So what is the condition number of the coefficient matrix $\begin{bmatrix}I&A\\ A^T&0\end{bmatrix}$ in terms of the singular values of A if A has full rank and is m-by-n?
I'm assuming we have to use the SVD of A. So I was able to deduce that(if I'm wrong about this please correct it) $\begin{bmatrix}I&U\Sigma V^T\\ V\Sigma U^T&0\end{bmatrix}=\begin{bmatrix}U&0\\ 0&V\end{bmatrix}\begin{bmatrix}I&\Sigma \\ \Sigma &0\end{bmatrix}\begin{bmatrix}U^T&0\\ 0&V^T\end{bmatrix}$. However, I'm not sure how to proceed further. Maybe I'm confusing what the books says in terms of coefficient matrix. The hint it provides says to use SVD of A
I suppose that $A$ has full column rank, otherwise the block matrix in question is singular. Let $A$ be $m\times n$ with $m\ge n$. If the induced $2$-norm is used, then $B:=\begin{bmatrix}I&A\\ A^T&0\end{bmatrix}$ and $\begin{bmatrix}I&\Sigma \\ \Sigma &0\end{bmatrix}$ have the same singular values. The latter matrix can be reindexed to $\begin{bmatrix}1&\sigma_1\\ \sigma_1&0\end{bmatrix}\oplus\cdots\oplus\begin{bmatrix}1&\sigma_n\\ \sigma_n&0\end{bmatrix}\oplus I_{m-n}$. Its $i$-th $2\times2$ sub-block have singular values $\frac12\left(\sqrt{1+4\sigma_i^2}\pm1\right)$. Therefore $$ \sigma_\max(M)=\frac12\left(\sqrt{1+4\sigma_1^2}+1\right), \ \sigma_\min(M)=\min\left\{\frac12\left(\sqrt{1+4\sigma_n^2}-1\right),1\right\} $$ and you may readily calculate $\kappa(M)=\sigma_\max(M)/\sigma_\min(M)$.