Definition

Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $\lambda$ be a scalar. A non-zero vector $x$ in $V$ is called a generalized eigenvector of $T$ corresponding to $\lambda$ if $(T - \lambda I)^p(x) = 0$ for some positive integer $p$.

Note here that if $x$ is a generalized eigenvector of $T$ corresponding to $\lambda$ and $p$ is the smallest positive integer for which $(T - \lambda I)^{p}(x) = 0$, then

$$ \begin{align*} (T - \lambda I)^{p-1}(x) &= y \neq 0 \\ (T - \lambda I)(T - \lambda I)^{p-1}(x) &= (T - \lambda I) y \\ (T - \lambda I)^{p}(x) &= T(y) - \lambda y \\ \bar{0} &= T(y) - \lambda y \\ T(y) &= \lambda y \end{align*} $$

So $y$ is an eigenvector of $T$.

Definition

Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $\lambda$ be an eigenvalue of $T$. The generalized eigenspace of $T$ corresponding to $\lambda$ denoted $K_{\lambda}$, is the subset of $V$ defined by $$ \begin{align*} K_{\lambda} = \{ x \in V: (T - \lambda I)^p(x) = 0 \quad \text{for some positive integer $p$} \} \end{align*} $$

Note that $K_{\lambda}$ consists of the zero vector and all generalized eigenvectors corresponding to $\lambda$.

Theorem 7.1

Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $\lambda$ be an eigenvalue of $T$. Then

$K_{\lambda}$ is a $T$-invariant subspace of $V$ containing $E_{\lambda}$ (the eigenspace of $T$ corresponding to $\lambda$)
For any eigenvalue $\mu$ of $T$ such that $\mu \neq \lambda$, $K_{\lambda} \cap K_{\mu} = \{0\}$
For any scalar $\mu \neq \lambda$, the restriction of $T - \mu I$ to $K_{\lambda}$ is one-to-one and onto.

Proof
(a) Showing that $K_{\lambda}$ is a subspace is straightforward. We need to show that $\bar{0} \in K_{\lambda}$ and that $K_{\lambda}$ is closed under scalar multiplication and addition.

To show that $K_{\lambda}$ is $T$-invariant. We need to show for any $x \in K_{\lambda}$, that $T(x) \in K_{\lambda}$. By definition, let be $p$ be a positive integer, then $(T - \lambda I)^p(x) = \bar{0}$. We know to show that $(T - \lambda I)^p(T(x)) = \bar{0}$. Observe that

$$ \begin{align*} (T - \lambda I)^p(T(x)) = T(T - \lambda I)^p(x) = T(\bar{0}) = \bar{0} \end{align*} $$

(b) TODO
(c) Let $\mu$ be a scalar such that $\mu \neq \lambda$. Let $x \in K_{\lambda}$. Let $p$ be the smallest integer such that $(T - \lambda I)^p x = 0$ and

$$ \begin{align*} W = \text{span}\{x, (T - \lambda I)(x),...,(T - \lambda I)^{p-1}(x)\} \end{align*} $$

Theorem 7.2

Let $T$ be a linear operator on a finite-dimensional vector space $V$ such that the characteristic polynomial splits. Suppose that $\lambda$ is an eigenvalue of $T$ with multiplicity $m$. Then

$\dim(K_{\lambda}) \leq m$
$K_{\lambda} = N((T - \lambda I)^m)$

Proof
(a) Let $W = K_{\lambda}$. Let the characteristic polynomial of $W$ be $h(t)$ We know by Theorem 7.1, that $W$ is a $T$-invariant subspace of $V$. Therefore, by Theorem 5.20, $h(t)$ divides the characteristic polynomial of $T$. From Theorem 7.1(b), we know that $\lambda$ is the only eigenvalue of $T_W$. Therefore $h(t) = (-1)^d(t - \lambda)^d$ where $d = \dim(W)$ and $d \leq m$.

(b) By definition, we know that $N(T)=\{x \in V \ | \ T(x) = 0\}$. We also know that $K_{\lambda} =\{ x \in V \ | \ (T - \lambda I)^p = 0 \ \text{ for }p > 0 \}$. So we can see that for any $x \in N((T - \lambda I)^m)$

$$ \begin{align*} (T - \lambda I)^m(x) = 0 \end{align*} $$

Since $m$ is the multiplicity of an eigenvalue, then it’s positive and so $x \in K_{\lambda}$ by definition. So $N((T - \lambda I)^m) \in K_{\lambda}$.

Now, suppose that $x \in K_{\lambda}$. We want to show that $x \in N((T - \lambda I)^m)$ where $m$ is the multiplicity of $\lambda$. Since the characteristic polynomial of $T$ splits, then let it be of the form $f(t) = (t - \lambda)^m g(t)$ where $g(t)$ is the product of powers of the form $t - \mu$ for eigenvalues $\mu \neq \lambda$. By Theorem 7.1, $g(T)$ is one-to-one on $K_{\lambda}$[TODO: WHY?]. It is also onto since $K_{\lambda}$ is finite dimensional. Since $x \in K_{\lambda}$, then there exists some $y$ such that $g(T)(x) = y$. [WHY?]. Hence

$$ \begin{align*} (T - \lambda I)^m(x) = (T - \lambda I)^m g(T)(y) = f(T)(y) = \bar{0} \end{align*} $$

Why? ….. [TODO]

Theorem 7.3

Let $T$ be a linear operator on a finite-dimensional vector space $V$ such that characteristic polynomial of $T$ splits, and let $\lambda_1,\lambda_2,..., \lambda_k$ be the distinct eigenvalues of $T$. Then, for every $x \in V$, there are exist unique vectors $v_i \in K_{\lambda}$, for $i = 1,2,...,k$ such that $$ \begin{align*} x = v_1 + v_2 + ... + v_k \end{align*} $$

Proof
TODO

Theorem 7.4

Let $T$ be a linear operator on a finite-dimensional vector space $V$ whose characteristic polynomial $(t - \lambda_1)^{m_1}(t - \lambda_2)^{m_2}...(t - \lambda_k)^{m_k}$ splits. For $i = 1,2,...,k$, let $\beta_i$ be an ordered basis for $K_{\lambda_i}$. Then

$\beta_i \cap \beta_j = \emptyset \text{ for } i \neq j$
$\beta = \beta_1 \cup \beta_2 \cup ... \cup \beta_k $ is an ordered basis for $V$
$\dim(K_{\lambda_j}) = m_i$ for all $i$

Proof
(a) Consequence of Theorem 7.1(b)

(b) We need to show that $\beta$ is linearly independent and that $\beta$ spans $V$. To see that $\beta$ is linearly independent.

(c)

Theorem 7.6

Let $T$ be a linear operator on a vector space $V$, and let $\lambda$ be an eigenvalue of $T$. Suppose that $\gamma_1, \gamma_2,...,\gamma_q$ are cycles of generalized eigenvectors of $T$ corresponding to $\lambda$ such that the initial vectors of the $\gamma_i$'s are distinct and form a linearly independent set. Then, the $\gamma_i$'s are disjoint and their union $\gamma = \bigcup\limits_{i=1}^q \gamma_i$ is linearly independent.

Proof
TODO

Corollary (7.6)

Every cycle of generalized eigenvectors of a linear operator is linearly independent

Theorem 7.7

Let $T$ be a linear operator on a finite-dimensional vector space $V$, and let $\lambda$ be an eigenvalue of $T$. Then $K_{\lambda}$ has an ordered basis consisting of a union of disjoint cycles of generalized eigenvectors corresponding to $\lambda$.

Proof
TODO

Corollary 1 (7.7)

Let $T$ be a linear operator on a finite-dimensional vector space $V$ whose characteristic polynomial splits. Then $T$ has a Jordan canonical form.

Proof
TODO

References

Linear Algebra 5th Edition