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Reducing Subspaces.

A pair of right and left reducing subspaces $\cal X$ and $\cal Y$ of $A - \lambda B$ satisfy $Ax \in \cal Y$ and $Bx \in \cal Y$ for all $x \in \cal X$, and furthermore ${\rm span}_{x \in {\cal X}} (Ax, Bx) = \cal Y$. Also, the dimension of $\cal X$ exceeds the dimension of $\cal Y$ by ${\rm dim}(N_r)$, the dimension of the right null space of $A - \lambda B$ over the field of all rational functions of $\lambda$ [119]. It is easy to express ${\rm dim}(N_r)$ in terms of the Kronecker canonical form, described below. There is still a correspondence between subsets of eigenvalues and reducing subspaces (as there was a correspondence between subsets of eigenvalue of invariant subspaces for single matrices), but reducing subspaces are no longer spanned by eigenvectors, which are no longer well defined.

Consider the singular pencil (2.6). It has a nontrivial right reducing subspace $\cal X$ spanned by the first two columns of the 3 by 3 identity matrix and corresponding left reducing subspace $\cal Y$ spanned by $[1,0]^T$. Here $N_r$ is spanned by $[1,\lambda,0]^T$.

Susan Blackford 2000-11-20