## 每週問題 April 18, 2016

Let $\mathcal{V}$ be an $n$-dimensional vector space, and $S_1, \ldots, S_k$ be subspaces in $\mathcal{V}$. If $\sum_{i=1}^k\dim S_i>n(k-1)$, show that $\bigcap_{1\le i\le k}S_i\neq\{\mathbf{0}\}$.

$\displaystyle \dim\left(\bigcap_{1\le i\le k}S_i\right)\ge \sum_{i=1}^k\dim S_i-n(k-1)$

$k=1$，命題顯然成立。假設命題於 $k\le p$ 時成立。令指標集合 $I=\{1,2,\ldots,p+1\}$。使用容斥定理

$\dim(U\cap W)=\dim U+\dim W-\dim(U+W)$

\displaystyle\begin{aligned} \dim\left( \bigcap_{i\in I}S_i\right)&=\dim \left(\left(\bigcap_{i\in I\setminus\{1\}}S_i\right)\bigcap \left(\bigcap_{i\in I\setminus\{2\}}S_i\right)\right)\\ &=\dim\left( \bigcap_{i\in I\setminus\{1\}}S_i\right)+\dim\left( \bigcap_{i\in I\setminus\{2\}}S_i\right)-\dim\left(\bigcap_{i\in I\setminus\{1\}}S_i+\bigcap_{i\in I\setminus\{2\}}S_i\right)\\ &=\dim\left( \bigcap_{i\in I\setminus\{1\}}S_i\right)+\dim\left(\left(\bigcap_{i\in I\setminus\{1,2\}}S_i\right)\bigcap S_1\right)-\dim\left(\bigcap_{i\in I\setminus\{1\}}S_i+\bigcap_{i\in I\setminus\{2\}}S_i\right)\\ &\ge \sum_{i\in I\setminus\{1\}}\dim S_i-n(p-1)+\dim\left(\bigcap_{i\in I\setminus\{1,2\}}S_i\right)+\dim S_1-n\\ &~~~~~-\dim\left(\bigcap_{i\in I\setminus\{1\}}S_i+\bigcap_{i\in I\setminus\{2\}}S_i\right)\\ &=\sum_{i=1}^{p+1}\dim S_i-np+\dim\left(\bigcap_{i\in I\setminus\{1,2\}}S_i\right)-\dim\left(\bigcap_{i\in I\setminus\{1\}}S_i+\bigcap_{i\in I\setminus\{2\}}S_i\right). \end{aligned}

$\bigcap_{i\in I\setminus\{1\}}S_i+\bigcap_{i\in I\setminus\{2\}}S_i\subseteq \bigcap_{i\in I\setminus\{1,2\}}S_i$，因此

$\displaystyle \dim\left( \bigcap_{1\le i\le p+1}S_i\right)\ge \sum_{i=1}^{p+1}\dim S_i-np$