Computed Schur decompositions and nonnormal spectra — Examples

Put $x=\mathrm{\mid }z{\mathrm{\mid }}^2$ and $M=zI-J=\left(\begin{array}{cc}z& -1\\ 0& z\end{array}\right)$. Then $M^{\ast }M=\left(\begin{array}{cc}x& -\bar{z}\\ -z& x+1\end{array}\right)$ has trace $1+2x$ and determinant $x^2$, so its smaller eigenvalue is $f(x)={\sigma }_{\min }(M{)}^2=(1+2x-\sqrt{1+4x})\mathrm{/}2$ by [F2].

For $y\ge x\ge 0$, $f(y)-f(x)=(y-x)(1-\frac{2}{\sqrt{1+4y}+\sqrt{1+4x}})\ge 0$. The parenthesized factor is strictly positive when $y>x$, since then $\sqrt{1+4y}+\sqrt{1+4x}>2$; thus $f$ is strictly increasing on $[0,\mathrm{\infty })$. Also $\sqrt{1+4(ϵ+{ϵ}^2)}=1+2ϵ$, hence $f(ϵ+{ϵ}^2)={ϵ}^2$. By strict monotonicity and [F1], $f(\mathrm{\mid }z{\mathrm{\mid }}^2)\le {ϵ}^2$ exactly when $\mathrm{\mid }z{\mathrm{\mid }}^2\le ϵ+{ϵ}^2$.

Direct multiplication gives $Jv=(\widehat{\lambda },0{)}^T\mathrm{/}\sqrt{1+{\widehat{\lambda }}^2}$ and $\widehat{\lambda }v=(\widehat{\lambda },{\widehat{\lambda }}^2{)}^T\mathrm{/}\sqrt{1+{\widehat{\lambda }}^2}$. Their difference has norm ${\widehat{\lambda }}^2\mathrm{/}\sqrt{1+{\widehat{\lambda }}^2}<{10}^{-8}$, and $\mathrm{\parallel }v{\mathrm{\parallel }}_2=1$.

Set $r=(J-\widehat{\lambda }I)v$ and $E=-r{v}^{\ast }$. For every unit vector $x$, $\mathrm{\parallel }Ex{\mathrm{\parallel }}_2=\mathrm{\parallel }r{\mathrm{\parallel }}_2\mathrm{\mid }{v}^{\ast }x\mathrm{\mid }\le \mathrm{\parallel }r{\mathrm{\parallel }}_2$, with equality at $x=v$, so $\mathrm{\parallel }E{\mathrm{\parallel }}_2=\mathrm{\parallel }r{\mathrm{\parallel }}_2<\delta$. Also $(J+E)v=\widehat{\lambda }v$. By [F2], $\widehat{\lambda }\in {\mathrm{\Lambda }}_{\delta }(J)$.

If $\lambda \mathrm{\ne }0$ and $(J-\lambda I)(x_1,x_2{)}^T=0$, the second row gives $-\lambda x_2=0$, hence $x_2=0$; the first row then gives $-\lambda x_1=0$, hence $x_1=0$. But $e_1$ is a nonzero eigenvector for $\lambda =0$. Thus [F1] gives $\sigma (J)=\{0\}$ and $\mathrm{dist}(\widehat{\lambda },\sigma (J))={10}^{-4}$, which is ${10}^4$ times the residual scale.

Since $Q=I$, $Q^{T}Q=I$ and $QT{Q}^T=T$. The only diagonal block is $T$, and its characteristic polynomial is $(\lambda -a{)}^2+b^2$, with discriminant $-4b^2<0$. Thus [F1] says $(Q,T)$ is a real Schur pair.

The roots of $(\lambda -a{)}^2+b^2$ are $\lambda =a\pm ib$. Equivalently, [F2] gives trace $2a$ and determinant $a^2+b^2$, yielding the same conjugate roots.