# Generalized Eigenvectors of Square Matrices

Recall from the Eigenvectors of Square Matrices that if $A$ is an $n \times n$ square matrix and $\lambda$ is an eigenvalue of $A$ then a nonzero vector $v$ is said to be a eigenvector of $A$ corresponding to the eigenvalue $\lambda$ if $(A – \lambda I)v = 0$.

We now define a more general type of eigenvector.

Definition: Let $A$ be an $n \times n$ matrix and let $\lambda$ be an eigenvalue of $A$. A Generalized Eigenvector of Rank $k$ corresponding to the eigenvalue $\lambda$ is a vector $v$ such that $(A – \lambda I)^k v = 0$ and $(A – \lambda I)^{k-1} v \neq 0$. |

It is important to note that regular eigenvectors are the same as generalized eigenvectors of rank $1$.

Suppose that $v$ is a generalized eigenvector of rank $k$ corresponding to the eigenvalue $\lambda$. Then $(A – \lambda I)^k v = 0$ and $(A – \lambda I)^{k-1}v \neq 0$.

Let:

(1)

For each $i \in \{ 1, 2, …, k \}$, $v_k$ is a generalized eigenvector of rank $i$ corresponding to the eigenvalue $\lambda$. To see that, let $i \in \{ 1, 2, …, k \}$. Then:

(2)

Therefore (using the fact that $(A – \lambda)^k v = 0$ we have that:

(3)

However (using the fact that $(A – \lambda)^{k-1}v \neq 0$) we have that:

(4)

Lemma 1: If $A$ is an $n \times n$ square matrix and $v$ is a generalized eigenvector of rank $k$ corresponding to an eigenvalue $\lambda$, then the set of vectors $\{ v_1, v_2, …, v_k \}$ defined above are linearly independent. |

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