4

Given a norm on a (finite dimensional) vector space $V$, we can define the induced norm on linear maps from $V \rightarrow V$, by $\|M\|_{I(V)} = \sup\limits_{\|x\|=1}\|Mx\|$.

It is also true that we can define a dual norm on the vector space $V$ by $\|x\|_{*} = \sup\limits_{\|y\|=1} |y^{t}x|$, and from here define the corresponding matrix norm $I(V^{*})$induced from this dual norm.

However, one can reverse the order of operations: First take the induced norm and then take the dual in the space of linear maps, i.e. $\|M\|_{I(V)^{*}} = \sup\limits_{\|N\| = 1} <M,N>$ where we treat the matrices as vectors and take their inner products.

My question is there any relation between these two constructions? For instance, say we take the norm on the underlying vector space to be the 2-norm. Then can we say that $\|M\|_{I(V)^{*}} = \|M\|_{I(V)}$, in the same way that the 2-norm is self-dual on the underlying vector space?

ttb
  • 1,127
  • 11
  • 19

1 Answers1

2

The answer is "no" in general, and I'll provide a counterexample. Take $V$ to be $\mathbb{R}^n$ equipped with the Euclidean norm $\|.\|_2$. Then the dual space $V^\ast$ with the usual dual norm $y\mapsto \sup\{y(x): x\in V, \|x\|\leq 1\}$ can be identified with the "primal", i.e., original space $(V,\|.\|_2)$. (The isomorphism is given by $V\to V^\ast$, $y\mapsto (x\mapsto y^\top x)$.)

For an bounded linear operator $M:X\to X$ acting on a normed space $(X,\|\cdot\|_X)$ you may define the operator norm $\|M\|=\sup\{\|Mx\|_X:\|x\|_X=1\}$.

On our example, we get the same operator norm $\|M\|$, no matter if we take $X=V$ or $X=V^\ast$, because the two options are the same. The result is what we call the spectral norm of the matrix $M$.

But the dual norm of the spectral norm is a different norm, called the nuclear norm, see, for instance, Show that the dual norm of the spectral norm is the nuclear norm

tjah
  • 632