Express the DUAL norm in terms of a Matrix

Question

Given a symmetric positive definite matrix $M$, define the norm $\|x\|_M = \sqrt{\|Mx\|^2} $.Express the dual norm in terms of $M$.

I know, to solve this problem need to consider the Langrangian. i.e : $L(x) = y.x + \frac{\lambda\|Mx\|^2}{2}$

But I don't know How to proceed.

Answer 1

I am assuming the space you work on is $\mathbb{R}^n$. Given a symmetric, positive definite matrix $M$, you have a complete space equipped with an inner product, a Hilbertspace.
I will denote the inner product with $$ \langle x,y \rangle_M=\langle Mx,My \rangle $$ where $\langle \cdot ,\cdot \rangle$ denotes the euclidean standard inner product.
Then, by the Riesz representation theorem, every functional/element in the dual space $D=(\mathbb{R}^n)^*$ is given by: $$ \phi(x)=\langle x,y \rangle_M $$ for some $y \in \mathbb{R}^n$. Using the defintion of the dual space, you get to: $$ ||\phi||_{D}=\sup_{x \in \mathbb{R}^n} \frac{|\phi(x)|}{||x||_M}=\frac{\langle Mx,My \rangle}{\sqrt{\langle Mx,Mx \rangle}} $$ At this point, it becomes hard to simplify. You could proabably get to another form by using the spectral decomposition of $M$, but it doesnt get much prettier.

Answer 2

You are given $\|x\|_M=\|Mx\|$ where $M$ is invertible. By the definition of the dual norm as the operator norm on the space of linear functionals, you get using $I=M^{-1}M$ $$ \|\alpha\|_{M,\rm dual}=\sup_{x\ne 0}\frac{|α(x)|}{\|x\|_M} =\sup_{x\ne 0}\frac{|α(M^{-1}(Mx))|}{\|Mx\|} =\sup_{y\ne 0}\frac{|α(M^{-1}y)|}{\|y\|} =\|(M^{-1})^*α\|_{\rm dual} $$ where the pull-back notation $F^*α=α\circ F$ was used.

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Minimizing the quoted Lagrangian uses the same trick, assuming that the original norm is Euclidean $$ L(x,y)=y^Tx+\fracλ2\|Mx\|^2=\fracλ2\|(λM)^{-1}y+Mx\|^2-\frac1{2λ}\|M^{-1}y\|^2 $$ where you get the minimum if the first term is zero, which is always possible for some $x$.