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)

\begin{align} \quad v_k &:= v \\ \quad v_{k-1} &:= (A - \lambda I)v \\ \quad v_{k-2} &:= (A - \lambda I)^2 v \\ \quad &\vdots \\ \quad v_1 &:= (A - \lambda I)^{k-1} v \end{align}

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)

\begin{align} \quad v_i = (A - \lambda)^{k - i}v \end{align}

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

(3)

\begin{align} \quad (A - \lambda)^i v_i &= (A - \lambda)^i (A - \lambda)^{k - i} v \\ &= (A - \lambda)^k v \\ & = 0 \end{align}

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

(4)

\begin{align} \quad (A - \lambda)^{i-1} v_i &= (A - \lambda)^{i - 1} (A - \lambda)^{k - i} v \\ &= (A - \lambda)^{k - 1}v \\ & \neq 0 \end{align}

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.