Eigenvalue is Instance of Generalized Eigenvalue
Jump to navigation
Jump to search
Theorem
Let $\mathbf A$ be a square matrix of order $n$.
Let $\lambda$ be an eigenvalue of $\mathbf A$.
Then:
- $\lambda$ is a generalized eigenvalue of $\mathbf A$
- the corresponding eigenvector $\mathbf x$ corresponding to $\lambda$ is the generalized eigenvector of $\mathbf A$ corresponding to $\lambda$.
Proof
By the definition of eigenvalue of $\mathbf A$:
- $\map \det {\mathbf I_n \mathbf x - \mathbf A} = 0$
for some non-zero vector $\mathbf x$.
Recall the definition of generalized eigenvalue of $\mathbf A$:
- $\mathbf A \mathbf x = \lambda \mathbf B \mathbf x$
for:
- some non-zero vector $\mathbf x$
- some square matrix $\mathbf B$ of order $n$.
Then from Generalized Eigenvalues as Roots of Equation:
- $\map \det {\mathbf A - \lambda \mathbf B} = 0$
where:
- $\mathbf B$ is another square matrix of order $n$
- $\det$ denotes the determinant.
Hence setting:
- $\mathbf B = \mathbf I_n$
it is seen that $\lambda$ is a generalized eigenvalue of $\mathbf A$
Hence, by definition, the eigenvector $\mathbf x$ corresponding to $\lambda$ is the generalized eigenvector of $\mathbf A$ corresponding to $\lambda$.
$\blacksquare$
Sources
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): generalized eigenvalue