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The subspace in question: $V=\{ \vec{u} \in \Bbb{R}^n : \vec{n}^T\vec{u}=\vec{0} \}$

I am assuming that $\vec{u} = \begin{bmatrix}x_0 \\ x_1 \\ \vdots \\ x_2\end{bmatrix}$.

The dimension of a vector space/subspace is equal to the number of linearly independent vectors in its basis. So it's either 1 or n. Does $\vec{u}$ count as 1 or as n?

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    (1) By definition, the vectors in a basis are linearly independent. (2) Why do you say "So it's either 1 or $n$"?2012-11-17
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    (1) Ok. (2) Because I don't know what the answer is, but I'm fairly certain that it's one of those.2012-11-17

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