Eigenvalues and Eigenvectors.

The singular case of $A - \lambda B$ corresponds to either

• $A - \lambda B$ is square and singular for all values of $\lambda$, or
• $A - \lambda B$ is rectangular.

Both cases arise in practice, and are significantly more challenging than the regular case. We outline the theory here and leave details to §8.7.

Consider

$$A - \lambda B = \bmat{cc} 1 & 0 \\ 0 & 0 \emat - \lambda \bmat{cc} 1 & 0 \\ 0 & 0 \emat.$$ (4)

Then $p(\lambda ) = {\rm det} (A - \lambda B) \equiv 0$ for all $\lambda$, so $A - \lambda B$ is singular. For any $\lambda$, $Ax = \lambda Bx = 0$ for $x = [0,1]^T$. But rather than calling all $\lambda$ eigenvalues, we only call $1$ an eigenvalue of this pencil, because $Ax = 1 \cdot Bx$ for $x = [1,0]^T$, and the rank of $A-1 \cdot B$ is 0, which is lower than the rank of $A - \lambda B$ for any other value of $\lambda$. In general, if $A - \mu B$ has a lower rank than the rank of $A - \lambda B$ for almost all other values of $\lambda$, then $\mu$ is an eigenvalue.

Eigenvalues are discontinuous functions of the matrix entries when the pencil is singular, which is one reason we have to be careful about definitions. This discontinuity is further discussed below.

Eigenvectors are also no longer so simply defined. For example, consider

$$A - \lambda B = \bmat{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \emat -\lambda \bmat{ccc} 1 & 0 & 0 \\ 0 & 0 & 1 \emat.$$ (5)

Then $1$ is an eigenvalue but $Ax = 1 \cdot Bx$ for any $x = [z,z,1]^T$ for any value of $z$. Instead we consider reducing subspaces, as defined below.

$$% latex2html id marker 1839For example, change $A_{22}$\ t... ...ussed in more detail in section~\ref{sec_chap2_GnHepCondSing}.$$ (6)