norm of a vector and linear transformation

Question

Let $A \in M_{n}(\mathbb{C})$ be nonsingular matrix. Then I m trying to find the dual norm of $\|x\|_{A} \triangleq\|Ax\|_{2}$.

My try:

\begin{align} \|y\|_{A}^{D} &= \max_{\|x\|_{A}=1} |y^*x| \\ &= \max _{\|Ax\|_{2}=1} |y^*x| \\ &= \max _{\|Ax\|=1} |y^*A^{-1}Ax| \\ &\leq \max _{\|Ax\|=1} \|y^*A^{-1}\|_{2} \|Ax\|_{2} \\ &= \|y^*A^{-1}\|_{2} \end{align}

but now I don't know how to continue! Any idea, comment? Thanks

Answer 1

I'm not sure why you wrote $\le$ when in fact we have equality.

$$\|y\|_2^D = \max_{\|x\|_A=1}\left|y^*x\right| = \max_{\|v\|_2=1}\left|(y^*A^{-1})v\right| = \|y^*A^{-1}\|_2$$

Assuming $\|\cdot\|_2$ is the euclidean norm on $\mathbb{C}^n$, we also have

$$\|y^*A^{-1}\|_2 = \|y^*(A^{-1})^{**}\|_2 = \|((A^{-1})^*y)^{*}\|_2 = \|(A^{-1})^*y\|_2$$

Answer 2

Let $x=\frac {A^{-1}y^{*}A^{-1} } a$ where $a=||y^{*}A^{-1}||$. Then $||Ax||= 1$. Hence $||y||_A^{D} \geq |y^{*}x|$. If you compute $y^{*}x$ for this particular $x$ you will see that $||y||_A^{D} \geq ||y^{*}A^{-1}||_2$. Combined with what you already proved we get $||y||_A^{D} = ||y^{*}A^{-1}||_2$.