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?
Best Answer
The answer is "no" in general. I give an example, where the dual norm of the operator norm norm with respect to the primal norm does not coincide with the operator norm with respect to the dual norm. Take $V$ to be $\mathbb{R}^n$ equipped with the Euclidean norm $\|.\|_2$. By a suitable identification of dual space and primal space (this works well in finite dimensions), you see that the dual space coincides with the primal space.
For an bounded linear operator $M:X\to X$ acting on a normed space $(X,\|\cdot\|_X)$ you may define the operator norm $\|M\|=\sup\left\{\|Mx\|_X:\|x\|_X=1\right\}$.
Here, $\|M\|$ does not depend on whether you take $\|\cdot\|_X$ to be the primal or the dual norm on the right hand side because the 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 the nuclear norm, see, for instance, Show that the dual norm of the spectral norm is the nuclear norm