## 每週問題 December 19, 2016

Let $\mathcal{V}$ be a vector space, $\dim\mathcal{V}=n$, and let $\mathbf{x}_1,\ldots,\mathbf{x}_m\in\mathcal{V}$. Prove that if $m\ge n+2$, then there exist scalars $c_1,\ldots,c_m$ not all of them equal to zero such that $\sum_{i=1}^mc_i\mathbf{x}_i=\mathbf{0}$ and $\sum_{i=1}^mc_i=0$.

(1) 若 $\mathbf{x}_{k+1}=\mathbf{0}$$\mathbf{x}_{k+2}=\mathbf{0}$，則 $\mathbf{x}_{k+1}-\mathbf{x}_{k+2}=\mathbf{0}$，如此便得證。

(2) 若 $\mathbf{x}_{k+1}\neq\mathbf{0}$$\mathbf{x}_{k+2}=\mathbf{0}$，存在不全為零的數組 $a_1,\ldots,a_k$ 使得

$a_1\mathbf{x}_1+\cdots+a_k\mathbf{x}_k+\mathbf{x}_{k+1}=\mathbf{0}$

$\displaystyle a_1\mathbf{x}_1+\cdots+a_k\mathbf{x}_k+\mathbf{x}_{k+1}-\left(\sum_{i=1}^ka_i+1\right)\mathbf{x}_{k+2}=\mathbf{0}$

(3) 若 $\mathbf{x}_{k+1}\neq\mathbf{0}$$\mathbf{x}_{k+2}\neq\mathbf{0}$，存在不全為零的數組 $a_1,\ldots,a_k$$b_1,\ldots,b_k$ 使得

\begin{aligned} a_1\mathbf{x}_1+\cdots+a_k\mathbf{x}_k+\mathbf{x}_{k+1}&=\mathbf{0}\\ b_1\mathbf{x}_1+\cdots+b_k\mathbf{x}_k+\mathbf{x}_{k+2}&=\mathbf{0}.\end{aligned}

$\sum_{i=1}^ka_i+1=0$$\sum_{i=1}^kb_i+1=0$，我們便得到所求。若 $\sum_{i=1}^ka_i+1\neq 0$$\sum_{i=1}^kb_i+1\neq 0$，則

\begin{aligned} \mathbf{0}&=\left(\sum_{i=1}^kb_i+1\right)(a_1\mathbf{x}_1+\cdots+a_k\mathbf{x}_k+\mathbf{x}_{k+1})-\left(\sum_{i=1}^ka_i+1\right)(b_1\mathbf{x}_1+\cdots+b_k\mathbf{x}_k+\mathbf{x}_{k+2})\\ &=\left(\left(\sum_{i=1}^kb_i+1\right)a_1-\left(\sum_{i=1}^ka_i+1\right)b_1\right)\mathbf{x}_1+\cdots\\ &~~~~+\left(\left(\sum_{i=1}^kb_i+1\right)a_k-\left(\sum_{i=1}^ka_i+1\right)b_k\right)\mathbf{x}_k\\ &~~~~+\left(\sum_{i=1}^kb_i+1\right)\mathbf{x}_{k+1}-\left(\sum_{i=1}^ka_i+1\right)\mathbf{x}_{k+2}, \end{aligned}

\begin{aligned} &\left(\left(\sum_{i=1}^kb_i+1\right)a_1-\left(\sum_{i=1}^ka_i+1\right)b_1\right)+\cdots+\left(\left(\sum_{i=1}^kb_i+1\right)a_k-\left(\sum_{i=1}^ka_i+1\right)b_k\right)\\ &+\left(\sum_{i=1}^kb_i+1\right)-\left(\sum_{i=1}^ka_i+1\right)\\ &=\left(\sum_{i=1}^kb_i+1\right)\sum_{j=1}^ka_j-\left(\sum_{i=1}^ka_i+1\right)\sum_{j=1}^kb_j+\sum_{i=1}^kb_i-\sum_{i=1}^ka_i=0.\end{aligned}

$\begin{bmatrix} [\mathbf{x}_1]_{\boldsymbol{\beta}}&\cdots&[\mathbf{x}_m]_{\boldsymbol{\beta}}\\ 1&\cdots&1 \end{bmatrix}\begin{bmatrix} c_1\\ \vdots\\ c_m \end{bmatrix}=\begin{bmatrix} \mathbf{0}\\ 0 \end{bmatrix}$