Condition number of $2 \times 2$ block matrix in terms of the singular values of the off-diagonal blocks If $A$ is $m \times n$ matrix such that $ m \geq n $ and $B$ is the block matrix
$$ B = \begin{bmatrix}I & A \\ A^T & 0 \end{bmatrix} $$
then what is the condition number of $B$ in terms of singular values of $A$?
 A: I assume you're using the condition number with respect to the Euclidean norm.
Presumably $A$ has real entries.  Then $B$ is symmetric.  $\pmatrix{u\cr v\cr}$ is an eigenvector of $B$ for eigenvalue $\lambda$ iff $A^T u = \lambda v$ and $u + A v = \lambda u$.
Then $A A^T u = \lambda A v = \lambda (\lambda - 1) u$ and $A^T A v = A^T (\lambda - 1) u
= \lambda (\lambda - 1) v$.  Conversely, if $A A^T$ has an eigenvector $u$ for eigenvalue
$ \lambda (\lambda - 1)$ with $\lambda \ne 0$ and we take $v = \lambda^{-1} A^T u$,
we get $u + A v = u + \lambda^{-1} A A^T u = \lambda u$, so $\pmatrix{u\cr v\cr}$ is an eigenvector of $B$ for eigenvalue $\lambda$.  If $A^T A$ has an eigenvector $v$ for eigenvalue $\lambda (\lambda - 1)$ with $\lambda \ne 1$ and we take $u = (\lambda - 1)^{-1} A v$, then $A^T u = (\lambda - 1)^{-1} A^T A v = \lambda v$, and again $\pmatrix{u\cr v\cr}$ is an eigenvector of $B$ for eigenvalue $\lambda$.  
By convention, one often omits the $0$ eigenvalues from the singular values of $A$, but 
we see that if $A$ has fewer than $n$ nonzero singular values (counted by multiplicity), $0$ is an eigenvalue of $A^T A$, implying that $0$ is an eigenvalue of $B$, and then the condition number of $B$ is $\infty$.   
If $0$ is an eigenvalue of $A A^T$, which is always the case if $m > n$, then $1$ is
an eigenvalue of $B$.  
Other than that, if $\sigma_1 \le \sigma_2 \le \ldots \le \sigma_n$ are the singular values of $A$, $\sigma_1^2 \le \sigma_2^2 \le \ldots \le \sigma_n^2$ are the eigenvalues of $A^T A$ and the nonzero eigenvalues of $A A^T$, and so the eigenvalues of $B$ (other than $1$) are 
the solutions of $\lambda (\lambda - 1) = \sigma_j^2$, i.e. $\dfrac{1 \pm \sqrt{1+4\sigma^2}}{2}$.  The largest of these in absolute value is $(1 + \sqrt{1+4 \sigma_n^2})/2$, so this is $\|B\|$.  The least in absolute value is $(1 - \sqrt{1+4\sigma_1^2})/2$ (or $1$ if that is less and $0$ is an eigenvalue of $A A^T$), and the reciprocal of that is the greatest in absolute value of the 
eigenvalues of $B^{-1}$.  Thus $\|B^{-1}\|$ is either  $2/(\sqrt{1+4\sigma_1^2}-1)$ or $1$.  We conclude
that the condition number of $\|B\| \|B^{-1}\|$ of $B$ is either $\dfrac{1+\sqrt{1+4\sigma_n^2}}{2}$ or  $\dfrac{1 + \sqrt{1+4\sigma_n^2}}{\sqrt{1+4\sigma_1^2}-1}$.
