Generalised Eigenstuff

Let $V$ be a vector space over a field $\mathbb K$ , and $T : V \to V$ be a linear transformation.

Recall that a nonzero vector $\mathbf v \in V$ is an eigenvector of $T : V \to V$ with eigenvalue $\lambda \in \mathbb K$ if $T(\mathbf v) = \lambda \mathbf v$ . Equivalently, $\mathbf v \in \ker(\lambda I_V - T)$ . If $V$ is finite dimensional with $\dim(V) = n$ , then $V$ has a basis $K$ of $n$ linearly independent vectors and $T$ is similar to the linear transformation $[T]_K : \mathbb K^n \to \mathbb K^n$ . Then $\mathbf \lambda$ is an eigenvalue of $T$ if and only if $c_T(\lambda) \equiv c_{[T]_K}(\lambda) = 0$ . We call $c_T$ the characteristic polynomial of $T$ . By the Cayley-Hamilton theorem, $c_T(T) = O_V$ .

Let $p \in \mathbb K[x]$ be a monic polynomial (i.e. the highest power of $x$ has coefficient $1$ ). If $p = c_T$ , then $p(T) = O_V$ . What is the “smallest” polynomial $p$ such that $p(T) = O_V$ ? We call this the minimal polynomial, and formulate it rigorously as follows.

Let $V$ be a finite-dimensional.

Theorem 1. For any linear transformation $T : V \to V$ , there exists a unique monic polynomial, denoted $m_T$ , such that $m_T(T) = O_V$ , and for any $p \in \mathbb K[x]$ such that $p(T) = O_V$ , $m_T \mid p$ .

Proof. By the Cayley-Hamilton theorem, there exists a monic polynomial $p = c_T \in \mathbb K[x]$ such that $p(T) = O_V$ . By the well-ordering principle, the set

$\{ n \in \mathbb N : (\exists p \in \mathbb K_n[x]\backslash \mathbb K_{n-1}[x]\quad\! p(T) = 0) \} \subseteq \mathbb N^+$

has a minimum element $m$ . Consider the monic polynomials $p_1 \in \mathbb K_{m}[x]\backslash \mathbb K_{m-1}[x], p_2 \in \mathbb K[x]$ such that $p_1(T) = p_2(T) = O_V$ . Use polynomial long division to compute $q \in \mathbb K[x]$ , $r \in \mathbb K_{m-1}[x]$ such that

$p_2 = q \cdot p_1 + r \quad \Rightarrow \quad r = p_2 - q \cdot p_1.$

Therefore, $r(T) = 0$ . By definition of $p_1,p_2$ , $r = 0$ . For uniqueness, if $p_2 \in \mathbb K_{m-1}[x]$ is monic, then $p_1 \mid p_2$ and $p_2 \mid p_1$ . Find polynomials $q_1, q_2 \in \mathbb K[x]$ such that

$p_2 = p_1 \cdot q_1\quad \text{and} \quad p_1 = p_2 \cdot q_2.$

Combining both decompositions, $q_1 \cdot q_2 = 1$ . Since $p_1,p_2$ are monic, so are $q_1, q_2$ , so that $q_1 = q_2 = 1$ , and hence, $p_1 = p_2$ .

Suppose $T$ has distinct eigenvalues $\lambda_1,\dots,\lambda_k$ .

Theorem 2. If $T$ has characteristic polynomial

$c_T(x) = (x - \lambda_1)^{\alpha_1} \cdot \cdots \cdot (x-\lambda_k)^{\alpha_k},$

then there exist unique integers $0 \leq \beta_i \leq \alpha_i$ such that $T$ has minimal polynomial

$m_T(x) = (x - \lambda_1)^{\beta_1} \cdot \cdots \cdot (x-\lambda_k)^{\beta_k},\quad 1 \leq \beta_i \leq \alpha_i.$

For each eigenvalue $\lambda_i$ , define the subspace $K_{\lambda_i} := \ker((T - \lambda_i)^{\beta_i})$ , where we abbreviate $T - \lambda I_V \equiv T - \lambda$ . Note that if $\beta_i = 1$ , then $E_{\lambda_i} = K_{\lambda_i}$ . Then

$V = K_{\lambda_1} \oplus \cdots \oplus K_{\lambda_k}.$

Proof. For the first claim, by definition, $m_T \mid c_T$ , and

$m_T(x) = (x - \lambda_1)^{\beta_1} \cdot \cdots \cdot (x-\lambda_k)^{\beta_k},\quad 0 \leq \beta_i \leq \alpha_i.$

It suffices to prove that $\beta_1 \geq 1$ without loss of generality. Let $\mathbf w \in E_{\lambda_1}$ and define $W := \mathrm{span}\{\mathbf w\}$ . Then $(T|_W - \lambda_1) = O_W$ . Hence,

$(x-\lambda_1) \mid m_{T|_W}(x) \quad \text{and}\quad m_{T|_W}(x) \mid (x - \lambda_1),$

so that $m_{T|_W}(x) = x-\lambda_1$ . Furthermore, it is not hard to see by definition that $m_{T|_W} \mid m_T$ . For the second claim, fix $\mathbf v \in V$ . Then $(m_T(T))(\mathbf v) = \mathbf 0$ . If $\mathbf v \notin K_{\lambda_i}$ for any $i$ , then $(m_T(T))(\mathbf v) \neq 0$ , a contradiction.

Corollary 1. Under the conditions in Theorem 2, the following propositions hold:

$E_{\lambda_i} \subseteq K_{\lambda_i}$ ,
$K_{\lambda_i}$ is $T$ -invariant,
$m_{ T|_{ K_{\lambda_i} } } = (x - \lambda_i)^{\beta_i}$ ,
$c_{ T|_{ K_{\lambda_i} } } = (x - \lambda_i)^{\alpha_i}$ ,
$\dim(K_{\lambda_i}) = \alpha_i$ .

Proof. Use an intermediate result that for $T$ -invariant subspaces $W_1, W_2 \subseteq V$ ,

$m_{T|_{W_1 + W_2}} = \mathrm{lcm}\{ m_{T|_{W_1}}, m_{T|_{W_2}} \}.$

Corollary 2. Under the conditions in Theorem 2, the following propositions are equivalent:

$T$ is diagonalisable,
$m_T(x) = (x-\lambda_1)\cdots (x - \lambda_k)$ ,
for each $i$ , $\dim(K_{\lambda_i}) = \alpha_i$ ,
$V = E_{\lambda_1} \oplus \cdots \oplus E_{\lambda_k}$ .

Recall that $E_{\lambda_i} = K_{\lambda_i}$ when $\beta_i = 1$ . In this regard, any vector in $K_{\lambda_i} = \ker( (T - \lambda_i)^{\beta_i} )$ is a generalised eigenvector of $T$ with generalised eigenvalue $\lambda_i$ . That is, $(T - \lambda_i)^{\beta_i}(\mathbf v) = 0$ , which reduces to an “actual” eigenvector precisely when $\beta_i = 1$ . Now, if

$\mathbb V = E_{\lambda_1} \oplus \cdots \oplus E_{\lambda_k},$

then $T$ is diagonalisable and vice-versa. That means $T$ is similar to a diagonal matrix $[T]_K$ for some basis $K$ of $V$ , which is a relatively simple transformation. On the other hand, in general, if $c_T$ can be factored into linear factors, then

$V = K_{\lambda_1} \oplus \cdots \oplus K_{\lambda_k}.$

Based on this decomposition, would there a suitably defined (and sufficiently simple) matrix (i.e. linear transformation) $[T]_K$ for some basis $K$ of $V$ such that $T$ is similar to $[T]_K$ ? The search for this answer will lead us to the climatic conclusion of this eigen-discussion—the Jordan normal form.

—Joel Kindiak, 11 Mar 25, 1615H

KindiakMath

Generalised Eigenstuff

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Generalised Eigenstuff

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