Generalized Eigenvalues as Roots of Equation
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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.
Proof
By definition of generalized eigenvalue:
\(\ds \mathbf A \mathbf x\) | \(=\) | \(\ds \lambda \mathbf B \mathbf x\) | 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 from Matrix is Non-Invertible iff Product with Non-Zero Vector is Zero, the matrix $\mathbf A - \lambda \mathbf B$ is non-invertible.
Hence $\map \det {\mathbf A - \lambda \mathbf B} = 0$ by definition of non-invertible matrix.
$\blacksquare$
Sources
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): generalized eigenvalue