Condition numbers of matrices with given spectrum

Abstract

This note surveys recent strategies to estimate the condition number $CN(T)=\Vert T\Vert \cdot \Vert T^{-1}\Vert$ of complex $n\times n$ matrices T with given spectrum. More precisely, we present a proof of the fact that if T acts on the Hilbert space $\mathbb {C}^{n}$, then the supremum of CN(T) over all contractions T with smallest eigenvalues of modulus $r>0$, is equal to $1/r^{n}$, and is achieved by an analytic Toeplitz matrix. The same question is treated for n-dimensional Banach spaces. These strategies provide with explicit and constructive solutions to the so-called Halmos and Schäffer’s problems, and are also shown to be effective in a closely related situation, namely considering Kreiss matrices instead of contractions.

Appendix: Proof of Lemma 3.1

We prove Lemma 3.1. Clearly, since J is an isometry, X and $\widetilde{X}$ have the same norm. In order to find the largest eigenvalue of $\widetilde{X}$, upon writing $\det (\widetilde{X}^{2}-z^{2})=\det (\widetilde{X}-z)\det (\widetilde{X}+z)$, we notice that it is enough to look for the largest eigenvalue of $\widetilde{X}^{2}$. Let us then search for the solution $z^{2}$ of $\det (\widetilde{X}^{2}-z^{2})=0$. Denoting by $C_{1},C_{2},\dots ,C_{n}$ the n columns of the matrix $\widetilde{X}^{2}-z^{2}$, and by $C_j(i)$ the i-th entry of $C_j$, we find

$$\begin{aligned} C_{j}=\left[ \begin{array}{c} \beta ^2\xi ^{j-1}(1+\xi ^2+\ldots +\xi ^{2(n-j-1)})+\beta \xi ^{2n-j-1}\\ \vdots \\ \beta ^2\xi ^{2j-3}(1+\xi ^2+\ldots +\xi ^{2(n-j-1)})+\beta \xi ^{2n-3}\\ -z^{2}+\beta ^{2}\xi ^{2j-2}(1+\xi ^{2}+\ldots +\xi ^{2(n-j-1)})+\xi ^{2n-2}\\ \beta ^2\xi ^{2j-1}(1+\xi ^{2}+\ldots +\xi ^{2(n-j-2)})+\beta \xi ^{2n-3}\\ \vdots \\ \beta ^2\xi ^{n+j-3}+\beta \xi ^{n+j-1}\\ \beta \xi ^{n+j-2}\\ \end{array} \right] , \end{aligned}$$

where the term $-z^{2}+\beta ^{2}\xi ^{2j-2}(1+\xi ^{2}+\cdots +\xi ^{2(n-j-1)})+\xi ^{2n-2}$ appears at line j. Note also that the sum $1+\xi ^2+\cdots +\xi ^{2(n-j-1)}$ is empty if $j=n$.

Then, in order to reduce $\det (\widetilde{X}^{2}-z^{2})$, we follow the three following steps:

(1) We divide the kth-column by $\xi ^{k-1}$ and the kth-row of $(\widetilde{X}^{2}-z^{2})$ by $\xi ^{k-1}$.

(2) To the determinant obtained in (1), we substract the $k^{th}$-column from $(k-1)$th-one and leave the $n^{th}$-one unchanged. E.g. for $n=5$ we obtain the determinant

$$\begin{aligned} \left| \begin{array}{ccccc} -z^{2}+\beta ^{2}\xi ^{6}-\beta \xi ^{6}+\xi ^{8} &{} \beta ^{2}\xi ^{4}+\beta \xi ^{6}-\beta \xi ^{4} &{} \beta ^{2}\xi ^{2}+\beta \xi ^{4}-\beta \xi ^{2} &{} \beta ^{2}+\beta \xi ^{2}-\beta &{} \beta \\ \frac{z^{2}}{\xi ^{2}}+\beta \xi ^{6}-\xi ^{6} &{} -\frac{z^{2}}{\xi ^{2}}+\xi ^{6}-\beta \xi ^{4}+\beta ^{2}\xi ^{4} &{} \beta ^{2}\xi ^{2}+\beta \xi ^{4}-\beta \xi ^{2} &{} \beta ^{2}+\beta \xi ^{2}-\beta &{} \beta \\ 0 &{} \frac{z^{2}}{\xi ^{4}}-\xi ^{4}+\beta \xi ^{4} &{} -\frac{z^{2}}{\xi ^{4}}+\xi ^{4}+\beta ^{2}\xi ^{2}-\beta \xi ^{2} &{} \beta ^{2}+\beta \xi ^{2}-\beta &{} \beta \\ 0 &{} 0 &{} \frac{z^{2}}{\xi ^{6}}-\xi ^{2}+\beta \xi ^{2} &{} -\frac{z^{2}}{\xi ^{6}}+\xi ^{2}+\beta ^{2}-\beta &{} \beta \\ 0 &{} 0 &{} 0 &{} \beta +\frac{z^{2}}{\xi ^{8}}-1 &{} -\frac{z^{2}}{\xi ^{8}}+1 \end{array}\right| . \end{aligned}$$

(3) To the determinant obtained in (2), we substract the $(k-1)^{th}$-row from the $k^{th}$-one and leave the $n^{th}$-one unchanged. At the end of this step, we obtain the determinant of a tridiagonal $n\times n$ matrix.

(4) In the determinant obtained in (3), we mutiply the $k^{th}$-row by $\frac{\xi ^{2k-1}}{z^{2}}$. At the end of this step, we obtain again the determinant of a tridiagonal matrix $n\times n$ matrix, which we will denote by A:

$$\begin{aligned} A=\left( \begin{array}{c@{\qquad }c@{\qquad }c@{\qquad }c@{\qquad }c} \gamma &{} \alpha ' &{} 0 &{} \cdots &{} 0\\ \alpha &{} \gamma &{} \alpha ' &{} \ddots &{} \vdots \\ 0 &{} \alpha &{} \ddots &{} \ddots &{} 0\\ \vdots &{} \ddots &{} \ddots &{} \gamma &{} \alpha '\\ 0 &{} \cdots &{} 0 &{} \alpha &{} y \end{array}\right) , \end{aligned}$$

where $\gamma$ and y are defined as in the statement of the lemma and where

$$\begin{aligned} \alpha =\xi +\frac{\xi ^{2n-1}}{z^{2}}(\beta -1)\text { and }\alpha '=\frac{1}{\xi ^{2}}\bigg (\xi +\frac{\xi ^{2n-1}}{z^{2}}(\beta -1)\bigg )=\frac{\alpha }{\xi ^{2}}. \end{aligned}$$

Thus one can get

$$\begin{aligned} \det A=\left| \begin{array}{c@{\qquad }c@{\qquad }c@{\qquad }c@{\qquad }c} \gamma &{} \frac{\alpha }{\xi } &{} 0 &{} \cdots &{} 0\\ \frac{\alpha }{\xi } &{} \gamma &{} \frac{\alpha }{\xi } &{} \ddots &{} \vdots \\ 0 &{} \frac{\alpha }{\xi } &{} \ddots &{} \ddots &{} 0\\ \vdots &{} \ddots &{} \ddots &{} \gamma &{} \frac{\alpha }{\xi }\\ 0 &{} \cdots &{} 0 &{} \frac{\alpha }{\xi } &{} y \end{array}\right| , \end{aligned}$$

which completes the proof of the lemma, setting $\mu =\frac{\alpha }{\xi }$.