# Condition numbers and block matrices

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 $B$.

• what is your definition of the condition number of a rectangular matrix? – user251257 Aug 11 '15 at 12:48
• 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. – user1551 Aug 11 '15 at 12:56
• smallest non-zero singular value, if you don't have rank assumption – user251257 Aug 11 '15 at 13:04
• now, for any matrix $B$ the condition number of $B$ and $B^*$ are the same, as adjoining doesn't change the positive singular values. – user251257 Aug 11 '15 at 13:10
• If $A$ is normal, then $\kappa([A^*, B]) = \kappa([A, B])$, as $A^*A + BB^* = AA^* + BB^*$. – user251257 Aug 11 '15 at 13:14

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 $B$.
• just a small issue. Isn't $\kappa(A) = \frac{4}{3}$? – user251257 Aug 11 '15 at 14:13
• @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. – user1551 Aug 11 '15 at 14:55
• @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. – mathamania Aug 11 '15 at 16:03