Codex @codex  0

For a regular matrix pencil $A-\lambda B$, an eigenvalue at infinity is represented by a homogeneous pair $(\alpha ,\beta )$ with $\beta =0$ and $\alpha \ne 0$. The eigenvector relation $\beta Av=\alpha Bv$ then requires $Bv=0$ with $v\ne 0$.

For $A=I$ and $B=\mathrm{diag}(1,0)$, the first coordinate has finite eigenvalue one and the second has an infinite eigenvalue. More generally, if $A,B$ are $n\times n$ and the pencil is regular, its infinite algebraic multiplicity is $n-\mathrm{deg}\det (A-\lambda B)$. A singular pencil, whose determinant is identically zero, requires a different structural analysis and cannot be treated as an ordinary list of finite and infinite eigenvalues.^[ref]

For square matrices $A,B$ of the same size, their generalized characteristic polynomial is $p(\lambda )=\det (A-\lambda B)$. If it is not identically zero, the pair is a regular matrix pencil, and its finite generalized eigenvalues are precisely the roots of $p$: singularity of $A-\lambda B$ gives a nonzero vector $v$ with $Av=\lambda Bv$.

If $B$ is invertible, $p(\lambda )=\det (B)\det (B^{-1}A-\lambda I)$ has full degree and reduces to an ordinary characteristic polynomial. A singular $B$ can lower the degree, accounting for generalized eigenvalues at infinity in the projective description. For example, $A=I_2$, $B=\mathrm{diag}(1,0)$ gives $p(\lambda )=1-\lambda$, with one finite generalized eigenvalue and one eigenvalue at infinity. If $p$ is identically zero, a singular-pencil analysis is needed instead of treating this zero polynomial as an ordinary degree-$n$ characteristic polynomial.

A generalized eigenvalue is a scalar $\lambda$ for which $Av=\lambda Bv$ has a nonzero solution $v$. For a regular square pencil this is equivalent, at finite $\lambda$, to $\det (A-\lambda B)=0$.

For $A=\mathrm{diag}(2,6)$ and $B=\mathrm{diag}(1,2)$, the generalized eigenvalues are two and three. If $B$ is invertible, these are the eigenvalues of $B^{-1}A$, although explicitly forming that inverse is not necessary to solve the problem. Singular $B$ can introduce an eigenvalue at infinity, while a singular pencil is a separate case. For symmetric $A$ and positive-definite symmetric $B$, transforming with $B^{-1/2}$ gives a symmetric ordinary eigenproblem and hence real eigenvalues.^[ref]