If it is known that all vectors in a set are perpendicular to some nonzero vector $\textbf{w}$, is that sufficient to show that the set is linearly dependent? Is it a necessary condition as well?
In two dimensions it seems obvious, because having two parallel vectors is equivalent to the existence of a vector perpendicular to both. I can't see any obvious reason why it would work with three dimensions.
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Well, the set $\{(0,1)\}$ has all vectors perpendicular to $(1,0)$, but it's not linearly dependent. Also, the empty set poses a problem. Or, the set of vectors $\{(0,1,0),(0,0,1)\}$ or, generally, $\{(0,1,0,0,\ldots),(0,0,1,0,\ldots),(0,0,0,1,\ldots),\ldots\}$.
The problem is that you could take any vector space add a new, perpendicular dimension - so it's meaningless to consider whether such a vector exists.
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Let $u, v \in \mathbb{R}^3$ be nonzero vectors.
Set $w = u \times v$. The vector $w$ is of course orthogonal, or perpendicular, to both $u$ and $v$, as is $2w$ and $3w$.
Consider the set $\{w,2w,3w\}$; each element of this set is perpendicular to the nonzero vector $v$ (and $u$) but the set is not linearly independent.
No, of course not. In 3 dimensions you can have two linearly independent vectors in a plane, which are both perpendicular to the normal of the plane.