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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 $n$"? – wj32 Nov 17 '12 at 10:13
(1) Ok. (2) Because I don't know what the answer is, but I'm fairly certain that it's one of those. – user1132363 Nov 17 '12 at 10:15

$\vec u$ is one vector, in itself determines $1$ dimension (unless $\vec u=\vec 0$), but lies in the (canonical) $n$ dimension space $\Bbb R^n$, that is: has $n$ coordinates.

If we have $k$ vectors: $\vec a_1,..,\vec a_k$ (doesn't matter from which vector space), then they can 'determine' at most $k$ dimension.

Intuitively, in the exercise, $V=\{\vec u \mid \vec u\perp\vec n\}$ for the fixed $\vec n$. Unless $\vec n=\vec 0$, geometrically viewed, it must be of dimension $n-1$. You can try to find a basis (to verify this claim), preferaribly starting out from an orthogonal basis of $\Bbb R^n$ which contains $\vec n$.

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It can be helpful to think of a concrete example. Suppose $n = 3$ and $\vec{n} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$. Then \begin{equation} V = \left \{ \begin{bmatrix} x \\ y \\ 0 \end{bmatrix} \mid x, y \in \mathbb R \right\}. \end{equation} In this example it seems clear that $V$ has dimension $2$.

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If I understand this correctly then the answer to my question is: "anywhere up to n" ? – user1132363 Nov 17 '12 at 12:44
Actually no, the answer is $n - 1$ (assuming $u \neq 0$). One way to prove this is to note that $V$ is the null space of $\vec{n}^T$, and then use the rank-nullity theorem. – littleO Nov 17 '12 at 20:04
I should have said "assuming $\vec{n} \neq 0$. (I was mistakenly writing $u$ instead of $\vec{n}$.) – littleO Nov 17 '12 at 20:56

$\vec{u} \in \Bbb{R}^n$, $\vec{u}\not=\vec{0}$

$U = \{ k\vec{u}, k \in \Bbb{R} \} \subset \Bbb{R}^n$

$V=\{ \vec{v} \in \Bbb{R}^n : \vec{u}.\vec{v}=0 \} \subset \Bbb{R}^n$

$U + V = \{ \vec{u} + \vec{v}, \vec{u} \in U, \vec{v} \in V \} \subset \Bbb{R}^n$

You need to prove $\Bbb{R}^n \subset U + V$

Chose any $\vec{x}\in \Bbb{R}^n$

$(\vec{x}-\cfrac{\vec{x}.\vec{u}}{\vec{u}.\vec{u}}\vec{u}).\vec{u} = \vec{x}.\vec{u} - \cfrac{\vec{x}.\vec{u}}{\vec{u}.\vec{u}}\vec{u}.\vec{u} = 0$

So $(\vec{x}-\cfrac{\vec{x}.\vec{u}}{\vec{u}.\vec{u}}\vec{u}).\vec{u} \in V$

So $\Bbb{R}^n \subset U + V$ and since we already have $U + V \subset \Bbb{R}^n$, $U + V = \Bbb{R}^n$

Then you need to prove $U\cap V=\{\vec{0}\}$

Chose any $\vec{x}\in U\cap V$

$\exists k\in\Bbb{R},\vec{x} = k \vec{u}$ and $0 = \vec{u}.\vec{x} = \vec{u}.k\vec{u} = k\|u\|^2$

So $0 = k$ because $\vec{u}\not= \vec{0}$

$\vec{x} = k \vec{u} = 0 \vec{u} = \vec{0}$

You know that $n = dim(\Bbb{R}^n) = dim(U+V) = dim( U ) + dim( V )- dim( U \cap V ) = 1 + dim( V ) - 0$

$n = 1 + dim( V )$

$dim( V ) = n - 1$

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