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Suppose we have $ \textbf{x}_\textbf{0},\textbf{x}_\textbf{1},\ldots,\textbf{x}_{n-1}$ such that $\textbf{x}_\textbf{1}-\textbf{x}_{n-0}, \ldots, \textbf{x}_{n-1}-\textbf{x}_0$ are linearly independent. How can you prove there is exactly one hyperplane containing all $\textbf{x}_i$?

My idea was to assume there are two hyperplanes $P= \{ \textbf{x} \mid \textbf{z} \cdot \textbf{x}= c \}$ and $P'= \{ \textbf{x} \mid \textbf{y} \cdot \textbf{x}= b \}$ and then conclude that $\textbf{z}$ and $\textbf{y}$ are scalar multiples of each other, and so are $b,c$.

Any suggestions?

1 Answers 1

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Your approach should work. Consider the following. I will use your notation for $P$ and $P'$.

Note that $P-x_0$ has dimension $n-1$ and it is spanned by $x_1-x_0, ..., x_{n-1}-x_0$. Thus $P-x_0 = span\{x_1-x_0, ..., x_{n-1}-x_0\}$.

Similarly, $P'-x_0$ has dimension $n-1$ and it is spanned by $x_1-x_0, ..., x_{n-1}-x_0$. Thus $P'-x_0 = span\{x_1-x_0, ..., x_{n-1}-x_0\}$.

This gives us $P - x_0 = P'-x_0$. Hence $P=P'$.