Generalized Eigenvalues as Roots of Equation

Theorem

Let $\mathbf A$ be a square matrix of order $n$.

Let $\lambda$ be a generalized eigenvalue of $\mathbf A$.

Then:

$\map \det {\mathbf A - \lambda \mathbf B} = 0$

where:

$\mathbf B$ is another square matrix of order $n$

$\det$ denotes the determinant.

$\ds \mathbf A \mathbf x$

$=$

$\ds \lambda \mathbf B \mathbf x$

for some non-zero vector $\mathbf x$

$\ds \leadsto \ \ $

$\ds \mathbf A \mathbf x - \lambda \mathbf B \mathbf x$

$=$

$\ds 0$

$\ds \leadsto \ \ $

$\ds \paren {\mathbf A - \lambda \mathbf B} \mathbf x$

$=$

$\ds 0$

We have that $\mathbf x$ is non-zero by hypothesis.

Hence $\map \det {\mathbf A - \lambda \mathbf B} = 0$ by definition of non-invertible matrix.

$\blacksquare$