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Is $(\mathbf{V} \cap \mathbf{W})^{\bot}=(\mathbf{V}^{\bot} \cap \mathbf{W}^{\bot})$? I tried element-chasing, but I am getting confused when trying to determine mutual containment.

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    Try examples. Try at least a few examples to see if it works.2011-02-21
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    And don't look back at this page before that, because someone is going to post an answer soon, thereby ruining this for you!2011-02-21
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    Guilty... Maybe I should temporarily hide it, but I suppose it's too late. Dear Jared, I hope that I have not ruined your joy of discovery.2011-02-21
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    By the way, $\LaTeX$'s way to say 'perpendicular' is `\perp` (which gives a $\perp$) and not `\bot` (which results in a $\bot$)2011-02-21
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    No. You should have union on the right hand side2014-04-23

2 Answers 2

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No. Try $W=\{0\}$, $V\neq\{0\}$.

What is true is that the perp of the span of two spaces is the intersection of the perps: $(V\vee W)^\perp=V^\perp\cap W^\perp$.

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If the ambient space has dimension $n$, then taking the orthogonal complement takes a subspace of dimension $k$ to a subspace of dimension $n-k$. Intersection in general reduces the dimension. So your equation looks suspect.

A nice analogy is from propositional logic: the orthogonal complement is similar to negation, and intersection is similar to logical and. So according to de Morgan's laws, if in one side of the equation you have $\cap$, in the other side you should have the dual operator, viz. $+$ (what Jonas uses $\lor$ for).