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Let us define the complex matrix $X \in \mathbb{C}^{2N \times M}$ where $N > M$. Additionally, the matrix ${X}$ consists of the following submatrices: $$ {X} = \left[ \begin{array}[c]. A \\ B\end{array} \right] $$ where $A \in \mathbb{C}^{N \times M}$ and $B \in \mathbb{C}^{N \times M}$ and their columns are linearly independent.

Is there a way I could show that if $A=B$, the condition number of $X$ would be larger than the condition number of $X'$ with $A \neq B$ (particularly if $A$ and $B$ are linearly independent)?

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  • $\begingroup$ How are you defining the condition number of a non-invertible matrix? Often this is just taken as $\infty$. $\endgroup$ – Robert Israel Mar 5 '20 at 15:35
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    $\begingroup$ @RobertIsrael if the columns are linearly independent, then not necessarily. With the $2$-norm, taking $\sigma_{\max}/\sigma_{\min}$ will work if the number of singular values is the number of columns. $\endgroup$ – Ben Grossmann Mar 5 '20 at 15:42
  • $\begingroup$ @Omnomnomnom Thanks for the comment, I added that the columns are independent. Indeed, even though $A$ is non-square matrix, its least squares estimate exists. So does the least squares estimate of $X$, (i.e. $(X^H X)^{-1}X^H$ exists). $\endgroup$ – Marco Mar 5 '20 at 15:53
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Here is an answer for the condition number relative to the $2$-norm. We have $$ \sigma_{\max}(X) \leq \sqrt{\sigma_{\max}^2(A) + \sigma_{\max}^2(B)}, \quad \sigma_{\min}(X) \geq \sqrt{\sigma_{\min}^2(A) + \sigma_{\min}^2(B)}. $$ From this, it follows that $$ \kappa(X) = \frac{\sigma_\max(X)}{\sigma_\min(X)} \leq \sqrt{\frac{\sigma_{\max}^2(A) + \sigma_{\max}^2(B)}{\sigma_{\min}^2(A) + \sigma_{\min}^2(B)}}. $$ Moreover, this inequality will be equality in the case that $A$ and $B$ are linearly dependent.

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  • $\begingroup$ Thanks for the answer. According to these inequalities if $A$ and $B$ are linearly dependent, the condition number would be smaller than the case where $A$ and $B$ are independent. If so, I would not agree with the result. But following your approach, I recognized that the first inequality, i.e. $\sigma_{\max}(X)$ should be written in a different way. We should have, $\sigma_{i+N}(X) \leq \sigma_{i}(A) \leq \sigma_{i}(X)$, where $\sigma_{i}(X)$ is in descending order. In this case, I would get the result I am expecting. $\endgroup$ – Marco Mar 5 '20 at 21:17

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