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Let $A$ and $B$ be a toeplitz and symmetric positive definite $NxN$ matrices. If $\kappa (A) > \kappa (B)$, how to show that: $\kappa (B^{-1}A) < \kappa (A)$ ? $\kappa $() is condition number of matrix.

Thank you!

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2 Answers 2

It's not true. Consider e.g. $$ A = \pmatrix{a & 0\cr 0 & 1/a}, \ B = \pmatrix{1/b & 0\cr 0 & b}$$ with $a> b > 1$. Then $\kappa(A) = a^2 > b^2 = \kappa(B)$ but $\kappa(B^{-1} A) = a^2 b^2 > \kappa(A)$.

What is true is that $\kappa(B^{-1} A) \le \kappa(A) \kappa(B)$.

EDIT: For an example with Toeplitz matrices, take

$$ A = \pmatrix{2 & 1\cr 1 & 2\cr},\ B = \pmatrix{3 & -1\cr -1 & 3\cr}, B^{-1} A = \pmatrix{7/8 & 5/8\cr 5/8 & 7/8\cr}$$ Then $\kappa(B^{-1} A) = 6 > \kappa(A) = 3 > \kappa(B) = 2$.

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Ok. I forgot to give the infomation that A and B are toeplitz and symmetric matrix. Is this can be true in this case? Thanks. @Robert Israel –  user60343 Jan 31 '13 at 2:58

Your assertion is not true. If it's true, then for $B=I$, we would have $\kappa(A)<\kappa(A)$ for any positive definite Toeplitz matrix $A\not=I$.

For a nontrivial ($B\not=I$) example, consider $A=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ and $B=\begin{pmatrix}1&-0.1\\-0.1&1\end{pmatrix}$. Both of them are positive definite and Toeplitz, but $\kappa(B^{-1}A)\approx 3.6667 > 3 = \kappa(A)>\kappa(B)=1.2222$.

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$B$ is not Toeplitz. –  Robert Israel Jan 31 '13 at 7:35
    
@RobertIsrael Thanks. What a silly mistake. The wrong counterexample is now deleted. –  user1551 Jan 31 '13 at 7:42
    
I see :( In my numerical simulation this was allways true, but I can' prove it. Also there is the relation sheep beetwen the matrices A i B (A is filtered autocorelation matrix of B - this is problem from digital signal proccessing ), but I havent't it in explicit form in this moment. I know my English is bad, I apologize fot it :) Thank you for this comments! @Robert Israel –  user60343 Jan 31 '13 at 11:37

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