## 每週問題 September 24, 2012

$T$ 為定義於向量空間 $\mathcal{V}$ 的線性變換，證明存在一正整數 $n$ 使得 $\mathcal{V}=R(T^n)\oplus N(T^n)$

Let $\mathcal{U}$ and $\mathcal{W}$ be two subspaces of a ﬁnite dimensional vector space $\mathcal{V}$. The sum $\mathcal{U} + \mathcal{W}$ is called a direct sum, denoted by $\mathcal{U}\oplus \mathcal{W}$, if every element $\mathbf{x}\in \mathcal{U}+\mathcal{W}$ can be uniquely written as $\mathbf{x} = \mathbf{u} + \mathbf{w}$, where $\mathbf{u}\in \mathcal{U}$ and $\mathbf{w}\in \mathcal{W}$. Let $T$ be a linear transformation on $\mathcal{V}$. Show that there exists a positive integer $n$ so that $\mathcal{V} = R(T^n)\oplus N(T^n)$, where $R(T^n)$ and $N(T^n)$ denote the range and kernel of $T^n$, respectively.

$T\mathbf{x} = \mathbf{0}$，則 $T^2\mathbf{x} = T(T\mathbf{x}) = T(\mathbf{0}) = \mathbf{0}$，即知 $N(T) \subseteq N(T^2)$，繼續此程序可推論

$N(T)\subseteq N(T^2) \subseteq\cdots\subseteq N(T^k) \subseteq N(T^{k+1}) \subseteq\cdots$

$\mathcal{V}$ 為一有限維向量空間，故必定存在一正整數 $n$ 使得對於所有正整數 $m$ 都有 $N(T^n) =N(T^{n+m})$。根據秩─零度定理 $\dim\mathcal{V} = \dim R(T^n) + \dim N(T^n)$，因此欲證明 $V =R(T^n)\oplus N(T^n)$，只須證明 $R(T^n) \cap N(T^n) = \{\mathbf{0}\}$ 即可。令 $\mathbf{x}\in R(T^n) \cap N(T^n)$，則存在 $\mathbf{y}$ 使得 $\mathbf{x} = T^n\mathbf{y}$$T^n\mathbf{x} = \mathbf{0}$。合併上面兩式，$\mathbf{0} = T^n\mathbf{x} = T^n(T^n\mathbf{y}) = T^{2n}\mathbf{y}$，這說明 $\mathbf{y}\in N(T^{2n}) = N(T^n)$，所以 $\mathbf{x} = T^n\mathbf{y} = \mathbf{0}$，得證。

PowSol-Sept-24-12