Condition number of block matrix Given a block matrix:
$$A=
\begin{bmatrix}
D & E^T \\
E & 0
\end{bmatrix} ,
$$
where $D$ is a diagonal, positive entry matrix and $E$ is an incidence matrix of a graph, how is the condition number of A related to that of D?
I tried to exploit the SVD decomposition of A to find a relation with D but I didn't find any interesting conclusion.
 A: If we remove the last a row $b$ from the matrix $A$. Then
$$A=\begin{pmatrix} B \\ b\end{pmatrix}$$
and
$$AA^*=\begin{pmatrix}BB^* & Bb^* \\ bB^* & bb^*\end{pmatrix}$$
It follows that $BB^*$ is a $n\times n$ submatrix of $AA^*$.
The Cauchy's interlacing theorem holds to  $AA^*$ and $BB^*$. See also this wikipedia text. It follows that
$$\lambda_{n+1}({A}{A}^*)\leq \lambda_n({B}{B}^*)\leq \cdots \leq \lambda_2({A}{A}^*)\leq \lambda_1({B}{B}^*)\leq  \lambda_1({A}{A}^*).$$
You can repeat the argument removing the lats column of $b_1$ of $B$, say
$$B=\begin{pmatrix} B_1&b_1\end{pmatrix},$$
if you consider the matrix $B^*B$ and the $(n-1)\times (n-1)$ matrix ${B_1}^*B_1$, you see that
$$\lambda_{n+1}({A}{A}^*)\leq \lambda_{n-1}({B_1}{B}_1^*)\leq \lambda_{n-1}({B}{B}^*)\leq \cdots \leq \lambda_1({B_1}{B}_1^*)\leq \lambda_1({B}{B}^*)\leq  \lambda_1({A}{A}^*).$$
You can repeat the argument and remove the entire matrix $E^T$, and use the equality $\sigma_i(A)=\lambda_i(A^*A)=\lambda_i(AA^*)$.
Remark: Please see this thread that I found with SearchOnMath when I search for "\(A^*A+a_1^*a_1\)  with \(a_1\) a row matrix singular values".
