Regular Versus Singular Problems

Let us start by considering the $2 \times 2$ generalized eigenvalue problem $Ax = \lambda Bx$, where

$$A = \left[ \begin{array}{cc} \alpha_1 & 0 \\ 0 & \alpha_2... ...y}{cc} \beta_1 & 0 \\ 0 & \beta_2 \\ \end{array} \right].$$

The eigenvalues of $A - \lambda B$ are the pairs $(\alpha_1$, $\beta_1$) and $(\alpha_2$, $\beta_2$) with the associated eigenvectors $x_1 = [1 \quad 0]^T$ and $x_2 = [0 \quad 1]^T$, respectively. If $\beta_i$ is nonzero, then $\lambda_i = \alpha_i/\beta_i$ is a finite eigenvalue. Otherwise, if $\beta_i$ is zero, then $\lambda_i = \infty$ is an eigenvalue of the matrix pair $\{A,B\}$. But what happens if, for example, $\alpha_2 = \beta_2 = 0$? Then ${\rm det}(A - \lambda B)$ is zero for all $\lambda$, which means that we have a singular eigenvalue problem. In this case we have $Ax_2 = Bx_2 = 0$; i.e., $A$ and $B$ have a common null space. We say that $x_2$ is an eigenvector for an indeterminate eigenvalue $0/0$. Note that the common null space is a sufficient but not necessary condition to have a singular eigenvalue problem.

The most common generalized eigenvalue problems $Ax = \lambda Bx$ are regular; i.e., $A$ and $B$ are square matrices and the characteristic polynomial $p_m(\lambda) = {\rm det}(A - \lambda B)$ is only vanishing for a finite number of values, where $m$ denotes the degree of the polynomial. The corresponding $A - \lambda B$ is called a regular matrix pencil. The $n$ eigenvalues of a regular pencil are points in the extended complex plane ${\cal C} \cup \infty$. The eigenvalues $\lambda_i$ are defined as the zeros of $p_m(\lambda)$ and $n - m$ additional $\infty$ eigenvalues.

An alternative representation of a matrix pencil is the cross product form: the set of matrices $\beta A-\alpha B$ where $(\alpha,\beta) \in {\cal C}^2$. The mapping $(\alpha,\beta) \mapsto \alpha/\beta$ shows the relation between the eigenvalues of $\beta A-\alpha B$ and $A - \lambda B$. For example, zero and infinite eigenvalues are represented as $(0, \beta)$ and $(\alpha, 0)$, respectively, and can be treated as any other points in ${\cal C}^2$.

If ${\rm det}(A - \lambda B)$ (and ${\rm det}(\beta A - \alpha B)$) is identically zero for all $\lambda$ (and pairs ($\alpha$, $\beta$)), then $A - \lambda B$ is called singular and $\{A,B\}$ is a singular matrix pair. Singularity of $\{A,B\}$ is signaled by some $\alpha = \beta = 0$. In the presence of roundoff, $\alpha$ and $\beta$ may be very small. In these situations, the eigenvalue problem is very ill-conditioned, and some of the other computed nonzero values of $\alpha$ and $\beta$ may be indeterminate. Such problems are further discussed and illustrated by examples in §8.7.4. Moreover, rectangular matrix pairs are singular and the corresponding $A - \lambda B$ is a singular pencil.