Generalized eigenvalue pencils and QZ methods — Examples

Direct calculation gives $\det (A-\lambda B)=1-\lambda$, a nonzero polynomial, so the pencil is regular although $\det B=0$.

The homogeneous determinant is $\det (\beta A-\alpha B)=\det \left(\begin{array}{cc}\beta -\alpha & 0\\ 0& \beta \end{array}\right)=\beta (\beta -\alpha )\mathrm{.}$ Its projective roots are $[1:1]$, the finite eigenvalue $1$, and $[1:0]$, the infinite eigenvalue.

For $(A,B)=(I_2,0)$, $\det (A-\lambda B)=\det I_2=1$, so the pencil is regular. Its homogeneous determinant is $\det (\beta I_2)={\beta }^2$, which vanishes only at $[1:0]$, the infinite eigenvalue pair.

For $(A,B)=(\mathrm{diag}(1,0),0)$, $\det (A-\lambda B)=\det A=0$ for every $\lambda$, so the determinant polynomial is identically zero and the pencil is singular.

The matrix $M$ is positive definite because $u^{T}Mu=2u_1^2+u_2^2>0$ for $u\mathrm{\ne }0$. Also $u^{T}Ku=2u_1^2+2(u_1-u_2{)}^2+u_2^2>0$ for $u\mathrm{\ne }0$. The determinant polynomial is $\det (K-\lambda M)=(4-2\lambda )(3-\lambda )-4=2(\lambda -1)(\lambda -4),$ so the pencil is regular and its eigenvalues are $1,4$.

For $x_1=(1,1{)}^T$, $K{x}_1=(2,1{)}^T=M{x}_1$. For $x_2=(1,-2{)}^T$, $K{x}_2=(8,-8{)}^T=4M{x}_2$. Their mass inner product is $x_1^{T}M{x}_2=2-2=0$. Thus they are distinct generalized eigenmodes and are $M$-orthogonal; [F1] gives the general real-spectrum and orthogonal-basis conclusion for this pencil.

For $x=e_1$ and $\mu =2$, the residual is $r=Kx-\mu Mx=(0,-2{)}^T$. Hence $\mathrm{\parallel }x{\mathrm{\parallel }}_M=\sqrt{2}$ and, since $M^{-1}=\mathrm{diag}(1\mathrm{/}2,1)$, $\mathrm{\parallel }r{\mathrm{\parallel }}_{M^{-1}}=2$. The theorem [F2] therefore bounds the distance from $2$ to $\{1,4\}$ by $2\mathrm{/}\sqrt{2}=\sqrt{2}$; indeed that distance is $1$.