For a non-zero vector $\mathbf{v} \in{\mathbb{E}^n}$, I need to show that the collection $W$ of vectors orthogonal to $\mathbf{v}$ forms an (n-1)-dimensional subspace of $\mathbb{E}^n$. I've been working with a spanning set $\alpha_1\mathbf{w}_1 + \dotsb + \alpha_{n-1}\mathbf{w}_{n-1}$, but I'm having trouble trying to wrap my head around how to prove this is linearly independent or why it has to have dimension of $(n-1)$. Thanks
Finding a basis for an $(n-1)$-dimensional subspace of $\mathbb{E}^n$
1
$\begingroup$
linear-algebra
-
1Do you have the [Gram-Schmidt process](http://en.wikipedia.org/wiki/Gram%E2%80%93Schmidt_process) available? – 2011-10-11
-
1Here's one way of approaching this: suppose $\{w_1, \ldots, w_k \}$ is a basis for $W$. Show that $\{w_1, \ldots , w_k, v\}$ is a basis for $\mathbb{E}^n$. Since every basis of $\mathbb{E}^n$ has $n$ elements, we have $k+1=n$. – 2011-10-11
-
0Unfortunately not, I've read about it but apparently this is solvable without using it... – 2011-10-11
-
0At least, then, skimming through Wikipedia's account of Gram-Schmidt ought to inspire an attack on the missing piece of Chris's hint. – 2011-10-11