# Proving the relative condition of a matrix is the same as the relative condition of its transpose

Let A $$\epsilon$$ $$R^{n*n}$$ be invertible and $$1\leq p,q\leq \infty$$ with $$1/p+1/q=1.$$ Show that $$k_p(A)=k_q(A^T)$$. To be clear $$k_p(A)$$ is the relative condition number with the Lp norm. Similarly $$k_q(A^T)$$ is the relative condition number with the Lq norm. I am not sure exactly where to go but I started by transforming the equation into the definition of the relative condition factor to get: $$||A||_p||A^{-1}||_p=||A^T||_q||{A^T}^{-1}||_q$$.

• I am working on this exact question now and still haven't figured it out. Hoping someone can elaborate! This question appears in Wendland's Numerical Linear Algebra and, to be honest, I am confused as all heck with it. – Carah Sep 12 '19 at 19:16

$$\|A\|_p = \sup_{\|x\|_p = 1} \|A x\|_p = \sup_{\|x\|_p = 1} \sup_{\|y\|_q = 1} |y^T A x|$$ and similarly for $$\|A^T\|_q$$. Same with $$A$$ replaced by $$A^{-1}$$.
• Would it be possible for you to elaborate at all? In the textbook I'm using where this exact question appears, there are no theorems/definitions of $||A||_p$ that involve two sup's... I'm pretty confused. Thanks for your input. – Carah Sep 12 '19 at 19:22