每週問題 August 22, 2016

Posted on 08/22/2016 by ccjou

證明一個直覺命題：若一個子空間與線性變換的零空間不交集，則該子空間的像 (image) 的維數等於子空間的維數。

Let \mathcal{V} and \mathcal{W} be finite dimensional vector spaces, and T:\mathcal{V}\to\mathcal{W} be a linear transformation. For a subspace \mathcal{X} of \mathcal{V}, the image T(\mathcal{X})=\{T(\mathbf{x})|\mathbf{x}\in\mathcal{X}\} of \mathcal{X} under T is a subspace of \mathcal{W}. Prove that if \mathcal{X}\cap N(T)=\{\mathbf{0}\}, then \dim T(\mathcal{X})=\dim\mathcal{X}. Note that N(T) denotes the nullspace (kernel) of T.

 
參考解答：

我們證明若 \{\mathbf{x}_1,\ldots,\mathbf{x}_k\} 是子空間 \mathcal{X} 的一組基底且 \mathcal{X}\cap N(T)=\{\mathbf{0}\}，則 \{T(\mathbf{x}_1),\ldots,T(\mathbf{x}_k)\} 是 T(\mathcal{X}) 的一組基底。對於任一 \mathbf{x}\in\mathcal{X}，寫出 \mathbf{x}=c_1\mathbf{x}_1+\cdots+c_k\mathbf{x}_k，則

T(\mathbf{x})=T(c_1\mathbf{x}_1+\cdots+c_k\mathbf{x}_k)=c_1T(\mathbf{x}_1)+\cdots+c_kT(\mathbf{x}_k)。

因此，T(\mathcal{X})=\hbox{span}\{T(\mathbf{x}_1),\ldots,T(\mathbf{x}_k)\}。接下來，我們證明 \{T(\mathbf{x}_1),\ldots,T(\mathbf{x}_k)\} 是一個線性獨立集：

\displaystyle \begin{aligned} \sum_{i=1}^kc_iT(\mathbf{x}_i)=\mathbf{0}&\Rightarrow T\left(\sum_{i=1}^kc_i\mathbf{x}_i\right)=\mathbf{0}\\ &\Rightarrow \sum_{i=1}^kc_i\mathbf{x}_i\in N(T)\\ &\Rightarrow \sum_{i=1}^kc_i\mathbf{x}_i=\mathbf{0}~~ (\mathcal{X}\cap N(T)=\{\mathbf{0}\})\\ &\Rightarrow c_1=\cdots=c_k=0.\end{aligned}

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