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I'm solving questions from am exam that I failed miserably, so I would love it if someone can take a look at my proof and make sure I'm not making any gross mistakes.

Question

Let $A$ a symmetric matrix. Which $\rho$-norm minimizes $A$'s condition number: $\kappa(A,\rho)$?

Edit

Well, my solution was clearly wrong. I would love any suggestion or direction for solution you might have.

Thanks! :)

Hila
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    I think, that this what you did only shows that $k(A,2)\leq \frac{M}{m}k(A,\rho)$, where $M\ge m$, i.e $\frac{m}{m}\ge 1$. Maybe you can use 1) that $|\lambda_i |\leq |A|$ for any norm and any eigenvalue $\lambda_i$ and 2) that $k(A,2)=\frac{max |\lambda_i |}{min |\lambda_j|}$, where $\lambda_i$ are the eigenvalues of $A$. – Svetoslav Jul 28 '15 at 18:08
  • Ouch! Back to the drawing board :( – Hila Jul 28 '15 at 18:42
  • Are you looking for any matrix norm (called $\rho$-norm) or for a particular instance of a $p$-norm? – Algebraic Pavel Jul 29 '15 at 10:21

1 Answers1

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$k(A,2)=\|A\|_2\|A^{-1}\|_2= \max | \lambda_A|.\max|\lambda_{A^{-1}}|\leq\|A\|\|A^{-1}\|=k(A,\|.\|)$ for each operator norm $\|.\|$.

That $\max|\lambda_A|\leq \|A\|$ follows from:

Let $x$ be an eigenvector of $A$ for the eigenvalue $\lambda$, where $\max\limits_{i}|\lambda_i|=|\lambda|$. Then for arbitrary matrix norm $\|.\|$, subordinate to the vector norm $\|.\|$, we have $\|A\|=\max\limits_{y\neq 0} \frac{\|Ay\|}{\|y\|}\ge \frac{\|Ax\|}{\|x\|}=\frac{\|\lambda x\|}{\|x\|}=|\lambda|$

Svetoslav
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