## 每週問題 August 22, 2016

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 $T$.

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

\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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