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Assume that $\kappa([A, B])$ is the condition number of a block matrix $[A, B]$. Given that, we also know, $$\kappa(C) < \kappa(A)$$

I am curious whether if the following assertion is true or when does that inequality may hold: $$\kappa([C,B]) < \kappa([A, B])$$

Both $A$ and $C$ are $n \times n$ symmetric square matrices and B is an $n\times m$ rectangular matrix.

The condition number of a rectangular matrix $B$ is defined as the ratio of largest and smallest nonzero singular values, $$\kappa(B) = \frac{\sigma_{max}(B)}{\sigma_{min}(B)}$$ where $\sigma_{max}(B)$ is the largest singular value of $B$ and $\sigma_{min}(B)$ is the smallest nonzero singular value of matrix $B$.

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    $\begingroup$ what is your definition of the condition number of a rectangular matrix? $\endgroup$ – user251257 Aug 11 '15 at 12:48
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    $\begingroup$ Hmm, what are the $\lambda$s? When the operator norm is used, the condition number of a rectangular matrix is usually defined as $\kappa(A)=\|A\|\|A^+\|$, where $A^+$ is the Moore-Penrose pseudo inverse of $A$. That is, it is the ratio of the largest singular value of $A$ to the smallest non-zero singular value of $A$. Your definition does not resemble the usual one, however. $\endgroup$ – user1551 Aug 11 '15 at 12:56
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    $\begingroup$ smallest non-zero singular value, if you don't have rank assumption $\endgroup$ – user251257 Aug 11 '15 at 13:04
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    $\begingroup$ now, for any matrix $B$ the condition number of $B$ and $B^*$ are the same, as adjoining doesn't change the positive singular values. $\endgroup$ – user251257 Aug 11 '15 at 13:10
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    $\begingroup$ If $A$ is normal, then $\kappa([A^*, B]) = \kappa([A, B])$, as $A^*A + BB^* = AA^* + BB^*$. $\endgroup$ – user251257 Aug 11 '15 at 13:14
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Your question is not very clear.

  1. If the $A^\ast$ in your question simply denotes some other matrix than $A$ (rather than the conjugate transpose), you may consider $$A_0=\pmatrix{3\\ &3},\ A=\pmatrix{4\\ &3},\ B=\pmatrix{0\\ 4}.$$ We have $\kappa(A_0)=1<\frac43=\kappa(A)$ but $\kappa([A_0,B])=\frac53>\frac54=\kappa([A,B])$.

  2. If $A^\ast$ does mean the conjugate transpose of $A$ (i.e. $\bar{A}^T$), then your statement that $\kappa(A^\ast)<\kappa(A)$ is always false because a matrix and its conjugate transpose always have identical singular values. Nevertheless, you may still ask if there exist $A$ and $B$ such that $\kappa([\bar{A}^T,B])$ is larger than or smaller than $\kappa([A,B])$. Since $\kappa([\bar{A}^T,B])$ and $\kappa([A,B])$ in practice are seldom the same (unless $A$ is normal, as pointed out by user251257 in a comment), if we interchange the roles of $A$ and $\bar{A}^T$ when necessary, we can almost always construct an example of $\kappa([\bar{A}^T,B])>\kappa([A,B])$ or $\kappa([\bar{A}^T,B])<\kappa([A,B])$ at will out of random samples of $A$ and $B$.

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  • $\begingroup$ Thanks @user1551, I clarified the main-post. I was implying the 1). So do you know of any restriction where that might hold? $\endgroup$ – mathamania Aug 11 '15 at 14:07
  • $\begingroup$ just a small issue. Isn't $\kappa(A) = \frac{4}{3}$? $\endgroup$ – user251257 Aug 11 '15 at 14:13
  • $\begingroup$ @user251257 Yes, of course. $\endgroup$ – user1551 Aug 11 '15 at 14:49
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    $\begingroup$ @mathamania One obvious sufficient condition is that $AA^\ast-CC^\ast$ is positive definite, but this condition is surely too strong. I'm not sure if there are any weaker but clean sufficient conditions. $\endgroup$ – user1551 Aug 11 '15 at 14:55
  • $\begingroup$ @user1551 thanks. So the squared singular values of $[A, B]$, can also be obtained from $AA^{\ast} + BB^{\ast}$ by using min-max theorem. But I couldn't go very far from that. $\endgroup$ – mathamania Aug 11 '15 at 16:03

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