pmatrix{|&|&&|\\ u&0&\cdots&0\\ |&|&&|} \right\|_O $$ What remains to be seen is under which conditions this defines a valid vector-norm.


In fact, the above must always be a valid vector-norm, by the definition of a matrix norm. It is necessary to check whether the vector norm produced above induces the operator norm that we started with.


By the above, here's a neat criterion for checking whether $\|\cdot\|_O$ is an operator norm:

We can state that $\|\cdot\|_O$ is an induced (matrix) norm if and only if for all $A \in \mathbb{F}^{n\times n}$, we have $$ \|A\|_O = \max_{x \neq 0} \frac{\|Ax e_1^*\|_O}{\|xe_1^*\|_O} $$ and, presumably, $e_1$ can be replaced by any convenient $v \in \mathbb{F}^n: v^*v = 1$. I guess I answered my own question then.

Is this a known theorem? This question is now a reference request. If anyone has seen something like this, please say so.

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2  
Take a good look at rank one operators … err, matrices … and their norm. –  Harald Hanche-Olsen Dec 1 '13 at 20:49
    
@HaraldHanche-Olsen so with rank one matrices and vector norm $\|\cdot \|$, we have $A = \alpha u v^*$ for $\alpha \in \mathbb{F}$ and vectors $u,v$ with $\|u\|=\|v\|= 1$. From there, we note that the induced norm $|\|\cdot\||$ satisfies $$\left|\|A\|\right| = \alpha v^*v \frac{\|u\|}{\|v\|} = \alpha v^*v$$ Is there anything I'm missing? –  Omnomnomnom Dec 1 '13 at 21:22
    
There might be something missing. First, if you plug in the Frobenius norm, you obtain that it is an operator norm, which is not true. Second, what would be a convenient $v$? Last, $|||A|||=\alpha v^*v\frac{\|u\|}{\|v\|}$ does not hold for any induced norm except the 2-norm (and Frobenius norm). Actually, $|||A|||=|\alpha|\|u\|\|v\|_*$, where $\|\cdot\|_*$ is dual to $\|\cdot\|$. –  Algebraic Pavel Dec 2 '13 at 10:12
    
@AlgebraicPavel for some reason I had made the assumption that for $A = \alpha u v^*, x = \frac{v}{\|v\|}$ would be the $x$ that attains the maximum $\max_{\|x\| = 1} \|A\|$. This is clearly not generally the case, certainly not in the infinite dimensional case when this supremum (as I should have it) is not always attained. I will have to rethink this at some point. Thank you for the input. –  Omnomnomnom Dec 2 '13 at 13:50
    
You seem to be assuming norms of square matrices that are induced by a single vector norm? More generally, matrix norms can be induced on arbitrary matrices by specifying separate vector norms on the domain and the codomain. –  joriki Jul 12 at 5:32

1 Answer 1

Let $V$ be a topological vector space. Let $O(V)$ be the space of continuous operators in $V$, endowed with the strong operator topology. Assume that $O(V)$ can be given a structure of normed space with the norm $\| \cdot \|$, with the norm generating the strong operator topology of $O(V)$.

Let $f \in V^* \setminus \{ 0 \}$ be a linear and continuous functional. It is easy to show that $v \otimes f$ is a continuous linear operator, since $f$ is so. Define $\| v \| _f = \| v \otimes f \|$. Again, it is easy to show that this is a norm.

Let us show that the topology of $\| \cdot \| _f$ is the same as the original one of $V$. Let $(v_i) _{i \in I}$ be a net convergent to $v$ in the topology of $V$. Then

$$\begin{align} v_i \underset {i \in I} \to v \Leftrightarrow \\ f(u) v_i \underset {i \in I} \to f(u) v \; \forall u \in V \Leftrightarrow \\ (v_i \otimes f) (u) \underset {i \in I} \to (v \otimes f) (u) v \; \forall u \in V \Leftrightarrow \\ v_i \otimes f \underset {i \in I} \to v \otimes f \Leftrightarrow \\ \|v_i \otimes f - v \otimes f \| \underset {i \in I} \to 0 \Leftrightarrow \\ \|(v_i - v) \otimes f \| \underset {i \in I} \to 0 \Leftrightarrow \\ \| v_i - v \| _f \underset {i \in I} \to 0 \; . \end{align}$$

Finally, let the operator norm $\| \cdot \| _1$ induced on $O(V)$ by $\| \cdot \| _f$ be given by $\| U \| _1 = \sup \{ M>0 | \; \| Uv \| _f \le M \| v \| _f \}$, as usual. Let $U_i \to U$ in this norm. We have $\| (U_i - U)v \| _f \le \|U_i - U \| _1 \| v \| _f$, so $\| (U_i - U)v \| _f \to 0$, so $U_i v \to Uv$ in the original topology of $V$, so $U_i \to U$ in the topology of $O(V)$ and, since this was assumed to be given by $\| \cdot \|$, we obtain $\| U_i - U \| \to 0$, so the topology of $\| \cdot \| _1$ is stronger than the topology of $\| \cdot \|$. I suspect, but cannot prove, that the two topologies do not coincide.

Furthermore, if $\| \cdot \|$ has the supplementary property that $\| UV \| \le \| U \| \| V \|$ then we can obtain even more: since $Uv \otimes f = U \circ (v \otimes f)$, then $\| Uv \| _f = \| Uv \otimes f \| = \| U \circ (v \otimes f) \| \le \| U \| \| v \otimes f \| = \| U \| \| v \| _f$, so $\| \cdot \| \le \| \cdot \| _1$. Again, I believe that the opposite inequality does not hold, but I do not know how to prove it.

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I dont follow the fourth step in your chain of equivalences: Why is $(v_i \otimes f) \to (v \otimes f)$ in the strong operator topology (i.e. pointwise) equivalent to $v_i \otimes f \to v\otimes f$? Do you assume that the norm induces the strong operator topology? Because above you just write "with the topology of O(V) given by ∥⋅∥". –  PhoemueX Jul 12 at 18:01
    
@PhoemueX: Yes, this is what I meant, but I had formulated it vaguely. I have reformulated it now. –  Alex M. Jul 12 at 18:20

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