Condition number of block matrix

Question

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.

Answer 1

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".