Let $A$ be an invertible real $n\times n$ matrix and suppose the singular values of $A$ are in the interval $[1,\kappa_2(A)]$, where $\kappa_2(A)$ denotes the (2-norm) condition number of $A$. Consider $$M=\left[\begin{array}{cc}I & A\\ A^T & 0\end{array}\right].$$ Results from various papers I've found claim that the eigenvalues of such augmented systems in general (where $I$ is replaced by an arbitrary square matrix) are contained within a union of one negative interval and one positive interval $[a,b]\cup[c,d]$, i.e., $a,b<0$ and $c,d>0$. These two intervals moreover can be written directly in terms of the condition number of $A$. How can I show that this is true? Applying a factorization trick, we can deduce that $$ \left[ \begin{array}{cc} I & A \\ A^T & 0 \end{array} \right] = \left[ \begin{array}{cc} I & 0 \\ A^T & I \end{array} \right] \left[ \begin{array}{cc} I & 0 \\ 0 & -A^TA \end{array} \right] \left[ \begin{array}{cc} I & A \\ 0 & I \end{array} \right] .$$ Does this help us see the range of the eigenvalues?
1 Answers
Suppose $A=USV^T$ is a singular value decomposition. Using the block matrix determinant formula, we get $$ \det(xI-M) = \det\left((x-1)xI-AA^T\right) = \det\left((x-1)xI-S^2\right). $$ Therefore the eigenvalues of $M$ are the roots of $x^2 - x - \sigma_i^2$, i.e. pairs of numbers of the form $\frac{1 \pm \sqrt{1+4\sigma_i^2}}2$, where $\sigma_1,\ldots,\sigma_n$ are the singular values of $A$. Hence the spectrum of $M$ lies inside $$ \left[\frac{1-\sqrt{1+4\sigma_1^2}}2,\ \frac{1-\sqrt{1+4\sigma_n^2}}2\right]\cup \left[\frac{1+\sqrt{1+4\sigma_n^2}}2,\ \frac{1+\sqrt{1+4\sigma_1^2}}2\right]. $$ Since $[\sigma_n,\sigma_1]\subseteq[1,\kappa(A)]$, the above union of intervals is contained in $$ \mathcal I=\left[\frac{1-\sqrt{1+4\kappa(A)^2}}2,\ \frac{1-\sqrt{5}}2\right]\cup \left[\frac{1+\sqrt{5}}2,\ \frac{1+\sqrt{1+4\kappa(A)^2}}2\right]. $$
