# Condition number of a matrix whose submatrices are linearly dependent

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)?

• How are you defining the condition number of a non-invertible matrix? Often this is just taken as $\infty$. – Robert Israel Mar 5 '20 at 15:35
• @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. – Ben Grossmann Mar 5 '20 at 15:42
• @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). – Marco Mar 5 '20 at 15:53

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.
• 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. – Marco Mar 5 '20 at 21:17