The Cyclic Subspace

The Cyclic Subspace

Definition 1. Let $T \in \mathcal{L}(V)$, and let a nonzero vector $x \in V$. The subspace $W = \langle x, T(x), T^2(x), ... \rangle$ is called the $T$-cyclic subspace of $V$ generated by $x$.

Theorem 1

Theorem 1. Let $T \in \mathcal{L}(V)$, and let $W$ be the $T$-cyclic subspace of $V$ generated by $\mathbf{0} \neq x \in V$. Then
(a) $W$ is $T$-invariant.
(b) Any $T$-invariant subspace of $V$ containing $x$ also contain $W$.

Proof. (a) $\forall u \in T(W), \exists y \in W$ such that $u = T(y)$. Then $y = a_0x + a_1T(x) + \cdots$ for some $a_0, a_1, ...$. Thus $u = T(y) = a_0T(x) + a_1T^2(x) + \cdots \in W$. Thus $T(W) \subseteq W$.
(b) Let $K$ be a $T$-invariant subspace of $V$. $\forall v \in W, v = a_0x + a_1T(x) + \cdots$ for some $a_0, a_1, ...$. Note that $T^n(x) \in K, \forall n \in \mathbb{N}$, because $x \in K$. Then $v \in K$. Thus $W \subseteq K$. $\blacksquare$

다시 말해 $T$-순환 부분공간은 $x$를 포함하는 $V$의 가장 작은 $T$-불변 부분공간이다.

Theorem 2

Theorem 2. Let $T \in \mathcal{L}(V)$, and let $W$ denote the $T$-cyclic subspace of $V$ generated by $\mathbf{0} \neq v \in V$. Let $k = \dim(W)$. Then
(a) $\{v, T(v), ..., T^{k-1}(v)\}$ is a basis for $W$.
(b) If $a_0v + a_1T(v) + \cdots + a_{k-1}T^{k-1}(v) + T^k(v) = \mathbf{0}$, then the characteristic polynomial of $T_W$ is $f(t) = (-1)^k(a_0 + a_1t + \cdots + a_{k-1}t^{k-1} + t^k)$.

Proof. (a) Since $v \neq \mathbf{0}, \{v\}$ is linearly independent. Let $j$ be the largest positive integer such that $\beta = \{v, T(v), ..., T^{j-1}(v)\}$ is linaerly independent. Let $Z = \langle \beta \rangle$. Then $\beta$ is a basis for $Z$.
Let $u \in Z$. Then $! \exists b_0, ..., b_{j-1} \in F$ such that $u = b_0v + \cdots + b_{j-1}T^{j-1}(v)$. Since $T^j(v) \in Z$, $T(u) = b_0T(v) + \cdots + b_{j-1}T^j(v)$. Thus $T(u) \in Z$, so $Z$ is $T$-invariant. 
Since $Z$ is a subspace of $V$ containing $v$, by Theorem 1 $W \subseteq Z$. Clearly $Z \subseteq W$. Thus $W = Z$, and $j = k$.
(b) Suppose that $a_0v + a_1T(v) + \cdots + a_{k-1}T^{k-1}(v) + T^k(v) = \mathbf{0}$. Then $T^k(v) = (-a_0)v + (-a_1)T(v) + \cdots + (-a_{k-1})T^{k-1}(v)$. 
Note that $[T_W]_{\beta} = \begin{pmatrix} 0 & 0 & \cdots & 0 & -a_0 \\ 1 & 0 & \cdots & 0 & -a_1 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & -a_{k-2} \\ 0 & 0 & \cdots & 1 & -a_{k-1} \end{pmatrix},$ where $\beta = \{v, T(v), ..., T^{k-1}(v)\}$. Then the characteristic polynomial of $T_W$ is $f(t) = \det([T_W]_{\beta} - tI_k) = \det \begin{pmatrix} -t & 0 & \cdots & 0 & -a_0 \\ 1 & -t & \cdots & 0 & -a_1 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & -t & -a_{k-2} \\ 0 & 0 & \cdots & 1 & -a_{k-1}-t \end{pmatrix}.$
We claim that $f(t) = (-1)^k(a_0 + a_1t + \cdots + a_{k-1}t^{k-1} + t^k)$. We use the induction on $k$. If $k = 2$, then $f(t) = \det \begin{pmatrix} -t & -a_0 \\ 1 & -a_1 - t \end{pmatrix}$ = $-t(a_1 - t) + a_0$ = $(-1)^2(a_0 + a_1t + t^2)$. 
Assume the statement is true for $k-1 \geq 2$. Then $$f(t) = \sum_{j=1}^k (-1)^{1+j}A_{1j}\det(\widetilde{A_{1j}}) \\ = (-t)\det \begin{pmatrix} -t & \cdots & 0 & -a_1 \\ \vdots & \ddots & \vdots & \vdots \\ 0 & \cdots & 1 & -a_{k-1} - t \end{pmatrix} + (-1)^{1+k}(-a_0) \det \begin{pmatrix} 1 & -t & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{pmatrix} \\ = (-t)(-1)^{k-1}(a_1 + \cdots + a_{k-1}t^{k-2} + t^{k-1}) + (-1)^{1+k}(-a_0) \\ = (-1)^k(a_1t + \cdots + a_{k-1}t^{k-1} + t^k) + (-1)^ka_0 \\ = (-1)^k(a_0 + a_1t + \cdots + a_{k-1}t^{k-1} + t^k).$$ Thus $f(t) = (-1)^k(a_0 + a_1t + \cdots + a_{k-1}t^{k-1} + t^k)$. $\blacksquare$