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Suppose I have $\sum_{i = 1}^{n} v_{i} = \sum_{i = 1}^{n} w_{i}$ where $\{v_{i}\}_{i = 1}^{n}$ and $\{w_{j}\}_{j = 1}^{n}$ are both sets of linearly independent vectors in $\mathbb{R}^{d}$ ($d$ possibly $\neq n$). Is it true that we must have $\{v_{i}\}_{i = 1}^{n} = \{w_{j}\}_{j = 1}^{n}$?

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$(0,1)+(1,0) = (1/2, 0)+(1/2, 1)$. – David Mitra Feb 21 '12 at 0:37
up vote 2 down vote accepted

No; for example, let $d=n=2$, and let $$v_1=(0,1),\;v_2=(2,1)\quad\text{ and }\quad w_1=(1,2),\; w_2=(1,0).$$ Then $\{v_1,v_2\}\neq\{w_1,w_2\}$ but $$v_1+v_2=(0,1)+(2,1)=(2,2)=(1,2)+(1,0)=w_1+w_2.$$

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No it is not. Take $n=2$. Take $\left\{\begin{bmatrix}1\\1\end{bmatrix},\begin{bmatrix}1\\-1\end{bmatrix}\right\}$ and $\left\{\begin{bmatrix}2\\1\end{bmatrix},\begin{bmatrix}0\\-1\end{bmatrix}\right\}$.

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